JEE Equilibrium Questions

The equilibrium reaction is: $$A(g) \rightleftharpoons B(g)$$

Given initial equilibrium concentrations at temperature T(K):
$$[A] = 0.5 \, \text{M}$$ and $$[B] = 0.375 \, \text{M}$$

The equilibrium constant $$K_c$$ is calculated as:
$$K_c = \frac{[B]}{[A]} = \frac{0.375}{0.5} = 0.75$$

Now, 0.1 moles of A are added to the 1 L flask. Since the volume is 1 L, this increases the concentration of A by 0.1 M. The initial amounts before the new equilibrium are:
Moles of A = initial moles + added moles = 0.5 + 0.1 = 0.6 moles
Moles of B = 0.375 moles (unchanged)

Thus, initial concentrations after addition (before new equilibrium) are:
$$[A]_0 = 0.6 \, \text{M}$$
$$[B]_0 = 0.375 \, \text{M}$$

Let $$x$$ be the concentration (in M) of A that reacts to form B. At the new equilibrium:
$$[A] = 0.6 - x$$
$$[B] = 0.375 + x$$

Since $$K_c$$ remains constant at the same temperature:
$$K_c = \frac{[B]}{[A]} = 0.75$$
So, $$\frac{0.375 + x}{0.6 - x} = 0.75$$

Solving for $$x$$:
$$0.375 + x = 0.75 \times (0.6 - x)$$
$$0.375 + x = 0.45 - 0.75x$$
$$x + 0.75x = 0.45 - 0.375$$
$$1.75x = 0.075$$
$$x = \frac{0.075}{1.75} = \frac{75}{1750} = \frac{3}{70} \approx 0.042857 \, \text{M}$$

New equilibrium concentrations:
$$[A] = 0.6 - \frac{3}{70} = \frac{42}{70} - \frac{3}{70} = \frac{39}{70} \approx 0.55714 \, \text{M} \approx 0.557 \, \text{M}$$
$$[B] = 0.375 + \frac{3}{70} = \frac{3}{8} + \frac{3}{70} = \frac{105}{280} + \frac{12}{280} = \frac{117}{280} \approx 0.41786 \, \text{M} \approx 0.418 \, \text{M}$$

Comparing with the options, the new equilibrium concentrations are approximately 0.557 M and 0.418 M, which corresponds to option A.

For the equilibrium

$$A(g)\rightleftharpoons B(g),$$

let (x) moles of (A) be converted into (B).

At equilibrium,

$$n_A=1-x,\qquad n_B=x,\qquad n_{He}=1.$$

Using the expression for the equilibrium constant,

$$K_c=\frac{[B]}{[A]}=\frac{x/V}{(1-x)/V}=\frac{x}{1-x}=4.$$

Solving,

$$x=4(1-x),$$

$$x=4-4x,$$

$$5x=4,$$

$$x=0.8.$$

Therefore, the equilibrium amounts are

$$n_A=0.2\ \text{mol},\qquad n_B=0.8\ \text{mol},\qquad n_{He}=1.0\ \text{mol}.$$

Using the ideal gas equation,

$$P=\frac{nRT}{V},$$

with

$$R=0.082\ \text{L atm K}^{-1}\text{mol}^{-1},\qquad T=400\ \text{K},\qquad V=10\ \text{L},$$

we obtain

$$\frac{RT}{V}=\frac{0.082\times400}{10}=3.28\ \text{atm mol}^{-1}.$$

Hence, the partial pressure of helium is

$$P_{He}=1.0\times3.28=3.28\ \text{atm}.$$

The partial pressure of (B(g)) is

$$P_B=0.8\times3.28=2.624\ \text{atm}.$$

Therefore, the equilibrium partial pressures are

$$P_{He}=3.28\ \text{atm},\qquad P_B=2.624\ \text{atm},$$

which corresponds to Option A.

We need to find which mixture gives a buffer with pH = 9.25.

Key Formula: For a basic buffer: $$pOH = pK_b + \log\frac{[\text{salt}]}{[\text{base}]}$$

Since pH = 9.25 and pH + pOH = 14: $$pOH = 14 - 9.25 = 4.75 = pK_b$$

This means $$\log\frac{[\text{salt}]}{[\text{base}]} = 0$$, so $$[\text{salt}] = [\text{base}]$$, i.e., equal moles of $$NH_4OH$$ and $$NH_4Cl$$ after the reaction.

Checking Option 4: 0.2 M $$NH_4OH$$ (0.5 L) + 0.1 M HCl (0.5 L)

Moles $$NH_4OH$$ = $$0.2 \times 0.5 = 0.10$$ mol

Moles HCl = $$0.1 \times 0.5 = 0.05$$ mol

After reaction: $$NH_4OH$$ remaining = $$0.10 - 0.05 = 0.05$$ mol, $$NH_4Cl$$ formed = 0.05 mol.

Moles of salt = moles of base = 0.05. Equal! ✓

$$pOH = 4.75 + \log(1) = 4.75$$, so $$pH = 9.25$$ ✓

The correct answer is Option 4.

We need to find the pH at which YS precipitates from a solution saturated with $$H_2S$$.

The concentration $$[S^{2-}]$$ required to precipitate YS follows from the solubility product: $$[S^{2-}] = \frac{K_{sp}(YS)}{[Y^{2+}]} = \frac{4 \times 10^{-16}}{0.01} = 4 \times 10^{-14} \text{ M}$$.

The dissociation constants of $$H_2S$$ are related by $$K_{a1} \times K_{a2} = \frac{[H^+]^2[S^{2-}]}{[H_2S]}$$, so with $$1.0 \times 10^{-21} = \frac{[H^+]^2 \times 4 \times 10^{-14}}{0.1}$$ one finds $$[H^+]^2 = \frac{10^{-22}}{4 \times 10^{-14}} = 2.5 \times 10^{-9}$$ and hence $$[H^+] = \sqrt{25 \times 10^{-10}} = 5 \times 10^{-5} \text{ M}$$.

Therefore, $$\text{pH} = -\log(5 \times 10^{-5}) = 5 - \log 5 = 5 - 0.70 = 4.30$$. Rounding to the nearest integer gives pH = 4.

We have the equilibrium $$X_2 + Y_2 \rightleftharpoons 2Z$$ in a 1 L flask with equilibrium moles: $$[X_2] = 3$$, $$[Y_2] = 3$$, $$[Z] = 9$$ mol. Then 10 mol of Z is added and we need the new equilibrium moles of Z.

To begin, we calculate the equilibrium constant $$K_c$$.

Since the volume is 1 L, concentrations equal the number of moles:

$$ K_c = \frac{[Z]^2}{[X_2][Y_2]} = \frac{9^2}{3 \times 3} = \frac{81}{9} = 9 $$

Next, we set up the ICE table after adding 10 mol of Z.

After adding 10 mol Z, the initial (non-equilibrium) concentrations are:

$$[X_2] = 3$$, $$[Y_2] = 3$$, $$[Z] = 9 + 10 = 19$$ mol.

The reaction quotient $$Q = \frac{19^2}{3 \times 3} = \frac{361}{9} \approx 40.1 > K_c = 9$$.

Since $$Q > K_c$$, the reaction will shift backward (to the left), consuming Z and producing $$X_2$$ and $$Y_2$$.

We then apply the equilibrium condition.

$$ K_c = \frac{(19 - 2x)^2}{(3 + x)(3 + x)} = \frac{(19 - 2x)^2}{(3 + x)^2} = 9 $$

We solve the resulting equation.

Taking the square root of both sides (both numerator and denominator are positive):

$$ \frac{19 - 2x}{3 + x} = 3 $$

$$ 19 - 2x = 3(3 + x) = 9 + 3x $$

$$ 19 - 9 = 3x + 2x $$

$$ 10 = 5x $$

$$ x = 2 $$

Finally, we find the new equilibrium moles of Z.

$$ [Z] = 19 - 2x = 19 - 4 = 15 $$

The answer is 15.

To find the value of $$x$$ for the equilibrium $$NH_3(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g)$$, let the initial moles of $$NH_3$$ be 1 and the degree of dissociation be $$\alpha$$. At equilibrium, the moles are $$NH_3 = 1 - \alpha$$, $$N_2 = \frac{\alpha}{2}$$, and $$H_2 = \frac{3\alpha}{2}$$, so the total moles equal $$1 - \alpha + \frac{\alpha}{2} + \frac{3\alpha}{2} = 1 + \alpha$$.

The total pressure $$P_T$$ is given as $$\sqrt{3}$$ atm, which leads to the partial pressures $$p_{NH_3} = \frac{1-\alpha}{1+\alpha} \cdot \sqrt{3}$$, $$p_{N_2} = \frac{\alpha/2}{1+\alpha} \cdot \sqrt{3}$$, and $$p_{H_2} = \frac{3\alpha/2}{1+\alpha} \cdot \sqrt{3}$$.

The expression for $$K_p$$ is $$K_p = \frac{p_{N_2}^{1/2} \cdot p_{H_2}^{3/2}}{p_{NH_3}} = 9 \text{ atm}$$, so substitution gives $$K_p = \frac{\left(\frac{\alpha\sqrt{3}}{2(1+\alpha)}\right)^{1/2} \cdot \left(\frac{3\alpha\sqrt{3}}{2(1+\alpha)}\right)^{3/2}}{\frac{(1-\alpha)\sqrt{3}}{1+\alpha}}$$.

In the numerator, $$\left(\frac{\alpha\sqrt{3}}{2(1+\alpha)}\right)^{1/2} \cdot \left(\frac{3\alpha\sqrt{3}}{2(1+\alpha)}\right)^{3/2} = \frac{\alpha^{1/2} \cdot 3^{1/4}}{2^{1/2}(1+\alpha)^{1/2}} \cdot \frac{3^{3/2}\alpha^{3/2} \cdot 3^{3/4}}{2^{3/2}(1+\alpha)^{3/2}} = \frac{3^{1/4+3/2+3/4} \cdot \alpha^2}{2^2 \cdot (1+\alpha)^2} = \frac{3^{5/2} \cdot \alpha^2}{4(1+\alpha)^2}$$.

Hence $$K_p = \frac{3^{5/2}\alpha^2}{4(1+\alpha)^2} \cdot \frac{(1+\alpha)}{(1-\alpha)\sqrt{3}} = \frac{3^{5/2}\alpha^2}{4(1+\alpha)(1-\alpha)\sqrt{3}} = \frac{3^2 \alpha^2}{4(1-\alpha^2)} = \frac{9\alpha^2}{4(1-\alpha^2)} = 9$$.

Solving $$\frac{\alpha^2}{4(1-\alpha^2)} = 1$$ gives $$\alpha^2 = 4 - 4\alpha^2$$ and hence $$5\alpha^2 = 4$$, so $$\alpha^2 = \frac{4}{5}$$ and $$\alpha = \frac{2}{\sqrt{5}}$$.

Since $$\alpha = (x \times 10^{-2})^{-1/2}$$, it follows that $$\alpha^2 = \frac{1}{x \times 10^{-2}} = \frac{100}{x}$$. Equating $$\frac{4}{5} = \frac{100}{x}$$ yields $$x = \frac{500}{4} = 125$$.

Therefore, the answer is $$\boxed{125}$$.

For the equilibrium
$$2\,\text{N}_2\text{O}_5(g)\;\rightleftharpoons\;2\,\text{N}_2\text{O}_4(g)\;+\;\text{O}_2(g)$$
let us start with $$2$$ mol of $$\text{N}_2\text{O}_5$$ (this matches the stoichiometric coefficient).

Extent of dissociation
Given 50 % dissociation, exactly $$1$$ mol of the initial $$2$$ mol reacts.
Using the stoichiometry $$2\text{N}_2\text{O}_5 \rightarrow 2\text{N}_2\text{O}_4 + \text{O}_2$$:

• $$\text{N}_2\text{O}_5$$ left = $$2 - 1 = 1$$ mol
• $$\text{N}_2\text{O}_4$$ formed = $$1 \times \frac{2}{2} = 1$$ mol
• $$\text{O}_2$$ formed = $$1 \times \frac{1}{2} = 0.5$$ mol

Total moles at equilibrium
$$n_{\text{tot}} = 1 + 1 + 0.5 = 2.5\ \text{mol}$$

The total pressure is given as $$P = 2\ \text{atm}$$, so the partial pressures are

$$P_{\text{N}_2\text{O}_5} = \frac{1}{2.5}\times 2 = 0.8\ \text{atm}$$
$$P_{\text{N}_2\text{O}_4} = \frac{1}{2.5}\times 2 = 0.8\ \text{atm}$$
$$P_{\text{O}_2} = \frac{0.5}{2.5}\times 2 = 0.4\ \text{atm}$$

Equilibrium constant $$K_p$$
For the reaction
$$2\text{N}_2\text{O}_5 \rightleftharpoons 2\text{N}_2\text{O}_4 + \text{O}_2$$

$$K_p = \frac{(P_{\text{N}_2\text{O}_4})^{2}\; P_{\text{O}_2}}{(P_{\text{N}_2\text{O}_5})^{2}}$$

Substituting the calculated partial pressures
$$K_p = \frac{(0.8)^2 \times 0.4}{(0.8)^2} = 0.4$$

Standard Gibbs free-energy change
The relation between $$\Delta G^{\circ}$$ and $$K_p$$ is
$$\Delta G^{\circ} = -RT\ln K_p$$

Temperature: $$T = 50^{\circ}\text{C} = 323\ \text{K}$$
Gas constant: $$R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$$

First find $$\ln 0.4$$.
Using base-10 logs provided:
$$\log_{10}0.4 = \log_{10}4 - 1 = 0.602 - 1 = -0.398$$
Convert to natural log:
$$\ln 0.4 = 2.303 \times (-0.398) \approx -0.916$$

Now compute $$\Delta G^{\circ}$$:
$$\Delta G^{\circ} = -(8.314)(323)(-0.916)$$
$$\quad\; = 8.314 \times 323 \times 0.916$$
$$\quad\; \approx 2685.4 \times 0.916 \approx 2.46 \times 10^{3}\ \text{J mol}^{-1}$$

Hence
$$|\Delta G^{\circ}| \approx 2461\ \text{J mol}^{-1}$$ (nearest integer).

Answer: 2461

We need to find the pH of a 10 mM solution of weak acid HA with $$pK_a = 4$$.

Since $$pK_a = 4 \implies K_a = 10^{-4}$$ and the concentration is $$C = 10 \text{ mM} = 0.01 \text{ M} = 10^{-2} \text{ M}$$.

The dissociation equilibrium is given by $$HA \rightleftharpoons H^+ + A^-$$, and the acid dissociation constant is $$K_a = \frac{[H^+][A^-]}{[HA]}$$.

For a weak acid where the degree of dissociation can be neglected (as stated in the problem), the hydrogen ion concentration is $$[H^+] = \sqrt{K_a \times C}$$. Taking the negative logarithm gives $$pH = -\log[H^+] = -\frac{1}{2}\log(K_a \times C) = \frac{1}{2}(pK_a - \log C)$$.

Substituting the values yields $$pH = \frac{1}{2}(pK_a - \log C) = \frac{1}{2}(4 - \log(10^{-2}))$$, which becomes $$= \frac{1}{2}(4 - (-2)) = \frac{1}{2}(4 + 2) = \frac{1}{2} \times 6 = 3$$.

As a verification, one finds $$[H^+] = \sqrt{10^{-4} \times 10^{-2}} = \sqrt{10^{-6}} = 10^{-3}$$ M and the degree of dissociation $$\alpha = 10^{-3}/10^{-2} = 0.1$$. Since $$\alpha = 0.1$$ is small enough relative to unity (and the problem states it can be neglected), this is consistent.

The answer is 3.

For the reaction

$$X_2Y_4(g)\rightleftharpoons2XY_2(g),$$

initially, (1) mole of ($$X_{2}Y_4$$) is present and (75%) of it dissociates at equilibrium.

Hence, the equilibrium moles are:

• ($$X_{2}Y_4$$ = 1-0.75 = 0.25)
• ($$XY_2$$ = 2$$\times$$0.75 = 1.5)

The total number of moles at equilibrium is

$$0.25+1.5=1.75.$$

Since the total pressure is (1) atm, the mole fractions are

$$x_{X_2Y_4}=\frac{0.25}{1.75}=\frac{1}{7},$$

and

$$x_{XY_2}=\frac{1.5}{1.75}=\frac{6}{7}.$$

Therefore, the partial pressures are

$$P_{X_2Y_4}=\frac{1}{7}\text{ atm}$$

and

$$P_{XY_2}=\frac{6}{7}\text{ atm}.$$

The equilibrium constant is

$$K_p=\frac{(P_{XY_2})^2}{P_{X_2Y_4}}=\frac{\left(\frac{6}{7}\right)^2}{\frac{1}{7}}=\frac{36}{49}\times7=\frac{36}{7}\approx5.14.$$

The standard Gibbs free energy change is given by

$$\Delta_rG^\ominus=-RT\ln K_p.$$

Using

$$R=8.3\ \text{J mol}^{-1}\text{K}^{-1},\qquad T=600\ \text{K},$$

and

$$\ln K_p=\ln\left(\frac{36}{7}\right)=2\ln6-\ln7,$$

with

$$\ln x=2.3\log x,$$

we obtain

$$\ln K_p=2.3,[2(\log2+\log3)-\log7]$$

$$=2.3,[2(0.3+0.48)-0.84]$$

$$=2.3,(1.56-0.84)$$

$$=2.3\times0.72$$

$$=1.656.$$

Substituting into the Gibbs free energy equation,

$$\Delta_rG^\ominus=-(8.3)\times600\times1.656$$

$$\approx-8251\ \text{J mol}^{-1}$$

$$\approx-8.25\ \text{kJ mol}^{-1}.$$

Rounding to the nearest integer, the magnitude of the standard Gibbs free energy change is ($$8\ \text{kJ mol}^{-1}$$).

For a buffer containing a weak base $$\text{B}$$ and its salt $$\text{BH}^+$$, the Henderson-Hasselbalch type relation is

$$\text{pOH} = \text{p}K_b + \log\frac{[\text{salt}]}{[\text{base}]}$$

Step 1 - Calculate moles of each component.

Base ($$NH_4OH)$$: $$0.1\;M \times 5\;{\text{mL}} = 0.1 \times \frac{5}{1000} = 5.0 \times 10^{-4}\;\text{mol}$$

Salt $$(NH_4Cl)$$: $$0.1\;M \times 250\;{\text{mL}} = 0.1 \times \frac{250}{1000} = 2.5 \times 10^{-2}\;\text{mol}$$

Step 2 - Find concentrations after mixing.

Total volume $$V = 5\;{\text{mL}} + 250\;{\text{mL}} = 255\;{\text{mL}} = 0.255\;L$$

$$[\text{NH}_4\text{OH}] = \frac{5.0 \times 10^{-4}}{0.255} \approx 1.96 \times 10^{-3}\;M$$

$$[\text{NH}_4^+] = \frac{2.5 \times 10^{-2}}{0.255} \approx 9.80 \times 10^{-2}\;M$$

Step 3 - Insert into the buffer equation.

Ratio $$\frac{[\text{salt}]}{[\text{base}]} = \frac{9.80 \times 10^{-2}}{1.96 \times 10^{-3}} \approx 50$$

Using the given logarithms:
$$\log 50 = \log(5 \times 10^1) = \log 5 + \log 10 = 0.70 + 1 = 1.70$$

Given $$\text{p}K_b(\text{NH}_4\text{OH}) = 4.74$$, so

$$\text{pOH} = 4.74 + 1.70 = 6.44$$

Step 4 - Find pH.

$$\text{pH} = 14 - \text{pOH} = 14 - 6.44 = 7.56$$

Step 5 - Express in the required form.

$$7.56 = 756 \times 10^{-2}$$
Nearest integer = 756.

Answer: 756

To find the magnitude of $$\ \frac{\triangle\ H^{\circ\ }}{R}$$, we use the integrated form of the van 't Hoff equation:

$$\log_{10}K_p=-\frac{\triangle H^{\circ\ }}{2.303RT}\ +$$ Constant

Step 1: Identify the slope

The equation is in the form of $$y=mx+c$$, 

where:

$$y=\log_{10}K_p$$

$$x=\frac{1}{T}$$

$$m\left(slope\right)=-\frac{\triangle H^{\circ\ }}{2.303R}\ $$

Using the first two points from the table:

$$\left(x_1,y_1\right)=\left(0.05,3.5\right)$$

$$\left(x_2,y_2\right)=\left(0.06,2.5\right)$$

Therefore, 

$$slope\left(m\right)=\frac{y_2-y_1}{x_2-x_1}=\frac{2.5-3.5}{0.06-0.05}=-\frac{0.1}{0.01}=-100$$

Step 2: Calculate $$\ \frac{\triangle\ H^{\circ\ }}{R}$$

Equating the calculated slop $$(m)$$

$$-100=-\frac{\triangle H^{\circ\ }}{2.303R}$$

$$100\times\ 2.303=\frac{\triangle H^{\circ\ }}{R}$$

$$230.3=\frac{\triangle H^{\circ\ }}{R}$$

Therefore, the nearest integer is 230.

For a heterogeneous equilibrium, the equilibrium constant expression ($$K_p$$) only includes the partial pressures of gaseous components, as the activities of pure solids are taken as $$1$$.

From Reaction 1:

$$K_{p1} = P_{\text{CO}_2}$$

Given that $$K_{p1} = 0.08\text{ atm}$$, we can directly state:

$$P_{\text{CO}_2} = 0.08\text{ atm}$$

Since both reactions occur simultaneously in the same container, the $$\text{CO}_2$$ gas produced in the first reaction is utilized in the second reaction. The system will adjust until both equilibria are satisfied at the same time.

From Reaction 2:

$$K_{p2} = \frac{(P_{\text{CO}})^2}{P_{\text{CO}_2}}$$

Substitute the values of $$K_{p2}$$ and $$P_{\text{CO}_2}$$ into this expression:

$$2 = \frac{(P_{\text{CO}})^2}{0.08}$$

$$(P_{\text{CO}})^2 = 2 \times 0.08$$

$$(P_{\text{CO}})^2 = 0.16$$

Taking the square root on both sides:

$$P_{\text{CO}} = \sqrt{0.16} = 0.4\text{ atm}$$

Converting $$0.4\text{ atm}$$ into scientific notation:

$$0.4\text{ atm} = 4 \times 10^{-1}\text{ atm}$$

The weak monobasic acid is represented as $$HA$$.

Dissociation in water:
$$HA \rightleftharpoons H^+ + A^-$$ with $$K_a = 4.00 \times 10^{-11}$$

Let the initial concentration of the acid be $$C = 1.00 \times 10^{-3}\;{\rm M}$$.
At equilibrium, suppose:

$$\begin{aligned} [H^+]_{\text{total}} &= x \;{\rm M}\\[2pt] \text{Degree of dissociation from the acid} &= y \;{\rm M}\\[2pt] \Rightarrow [A^-] &= y \;{\rm M}\\[2pt] [HA] &= C - y \end{aligned}$$

Water also dissociates: $$H_2O \rightleftharpoons H^+ + OH^-$$ with $$K_w = 1.00 \times 10^{-14}$$.
Hence $$[OH^-] = \dfrac{K_w}{x}$$.

The $$H^+$$ that comes only from water equals the $$OH^-$$ formed, so

$$[H^+]_{\text{from water}} = \dfrac{K_w}{x}$$

Therefore the $$H^+$$ contributed by the acid is

$$y = x - \dfrac{K_w}{x}\qquad -(1)$$

Write the $$K_a$$ expression:

$$K_a = \dfrac{[H^+][A^-]}{[HA]} = \dfrac{x \, y}{C - y}\qquad -(2)$$

Because the acid is very weak, $$y \ll C$$ (numerically, we will soon see $$y \approx 2 \times 10^{-7}\,{\rm M}$$ while $$C = 10^{-3}\,{\rm M}$$).
Hence we can put $$C - y \approx C$$ in $$(2)$$:

$$K_a \approx \dfrac{x\,y}{C}\qquad -(3)$$

Substituting $$y$$ from $$(1)$$ into $$(3)$$:

$$K_a \approx \dfrac{x\left(x - \dfrac{K_w}{x}\right)}{C} = \dfrac{x^2 - K_w}{C}$$

Re-arranging gives

$$x^2 = K_a C + K_w\qquad -(4)$$

Insert the numerical data:

$$\begin{aligned} K_a C &= (4.00 \times 10^{-11})(1.00 \times 10^{-3}) = 4.00 \times 10^{-14}\\[4pt] K_a C + K_w &= 4.00 \times 10^{-14} + 1.00 \times 10^{-14} = 5.00 \times 10^{-14}\\[4pt] x &= \sqrt{5.00 \times 10^{-14}} = (\sqrt{5}) \times 10^{-7}\,{\rm M}\\[4pt] \sqrt{5} &\approx 2.236 \end{aligned}$$

Thus

$$[H^+] = 2.236 \times 10^{-7}\;{\rm M}$$

Comparing with the required form $$X \times 10^{-7}\;{\rm M}$$, we have

$$X \approx 2.24$$

Any value in the range 2.2-2.3 is therefore correct.

The dissolution equilibrium for barium iodate is
$$Ba(IO_3)_2(s) \rightleftharpoons Ba^{2+}+2\,IO_3^-$$
and its solubility-product constant is
$$K_{sp}=1.58\times10^{-9}= [Ba^{2+}][IO_3^-]^2$$

Step 1: Initial ion concentrations after mixing
Moles of $$Ba^{2+}$$ added: $$0.200\text{ L}\times0.010\text{ M}=2.0\times10^{-3}\,\text{mol}$$
Moles of $$IO_3^-$$ added: $$0.100\text{ L}\times0.10\text{ M}=1.0\times10^{-2}\,\text{mol}$$
Total volume after mixing: $$0.200+0.100=0.300\text{ L}$$

Initial concentrations
$$[Ba^{2+}]_0=\frac{2.0\times10^{-3}}{0.300}=6.67\times10^{-3}\,\text{M}$$
$$[IO_3^-]_0=\frac{1.0\times10^{-2}}{0.300}=3.33\times10^{-2}\,\text{M}$$

Step 2: Check for precipitation
Ionic product $$Q=[Ba^{2+}]_0[IO_3^-]_0^{\,2}$$
$$Q=6.67\times10^{-3}\times(3.33\times10^{-2})^{2}=6.67\times10^{-3}\times1.11\times10^{-3}=7.41\times10^{-6}$$
Since $$Q\;(7.41\times10^{-6}) \gg K_{sp}\;(1.58\times10^{-9})$$, the solution is supersaturated and $$Ba(IO_3)_2$$ will precipitate until equilibrium is attained.

Step 3: Let $$x$$ M of $$Ba(IO_3)_2$$ precipitate
Every mole of precipitate removes $$1\,Ba^{2+}$$ and $$2\,IO_3^-$$.
Final concentrations:
$$[Ba^{2+}]=6.67\times10^{-3}-x$$
$$[IO_3^-]=3.33\times10^{-2}-2x$$

Step 4: Apply $$K_{sp}$$ at equilibrium
$$(6.67\times10^{-3}-x)\,(3.33\times10^{-2}-2x)^{2}=1.58\times10^{-9}\;-(1)$$

Step 5: Simplify using the small-quantity approximation
Because $$K_{sp}$$ is extremely small, the final $$[Ba^{2+}]$$ will be much smaller than the initial $$6.67\times10^{-3}$$ M. Let $$z=[Ba^{2+}]_{eq}=6.67\times10^{-3}-x\ll6.67\times10^{-3}\;,$$ so $$x\approx6.67\times10^{-3}$$.

Then $$[IO_3^-]_{eq}=3.33\times10^{-2}-2x=3.33\times10^{-2}-2(6.67\times10^{-3}-z)=0.020+2z$$ Since $$z\ll0.020$$, take $$[IO_3^-]_{eq}\approx0.020\;\text{M}$$.

Insert these approximations into $$(1)$$:

$$z\,(0.020)^{2}=1.58\times10^{-9}$$
$$\Rightarrow\;z=\frac{1.58\times10^{-9}}{4.00\times10^{-4}}=3.95\times10^{-6}\,\text{mol dm}^{-3}$$

Step 6: Interpret the result
The equilibrium concentration of dissolved $$Ba^{2+}$$ equals the molar solubility of $$Ba(IO_3)_2$$ in the given medium. Therefore, the solubility is $$3.95\times10^{-6}\,\text{mol dm}^{-3}$$.

Expressed as $$X\times10^{-6}$$ mol dm$$^{-3}$$, $$X=3.95\;(\text{any value in }3.85-4.15\text{ is accepted}).$$

The Freundlich adsorption isotherm is written as
$$\frac{x}{m}=k\,C^{1/n}$$
where $$\tfrac{x}{m}$$ is the mass of phenol adsorbed per gram of fly-ash and $$C$$ is the equilibrium concentration of phenol in the solution (both here in mg g$$^{-1}$$).

Taking common logarithm (base 10):
$$\log\!\left(\frac{x}{m}\right)=\log k+\frac{1}{n}\,\log C \quad -(1)$$

The two experimental data pairs are

$$C_1=10\;\text{mg g}^{-1},\;\; \left(\frac{x}{m}\right)_1 = 4\;\text{mg g}^{-1}$$
$$C_2=16\;\text{mg g}^{-1},\;\; \left(\frac{x}{m}\right)_2 = 10\;\text{mg g}^{-1}$$

Substituting in equation $$-(1)$$:

For the first pair:
$$\log 4 = \log k + \frac{1}{n}\,\log 10 \quad -(2)$$

For the second pair:
$$\log 10 = \log k + \frac{1}{n}\,\log 16 \quad -(3)$$

Using the given value $$\log 2 = 0.3$$:

$$\log 4 = \log(2^2)=2\times0.3 = 0.6$$
$$\log 10 = 1$$
$$\log 16 = \log(2^4)=4\times0.3 = 1.2$$

Equations $$-(2)$$ and $$-(3)$$ become

$$0.6 = \log k + \frac{1}{n}(1) \quad -(4)$$
$$1.0 = \log k + \frac{1}{n}(1.2) \quad -(5)$$

Subtract $$-(4)$$ from $$-(5)$$:

$$1.0 - 0.6 = \frac{1}{n}(1.2 - 1.0)$$
$$0.4 = \frac{1}{n}\times0.2$$
$$\frac{1}{n} = 2 \;\;\Longrightarrow\;\; n = 0.5$$

From equation $$-(4)$$:
$$\log k = 0.6 - \frac{1}{n}(1) = 0.6 - 2 = -1.4$$
Hence $$k = 10^{-1.4}$$

For a new solution with concentration $$C_3 = 20\;\text{mg g}^{-1}$$,
$$\log C_3 = \log(2\times10)=\log 2 + \log 10 = 0.3 + 1 = 1.3$$

Using equation $$-(1)$$:
$$\log\!\left(\frac{x}{m}\right)_3 = -1.4 + 2 \times 1.3 = -1.4 + 2.6 = 1.2$$

Therefore,
$$\left(\frac{x}{m}\right)_3 = 10^{1.2}$$

An antilog of $$1.2$$ is
$$10^{1.2} = 10^{1}\times10^{0.2} \approx 10 \times 1.58 \approx 15.8\;\text{mg g}^{-1}$$

Thus, the concentration of phenol adsorbed from a 20 mg g$$^{-1}$$ solution is approximately $$15.8\;\text{mg g}^{-1}$$, which lies in the required range 15.5 - 16.5.

The conductometric titration curve represents the titration of a mixture of a strong acid (HCl) and a weak acid ($CH_3COOH$) with a strong base (NaOH).

The conductance of the solution depends on the number of ions present and their ionic mobility. Since the mobility of $$H^+$$ ions is exceptionally high, the conductance changes characteristically during the titration.

Initially, in segment AB, HCl is neutralized first because it is completely dissociated in solution. The highly mobile $$H^+$$ ions are replaced by the comparatively less mobile $$Na^+$$ ions from NaOH, causing a sharp decrease in conductance.

Point B, corresponding to $$2.0\ \text{mL}$$ of NaOH added, represents the equivalence point of HCl.

Beyond point B, in segment BC, NaOH starts neutralizing acetic acid:

$$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O.$$

Since acetic acid is only weakly ionized whereas sodium acetate is completely dissociated, the number of ions in solution increases, producing a gradual increase in conductance.

Point C, corresponding to $$5.0\ \text{mL}$$ of NaOH added, represents the complete neutralization of acetic acid.

After point C, any additional NaOH contributes free $$OH^-$$ ions to the solution. Because $$OH^-$$ ions possess high mobility, the conductance rises sharply in segment CD.

Hence:

• Option (C) is incorrect because HCl is neutralized before $$CH_3COOH$$.
• Option (D) is incorrect because point C corresponds to the equivalence point of $$CH_3COOH$$, while HCl is neutralized at point B.

Now calculate the concentrations.

Given,

$$V_{\text{mixture}}=40\ \text{mL},$$

$$M_{\text{NaOH}}=0.1\ \text{M}.$$

For HCl,

$$M_{\text{HCl}}\times40=0.1\times2.0,$$

so

$$M_{\text{HCl}}=\frac{0.2}{40}=0.005\ \text{M}.$$

For acetic acid, the volume of NaOH consumed is

$$5.0-2.0=3.0\ \text{mL}.$$

Therefore,

$$M_{CH_3COOH}\times40=0.1\times3.0,$$

giving

$$M_{CH_3COOH}=\frac{0.3}{40}=0.0075\ \text{M}.$$

Thus,

• $$[HCl]=0.005\ \text{M}$$
• $$[CH_3COOH]=0.0075\ \text{M}$$

Hence, the correct statement is that the concentration of HCl is

$$0.005\ \text{M},$$

so the correct answer is option (B).

The reaction: $$CO(g) + 3H_2(g) \rightleftharpoons CH_4(g) + H_2O(g)$$

Moles of gas: Reactants = 4, Products = 2. So $$\Delta n_g = -2$$.

When pressure is doubled at constant temperature:

(A) Concentration of reactants and products increases. Since volume decreases (due to increased pressure), concentrations increase. TRUE.

(B) Equilibrium will shift in forward direction. By Le Chatelier's principle, increasing pressure shifts equilibrium towards the side with fewer moles of gas (products side). TRUE.

(C) Equilibrium constant increases. $$K_c$$ depends only on temperature. It does NOT change with pressure. FALSE.

(D) Equilibrium constant remains unchanged as concentration of reactants and products remain same. $$K_c$$ does remain unchanged, but the reasoning is wrong - concentrations DO change. FALSE (the statement as a whole is false because of the incorrect reasoning).

Only statements A and B are correct.

The correct answer is Option 2: (A) and (B) only.

For $$AB_2 \rightleftharpoons AB + \frac{1}{2}B_2$$ with small degree of dissociation $$x$$:

Starting with 1 mole at pressure $$p$$, at equilibrium: $$AB_2$$: $$1-x$$, $$AB$$: $$x$$, $$B_2$$: $$x/2$$.

Total moles $$\approx 1$$ (since $$x \ll 1$$). Partial pressures: $$p_{AB} \approx xp$$, $$p_{B_2} \approx xp/2$$, $$p_{AB_2} \approx p$$.

$$K_p = \frac{p_{AB} \cdot p_{B_2}^{1/2}}{p_{AB_2}} \approx \frac{xp \cdot (xp/2)^{1/2}}{p} = \frac{x^{3/2} \sqrt{p}}{\sqrt{2}}$$

Squaring: $$K_p^2 = \frac{x^3 p}{2}$$, so $$x^3 = \frac{2K_p^2}{p}$$.

$$x = \left(\frac{2K_p^2}{p}\right)^{1/3} = \sqrt[3]{\frac{2K_p^2}{p}}$$

The correct answer is Option 3.

We need to evaluate two statements about catalysts and chemical equilibrium.

Statement I: A catalyst cannot alter the equilibrium constant $$(K_c)$$ of the reaction, temperature remaining constant.

A catalyst works by providing an alternative reaction pathway with lower activation energy. It speeds up both the forward and reverse reactions equally.

The equilibrium constant $$K_c$$ depends only on the temperature (through the relation $$\Delta G° = -RT \ln K$$). Since a catalyst does not change the Gibbs free energy of the reaction, it cannot change $$K_c$$ at constant temperature.

Therefore, Statement I is true.

Statement II: A homogeneous catalyst can change the equilibrium composition of a system, temperature remaining constant.

Since a catalyst speeds up both the forward and reverse reactions equally, it does not shift the position of equilibrium. The equilibrium concentrations of reactants and products remain the same whether or not a catalyst is present.

A catalyst only helps the system reach equilibrium faster, but it does not alter what the equilibrium composition is.

Note: This applies to both homogeneous and heterogeneous catalysts. Neither type can change the equilibrium composition at constant temperature.

Therefore, Statement II is false.

Statement I is true and Statement II is false.

Hence, the correct answer is Option D.

For an aqueous solution, $$\text{pH} = -\log_{10}[H^{+}]$$.

Initial pH of the HCl solution is given as $$1.0$$.
Therefore the initial hydrogen-ion concentration is $$[H^{+}]_1 = 10^{-1}\,\text{M} = 0.1\,\text{M}$$.

The solution is now diluted by adding an equal volume of water.
For any solute, doubling the total volume (while keeping the amount of solute the same) halves its concentration:

$$[H^{+}]_2 = \frac{[H^{+}]_1}{2} = \frac{0.1}{2} = 0.05\,\text{M}$$.

Calculate the new pH:

$$\text{pH}_2 = -\log_{10}(0.05)$$.

Write $$0.05$$ as $$5 \times 10^{-2}$$, then use the logarithm law $$\log(ab) = \log a + \log b$$:

$$\text{pH}_2 = -\left[\log_{10}(5) + \log_{10}(10^{-2})\right]$$.

Since $$\log_{10}(10^{-2}) = -2$$, we get

$$\text{pH}_2 = -\left[\log_{10}(5) - 2\right] = 2 - \log_{10}(5).$$

Using $$\log_{10}(5) \approx 0.699$$,

$$\text{pH}_2 \approx 2 - 0.699 = 1.301 \approx 1.3.$$

Thus the pH increases from $$1.0$$ to approximately $$1.3$$.

Hence, the correct option is Option B: increase to 1.3.

For the reaction $$X_2Y(g) \rightleftharpoons X_2(g) + \frac{1}{2}Y_2(g)$$, we want to find the relationship between the degree of dissociation $$x$$ and $$K_p$$.

We begin by setting up the equilibrium composition. Initially, we have 1 mol of X₂Y and no products, so the initial moles are 1 mol of X₂Y, 0, 0. At equilibrium, if the degree of dissociation is $$x$$, the amounts become $$(1-x)$$ for X₂Y, $$x$$ for X₂, and $$x/2$$ for Y₂. The total number of moles is therefore

$$1 - x + x + \tfrac{x}{2} = 1 + \tfrac{x}{2}.$$

Since $$x$$ is very small, we approximate the total moles as 1.

Next, with a total pressure of $$P$$, the partial pressures are given by

$$P_{X_2Y} = (1-x)P \approx P,$$

$$P_{X_2} = xP,$$

$$P_{Y_2} = \tfrac{x}{2}P.$$

The equilibrium constant $$K_p$$ is written in terms of these partial pressures as:

$$K_p = \frac{P_{X_2} \cdot P_{Y_2}^{1/2}}{P_{X_2Y}} = \frac{xP \cdot \left(\frac{x}{2}P\right)^{1/2}}{P}$$

$$= xP \cdot \frac{x^{1/2}}{2^{1/2}} \cdot \frac{P^{1/2}}{P}$$

$$= \frac{x^{3/2} P^{1/2}}{2^{1/2}}.$$

To solve for $$x$$, we rewrite the expression as:

$$K_p = \frac{x^{3/2}\sqrt{P}}{\sqrt{2}}$$

$$x^{3/2} = \frac{K_p\sqrt{2}}{\sqrt{P}}$$

$$x^3 = \frac{2K_p^2}{P}$$

$$x = \left(\frac{2K_p^2}{P}\right)^{1/3}.$$

The correct answer is Option 2: x = $$\sqrt[3]{\frac{2K_p^2}{P}}$$.

Find the expression for $$(pH - pK_a)$$ for a weak acid HA with degree of dissociation $$x$$. Let the initial concentration of HA be $$c$$. At equilibrium for HA $$\rightleftharpoons$$ H$$^+$$ + A$$^-$$, the concentrations are $$[HA] = c(1-x)$$, $$[H^+] = cx$$ and $$[A^-] = cx$$.

The acid dissociation constant is given by $$K_a = \frac{[H^+][A^-]}{[HA]} = \frac{cx \cdot cx}{c(1-x)} = \frac{cx^2}{1-x}$$.

Since $$pH = -\log[H^+] = -\log(cx)$$ and $$pK_a = -\log K_a = -\log\left(\frac{cx^2}{1-x}\right)$$, their difference follows from the logarithm rule $$\log A - \log B = \log\frac{A}{B}$$.

Thus, $$pH - pK_a = -\log(cx) - \left(-\log\frac{cx^2}{1-x}\right) = -\log(cx) + \log\frac{cx^2}{1-x} = \log\frac{cx^2/(1-x)}{cx} = \log\frac{x}{1-x}$$.

The correct answer is Option D: $$\log\left(\frac{x}{1-x}\right)$$.

$$N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2$$. Find final pressure when 50% reacts.

Initial: P₀ (N₂O₅ only)

At 50% decomposition: N₂O₅ = P₀/2, NO₂ = 2(P₀/2) = P₀, O₂ = ½(P₀/2) = P₀/4

Total = P₀/2 + P₀ + P₀/4 = 7P₀/4

The correct answer is Option 4: $$\frac{7}{4}$$ times initial pressure.

Zirconium phosphate has the formula $$(Zr^{4+})_3(PO_4^{3-})_4$$.

The dissociation equilibrium is:

$$Zr_3(PO_4)_4 \rightleftharpoons 3Zr^{4+} + 4PO_4^{3-}$$

If the molar solubility is $$s$$, then:

$$[Zr^{4+}] = 3s$$ and $$[PO_4^{3-}] = 4s$$

The solubility product expression is:

$$K_{sp} = [Zr^{4+}]^3 \cdot [PO_4^{3-}]^4 = (3s)^3 \cdot (4s)^4$$

$$K_{sp} = 27s^3 \cdot 256s^4 = 6912 \cdot s^7$$

Solving for $$s$$:

$$s = \left(\frac{K_{sp}}{6912}\right)^{\frac{1}{7}}$$

The correct answer is Option B: $$\left(\frac{K_{sp}}{6912}\right)^{\frac{1}{7}}$$.

When the salt $$AB_2$$ dissolves, it produces one divalent cation $$A^{2+}$$ and two anions $$B^-$$:
$$AB_2 \rightarrow A^{2+} + 2B^-$$

The salt $$XY$$ produces one monovalent cation $$X^{+}$$ and one anion $$Y^-$$:
$$XY \rightarrow X^{+} + Y^-$$

The required precipitate is $$AY_2$$, which dissociates as
$$AY_2 \rightleftharpoons A^{2+} + 2Y^-$$

Hence, the solubility-product expression is
$$K_{sp} = [A^{2+}][Y^-]^{2}$$

Let the initial molarities of $$AB_2$$ and $$XY$$ be $$C_1$$ and $$C_2$$ respectively (as given in each option). The two solutions are mixed in equal volumes, so the total volume becomes double. Every concentration therefore becomes half.

After mixing:
$$[A^{2+}] = \dfrac{C_1}{2}$$ (comes only from $$AB_2$$)
$$[Y^-] = \dfrac{C_2}{2}$$ (comes only from $$XY$$)

The ionic product (IP) for $$AY_2$$ just after mixing is
$$\text{IP} = \left(\dfrac{C_1}{2}\right)\left(\dfrac{C_2}{2}\right)^{2} = \dfrac{C_1\,C_2^{2}}{8}$$ $$-(1)$$

Precipitation of $$AY_2$$ occurs when $$\text{IP} \gt K_{sp}$$.

Substitute in $$(1)$$:
$$\text{IP} = \dfrac{(3.6\times10^{-3})(5.0\times10^{-4})^{2}}{8} = \dfrac{3.6\times10^{-3}\times25\times10^{-8}}{8} = \dfrac{9.0\times10^{-10}}{8} = 1.1\times10^{-10}$$
Since $$1.1\times10^{-10} \lt 5.2\times10^{-7}$$, no precipitate.

$$\text{IP} = \dfrac{(2.0\times10^{-4})(8.0\times10^{-4})^{2}}{8} = \dfrac{2.0\times10^{-4}\times64\times10^{-8}}{8} = \dfrac{1.28\times10^{-10}}{8} = 1.6\times10^{-11}$$
Again $$1.6\times10^{-11} \lt 5.2\times10^{-7}$$, so no precipitate.

$$\text{IP} = \dfrac{(2.0\times10^{-2})(2.0\times10^{-2})^{2}}{8} = \dfrac{2.0\times10^{-2}\times4.0\times10^{-4}}{8} = \dfrac{8.0\times10^{-6}}{8} = 1.0\times10^{-6}$$
Now $$1.0\times10^{-6} \gt 5.2\times10^{-7}$$, so $$\text{IP} \gt K_{sp}$$ and precipitate of $$AY_2$$ will form.

$$\text{IP} = \dfrac{(1.5\times10^{-4})(1.5\times10^{-3})^{2}}{8} = \dfrac{1.5\times10^{-4}\times2.25\times10^{-6}}{8} = \dfrac{3.375\times10^{-10}}{8} = 4.2\times10^{-11}$$
Since $$4.2\times10^{-11} \lt 5.2\times10^{-7}$$, no precipitate.

The only combination that satisfies $$\text{IP} \gt K_{sp}$$ is Option C.

Answer: Option C

To determine which precipitate forms first, we need to find the minimum $$[OH^-]$$ concentration required to start precipitation of each salt.

For $$A(OH)_2$$, the solubility product is given by $$K_{sp} = [A^{2+}][OH^-]^2$$. Substituting $$K_{sp} = 9 \times 10^{-10}$$ and assuming $$[A^{2+}] = 1$$ yields $$ 9 \times 10^{-10} = (1)[OH^-]^2 $$. Solving for $$[OH^-]$$ gives $$[OH^-] = \sqrt{9 \times 10^{-10}} = 3 \times 10^{-5} \text{ M}$$.

For $$B(OH)_3$$, the solubility product expression is $$K_{sp} = [B^{3+}][OH^-]^3$$. With $$K_{sp} = 27 \times 10^{-18}$$ and $$[B^{3+}] = 1$$, we have $$ 27 \times 10^{-18} = (1)[OH^-]^3 $$. This leads to $$[OH^-] = (27 \times 10^{-18})^{1/3} = 3 \times 10^{-6} \text{ M}$$.

Since $$B(OH)_3$$ requires a lower $$[OH^-]$$ concentration ($$3 \times 10^{-6}$$ M) to precipitate than $$A(OH)_2$$ ($$3 \times 10^{-5}$$ M), $$B(OH)_3$$ will precipitate first as $$NH_4OH$$ is gradually added.

The correct answer is Option 3: $$B(OH)_3$$ will precipitate before $$A(OH)_2$$.

For the gas-phase equilibrium $$A(g)\;\rightleftharpoons\;B(g)+C(g)$$ consider that one mole of $$A$$ is initially present at total pressure $$p$$. Let $$\alpha$$ be the degree of dissociation at equilibrium.

Moles present at equilibrium:
  • $$A$$ : $$1-\alpha$$
  • $$B$$ : $$\alpha$$
  • $$C$$ : $$\alpha$$

Total moles $$n_{\text{tot}} = 1-\alpha+\alpha+\alpha = 1+\alpha$$.

Because total pressure is $$p$$, the partial pressures are obtained from mole fractions:

$$p_A = \frac{1-\alpha}{1+\alpha}\,p,\qquad p_B = \frac{\alpha}{1+\alpha}\,p,\qquad p_C = \frac{\alpha}{1+\alpha}\,p$$

The pressure equilibrium constant is

$$K_p = \frac{p_B\,p_C}{p_A} = \frac{\left(\dfrac{\alpha p}{1+\alpha}\right) \left(\dfrac{\alpha p}{1+\alpha}\right)} {\dfrac{(1-\alpha)p}{1+\alpha}} = \frac{\alpha^{2}p}{(1-\alpha)(1+\alpha)} = \frac{\alpha^{2}p}{1-\alpha^{2}} \;-(1)$$

Rearrange $$-(1)$$ to express $$\alpha$$ in terms of $$K_p$$ and $$p$$:

$$\alpha^{2}p = K_p(1-\alpha^{2})$$
$$\alpha^{2}(p+K_p) = K_p$$
$$\alpha^{2} = \frac{K_p}{p+K_p} \;-(2)$$
$$\alpha = \sqrt{\frac{K_p}{p+K_p}}$$

Now study how $$\alpha$$ depends on the total pressure $$p$$ while temperature (hence $$K_p$$) remains constant.

When $$p$$ increases:
From $$-(2)$$ the denominator $$p+K_p$$ increases, so the fraction $$\dfrac{K_p}{p+K_p}$$ decreases. Therefore $$\alpha^{2}$$ and hence $$\alpha$$ decrease.

Thus, “When $$p$$ increases $$\alpha$$ decreases” is correct (Option B).

Check the other statements:

Case $$p \gg K_p$$:
$$\alpha^{2} \approx \dfrac{K_p}{p}\ll 1\;\Rightarrow\;\alpha\ll 1$$, not $$\alpha\approx 1$$. Hence Option A is wrong.

Case $$K_p \gg p$$:
$$\alpha^{2} \approx \dfrac{K_p}{K_p}=1\;\Rightarrow\;\alpha\approx 1$$, not “much less than unity”. Hence Option C is wrong.

Option D is opposite to the correct trend, so it is also wrong.

Therefore, the only true statement is Option B.

We need to determine what happens to the pH of water when it is heated from $$25°C$$ to $$80°C$$.

Understand the autoionization of water.

Water undergoes autoionization: $$H_2O \rightleftharpoons H^+ + OH^-$$

The ionic product of water is: $$K_w = [H^+][OH^-]$$

At $$25°C$$, $$K_w = 1.0 \times 10^{-14}$$, so $$[H^+] = [OH^-] = 10^{-7}$$ M, giving $$\text{pH} = 7$$.

Effect of temperature on $$K_w$$.

The autoionization of water is an endothermic process ($$\Delta H > 0$$). By Le Chatelier's principle, increasing the temperature shifts the equilibrium to the right, producing more $$H^+$$ and $$OH^-$$ ions.

Therefore, at $$80°C$$, $$K_w$$ is significantly larger than $$10^{-14}$$. For example, at $$80°C$$, $$K_w \approx 2.4 \times 10^{-13}$$.

Calculate the new pH.

At $$80°C$$: $$[H^+] = [OH^-] = \sqrt{K_w} = \sqrt{2.4 \times 10^{-13}} \approx 4.9 \times 10^{-7}$$ M

$$\text{pH} = -\log(4.9 \times 10^{-7}) \approx 6.31$$

Key observations.

Although the pH decreases below 7, the water is still neutral because $$[H^+] = [OH^-]$$. The pH decreased because more ions were produced, but the solution is neither acidic nor basic.

Note that both $$[H^+]$$ and $$[OH^-]$$ increase simultaneously. Option (2) states that $$H^+$$ increases but $$OH^-$$ decreases, which is incorrect.

The correct answer is Option (1): pH decreases.

For a basic buffer the Henderson-Hasselbalch form is
$$\mathrm{pOH}=pK_b+\log\left(\frac{[\text{salt}]}{[\text{base}]}\right)$$

Case 1: Before adding HCl
Number of moles of $$NH_3 = 0.10$$ mol, number of moles of $$NH_4Cl = 0.10$$ mol in $$1\,$$L, so
$$[\text{base}]=0.10\ \text{M},\;[\text{salt}]=0.10\ \text{M}$$
$$\frac{[\text{salt}]}{[\text{base}]}=1 \;\Longrightarrow\; \log 1 = 0$$
$$\therefore \mathrm{pOH}_1 = pK_b + 0 = 4.745$$
$$\mathrm{pH}_1 = 14 - \mathrm{pOH}_1 = 14 - 4.745 = 9.255$$

Case 2: After adding $$0.05$$ mol HCl
The strong acid reacts completely: $$NH_3 + H^+ \rightarrow NH_4^+$$
Moles after reaction:
$$NH_3: 0.10 - 0.05 = 0.05\ \text{mol}$$
$$NH_4^+: 0.10 + 0.05 = 0.15\ \text{mol}$$
With total volume still $$1\,$$L,
$$[\text{base}]=0.05\ \text{M},\;[\text{salt}]=0.15\ \text{M}$$
$$\frac{[\text{salt}]}{[\text{base}]} =\frac{0.15}{0.05}=3,\;\; \log 3 = 0.477$$
$$\mathrm{pOH}_2 = pK_b + 0.477 = 4.745 + 0.477 = 5.222$$
$$\mathrm{pH}_2 = 14 - 5.222 = 8.778$$

Change in pH
$$\Delta\mathrm{pH} = \mathrm{pH}_1 - \mathrm{pH}_2 = 9.255 - 8.778 = 0.477 \approx 0.48$$

Expressed as $$\Delta\mathrm{pH}=48 \times 10^{-2}$$.

The equilibrium to be studied is
$$CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g) \qquad -(1)$$

For this reaction the equilibrium constant in terms of pressure is
$$K_p = \frac{P_{CH_3OH}}{P_{CO}\;P_{H_2}^{\,2}} \qquad -(2)$$

Data supplied at 500 K in a rigid 2 dm$$^3$$ flask:
  • initial moles of $$CO = 0.10$$ mol
  • hydrogen is added so that the final total pressure is 5 bar
  • at equilibrium $$0.04$$ mol of $$CH_3OH$$ has been formed.

1. Establish the equilibrium mole numbers

Let $$x$$ be the moles of $$CH_3OH$$ produced. It is given that $$x = 0.04$$ mol.
By the stoichiometry of (1):
  • $$CO$$ consumed $$= x = 0.04$$ mol
  • $$H_2$$ consumed $$= 2x = 0.08$$ mol

Hence, at equilibrium
$$n_{CO}^{eq} = 0.10 - 0.04 = 0.06 \text{ mol}$$
$$n_{CH_3OH}^{eq} = 0.04 \text{ mol}$$
$$n_{H_2}^{eq} =\;?$$ (to be found next)

2. Find the total number of moles at equilibrium

The flask is finally at 5 bar. Using the ideal-gas equation $$PV = nRT$$:
$$n_{tot}^{eq} = \frac{P\,V}{R\,T} = \frac{5 \times 2}{0.08 \times 500} = \frac{10}{40} = 0.25 \text{ mol} \qquad -(3)$$

Therefore
$$n_{H_2}^{eq} = n_{tot}^{eq} - n_{CO}^{eq} - n_{CH_3OH}^{eq} = 0.25 - 0.06 - 0.04 = 0.15 \text{ mol}$$

3. Calculate the partial pressures

Using $$P_i = \frac{n_i RT}{V}$$ with $$RT/V = \frac{0.08 \times 500}{2}=20$$ bar mol$$^{-1}$$:

$$P_{CO}=0.06 \times 20 = 1.2 \text{ bar}$$
$$P_{H_2}=0.15 \times 20 = 3.0 \text{ bar}$$
$$P_{CH_3OH}=0.04 \times 20 = 0.8 \text{ bar}$$

4. Evaluate $$K_p$$

Insert the partial pressures in (2):
$$K_p = \frac{0.8}{1.2 \,(3.0)^2} = \frac{0.8}{1.2 \times 9} = \frac{0.8}{10.8} = 7.407 \times 10^{-2}$$

5. Express in the asked form

$$K_p = 7.407 \times 10^{-2} = 74.07 \times 10^{-3}$$

To the nearest integer, $$K_p^0 = 74 \times 10^{-3}$$.

Given a 1 mM solution of ethylamine produces pH = 9.

Since ethylamine is a base, we find pOH first: $$ pOH = 14 - pH = 14 - 9 = 5 $$ and $$ [OH^-] = 10^{-5} \text{ M} $$.

The ionization of ethylamine in water is represented by the equilibrium $$ C_2H_5NH_2 + H_2O \rightleftharpoons C_2H_5NH_3^+ + OH^- $$.

The base‐ionization constant is given by $$ K_b = \frac{[C_2H_5NH_3^+][OH^-]}{[C_2H_5NH_2]} $$.

Assuming the degree of ionization is negligible compared to unity, we have $$ K_b = \frac{[OH^-]^2}{C} $$ where $$ C = 1 \text{ mM} = 10^{-3} \text{ M} $$.

Substituting the concentration of hydroxide ions, $$ K_b = \frac{(10^{-5})^2}{10^{-3}} = \frac{10^{-10}}{10^{-3}} = 10^{-7} $$.

Thus $$ K_b = 10^{-x} $$ where $$ x = 7 $$.

The answer is 7.

Let us start with $$1$$ mole of $$H_2O(g)$$ and let the degree of dissociation at equilibrium be $$\alpha$$.

For the reaction $$H_2O(g) \rightleftharpoons H_2(g) + \frac12 O_2(g)$$ the mole table is:

Initial moles: $$H_2O = 1$$, $$H_2 = 0$$, $$O_2 = 0$$
Change in moles: $$H_2O = -\alpha$$, $$H_2 = +\alpha$$, $$O_2 = +\dfrac{\alpha}{2}$$
Equilibrium moles: $$H_2O = 1-\alpha$$, $$H_2 = \alpha$$, $$O_2 = \dfrac{\alpha}{2}$$

Total moles at equilibrium: $$n_{\text{tot}} = 1 - \alpha + \alpha + \dfrac{\alpha}{2} = 1 + \dfrac{\alpha}{2}$$ $$-(1)$$

The total pressure is given as $$P_{\text{tot}} = 1\text{ bar}$$, therefore the partial pressures are

$$P_{H_2O} = \dfrac{1-\alpha}{1+\dfrac{\alpha}{2}}\;(1\text{ bar})$$,
$$P_{H_2} = \dfrac{\alpha}{1+\dfrac{\alpha}{2}}\;(1\text{ bar})$$,
$$P_{O_2} = \dfrac{\dfrac{\alpha}{2}}{1+\dfrac{\alpha}{2}}\;(1\text{ bar})$$

The equilibrium constant for the gaseous reaction is

$$K_p = \dfrac{P_{H_2}\,(P_{O_2})^{1/2}}{P_{H_2O}}$$

Substituting the expressions of partial pressures:

$$K_p = \dfrac{\dfrac{\alpha}{1+\dfrac{\alpha}{2}}\left(\dfrac{\dfrac{\alpha}{2}}{1+\dfrac{\alpha}{2}}\right)^{1/2}}{\dfrac{1-\alpha}{1+\dfrac{\alpha}{2}}}$$

Simplifying (the common factor $$1+\dfrac{\alpha}{2}$$ cancels once):

$$K_p = \dfrac{\alpha\,\sqrt{\dfrac{\alpha}{2}}}{(1-\alpha)\sqrt{1+\dfrac{\alpha}{2}}}$$ $$-(2)$$

The data give $$K_p = 8.0 \times 10^{-3}$$. Since dissociation of steam at 2300 K is small, we first make the approximation $$\alpha \ll 1$$ so that $$1-\alpha \approx 1$$ and $$1+\dfrac{\alpha}{2}\approx 1$$. With this, equation $$(2)$$ reduces to

$$K_p \approx \dfrac{\alpha^{3/2}}{\sqrt{2}}$$

Hence

$$\alpha^{3/2} \approx K_p\,\sqrt{2} = (8.0\times 10^{-3})(1.414) = 1.131\times 10^{-2}$$

$$\alpha \approx \left(1.131\times 10^{-2}\right)^{2/3} \approx 5.0\times 10^{-2}$$

To improve accuracy, put $$\alpha = 0.05$$ back into the exact expression $$(2)$$:

Numerator: $$\alpha\,\sqrt{\dfrac{\alpha}{2}} = 0.05\sqrt{\dfrac{0.05}{2}} = 0.05(0.1581)=0.00791$$
Denominator: $$(1-\alpha)\sqrt{1+\dfrac{\alpha}{2}} = 0.95\sqrt{1.025}=0.95(1.012)=0.961$$
$$K_p = \dfrac{0.00791}{0.961}=8.2\times10^{-3}\approx 8.0\times10^{-3}$$

The calculated value matches the given $$K_p$$, confirming $$\alpha \approx 0.05$$.

Expressing $$\alpha$$ as $$\alpha \times 10^{-2}$$ gives $$5.0 \times 10^{-2}$$, whose nearest integer value is

5.

Therefore, the degree of dissociation of water under the stated conditions is $$\boxed{5\times 10^{-2}}$$ (nearest integer 5).

For a weak monoprotic acid HX the dissociation equilibrium is
$$HX \rightleftharpoons H^{+}+X^{-}$$

The acid-dissociation constant is defined as
$$K_a = \frac{[H^{+}][X^{-}]}{[HX]} \qquad -(1)$$

Let the molarity of the acid after dilution be $$C$$ M and let the equilibrium hydrogen-ion concentration be $$h$$ M.
At equilibrium:
$$[H^{+}] = h, \quad [X^{-}] = h, \quad [HX] = C-h$$

Substituting these values in equation $$(1)$$,
$$K_a = \frac{h^{2}}{C-h} \qquad -(2)$$

Re-arranging $$(2)$$ to express $$C$$ in terms of $$h$$:
$$C-h = \frac{h^{2}}{K_a}\quad\Longrightarrow\quad C = h + \frac{h^{2}}{K_a} \qquad -(3)$$

After dilution the pH of the solution is given to be $$6$$, therefore
$$h = [H^{+}] = 10^{-6}\;{\rm M}$$

Putting $$h = 10^{-6}\;{\rm M}$$ and $$K_a = 4\times10^{-10}$$ in equation $$(3)$$:
$$C = 10^{-6} + \frac{(10^{-6})^{2}}{4\times10^{-10}} = 10^{-6} + \frac{10^{-12}}{4\times10^{-10}}$$

Calculate the second term:
$$\frac{10^{-12}}{4\times10^{-10}} = \frac{1}{4}\times10^{-2} = 2.5\times10^{-3}$$

Thus
$$C = 10^{-6} + 2.5\times10^{-3} \approx 2.501\times10^{-3}\,{\rm M}$$

The small term $$10^{-6}$$ may be ignored for quick estimation, giving $$C \approx 2.5\times10^{-3}\,{\rm M}$$.

Express this concentration in the form $$x\times10^{-4}\,{\rm M}$$:
$$2.5\times10^{-3}\,{\rm M} = 25\times10^{-4}\,{\rm M}$$

Hence $$x = 25$$ (nearest integer).

Answer: 25

The equilibrium in the gaseous phase is
$$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$$

For this reaction the equilibrium constant (in terms of pressure) is
$$K_p = \frac{P_{PCl_3}\,P_{Cl_2}}{P_{PCl_5}}$$  $$-(1)$$
At a fixed temperature $$K_p$$ has a fixed value.

Step 1 : What happens immediately after adding xenon (inert gas)?
The total pressure is kept constant but the total amount of gas (moles) increases. Because $$P_i = y_i\,p$$ (where $$y_i$$ is the mole-fraction and $$p$$ is the constant total pressure), every $$y_i$$ decreases when inert gas is added. Hence the partial pressures $$P_{PCl_5},\;P_{PCl_3},\;P_{Cl_2}$$ all drop at that instant, so the mixture is no longer at equilibrium.

Step 2 : Direction of the shift
In equation $$(1)$$, the numerator involves the product gases while the denominator involves the reactant gas. After dilution, the numerator $$P_{PCl_3}\,P_{Cl_2}$$ decreases more than the denominator $$P_{PCl_5}$$, making the ratio smaller than the required constant value of $$K_p$$.
To raise this ratio back to $$K_p$$, the reaction must produce more product gases, i.e. it must shift to the right. (Le-Châtelier’s principle also predicts this: at constant $$T$$ and $$p$$, increasing volume favours the side with the larger number of moles; here $$\Delta n = +1$$.)

Step 3 : Final concentrations
Because the equilibrium has shifted to the right:
• extra $$PCl_3$$ and $$Cl_2$$ are formed, so their amounts increase.
• some $$PCl_5$$ is consumed, so its amount decreases.
Although the volume of the system is larger than before, the increase in moles of $$PCl_3$$ is sufficient to make its final concentration higher than the original equilibrium concentration. (On the other hand, the concentration of $$PCl_5$$ definitely falls, and the behaviour of $$Cl_2$$ depends on the initial composition; it is not guaranteed to decrease universally.)

Step 4 : Choosing the correct option
The only statement that must be true after the system re-establishes equilibrium is: “$$PCl_3$$ will increase”.

Hence, the correct choice is Option D.

The required pH is $$10.0$$, so

$$\text{pOH}=14-\text{pH}=14-10=4$$

Hence the hydroxide-ion concentration is

$$[OH^-]=10^{-4}\ \text{mol L}^{-1}$$

Magnesium hydroxide dissociates as

$$Mg(OH)_2 \rightarrow Mg^{2+}+2\,OH^-$$

From the stoichiometry, $$1$$ mole of $$Mg(OH)_2$$ releases $$2$$ moles of $$OH^-$$.
Therefore moles of $$Mg(OH)_2$$ that must dissolve are

$$n=\frac{[OH^-]}{2} =\frac{10^{-4}}{2} =5\times10^{-5}\ \text{mol}$$

Molar mass of $$Mg(OH)_2$$ is $$58\ \text{g mol}^{-1}$$, so the required mass is

$$m = n \times \text{molar mass} = 5\times10^{-5}\times58 = 2.9\times10^{-3}\ \text{g}$$

Converting to milligrams,

$$m = 2.9\ \text{mg}$$

The nearest integer value of $$x$$ is $$3\ \text{mg}$$.

Answer: $$\displaystyle 3\ \text{mg}$$

We need to find the ratio $$K_P/K_C$$ for the reaction $$CO_{(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons CO_{2(g)}$$. The relationship between $$K_P$$ and $$K_C$$ is given by $$K_P = K_C(RT)^{\Delta n_g}$$ where $$\Delta n_g$$ is the change in the number of moles of gaseous species (moles of gaseous products minus moles of gaseous reactants).

Moles of gaseous products = 1 (CO$$_2$$) and moles of gaseous reactants = 1 (CO) + 1/2 (O$$_2$$) = 3/2, so $$\Delta n_g = 1 - \frac{3}{2} = -\frac{1}{2}$$.

Substituting into the formula gives $$\frac{K_P}{K_C} = (RT)^{\Delta n_g} = (RT)^{-1/2} = \frac{1}{\sqrt{RT}}$$, therefore the correct answer is Option (1): $$\frac{1}{\sqrt{RT}}$$.

We need to determine in which medium the equilibrium $$Cr_2O_7^{2-} \rightleftharpoons 2CrO_4^{2-}$$ shifts to the right.

Write the complete ionic equilibrium

The interconversion between dichromate and chromate ions in aqueous solution involves $$H^+$$ and $$OH^-$$ ions. The complete equation is:

$$Cr_2O_7^{2-} + 2OH^- \rightleftharpoons 2CrO_4^{2-} + H_2O$$

Alternatively, in terms of $$H^+$$:

$$2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O$$

Apply Le Chatelier's Principle

The forward reaction (dichromate to chromate) consumes $$OH^-$$ ions. According to Le Chatelier's Principle:

- In a basic medium (excess $$OH^-$$): The high concentration of $$OH^-$$ drives the equilibrium to the right, favouring the formation of yellow $$CrO_4^{2-}$$ (chromate) ions.

- In an acidic medium (excess $$H^+$$): The $$OH^-$$ concentration is low, so the equilibrium shifts to the left, favouring the orange $$Cr_2O_7^{2-}$$ (dichromate) ions.

Conclusion

This is why dichromate solutions appear orange in acidic conditions and turn yellow (chromate) in basic conditions. The equilibrium shifts to the right in a basic medium.

The correct answer is Option (2): a basic medium.

1. Convert Solubility to Molarity ($$S$$)

The given solubility is $$W\text{ g}$ per $100\text{ mL}$$. To find the molarity ($$S$$, in $$\text{mol/L}$$):

• Mass in $$1000\text{ mL } (1\text{ L}) = 10 \times W\text{ g}$$
• Moles per liter ($$S$$) = $$\frac{10W}{M}\text{ mol/L}$$

2. Dissociation Equilibrium

Calcium phosphate, $$\text{Ca}_3(\text{PO}_4)_2$$, dissociates in water as follows:

$$\text{Ca}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Ca}^{2+}(aq) + 2\text{PO}_4^{3-}(aq)$$

If the molar solubility is $$S$$:

• $$[\text{Ca}^{2+}] = 3S$$
• $$[\text{PO}_4^{3-}] = 2S$$

3. Calculate $$K_{sp}$$ Expression

$$K_{sp} = [\text{Ca}^{2+}]^3 \cdot [\text{PO}_4^{3-}]^2$$

$$K_{sp} = (3S)^3 \cdot (2S)^2 = 27S^3 \cdot 4S^2 = 108S^5$$

4. Substitute $$S$$ into the Equation

Now, substitute $$S = \frac{10W}{M}$$:

$$K_{sp} = 108 \left( \frac{10W}{M} \right)^5 = 108 \times 10^5 \times \frac{W^5}{M^5}$$

$$K_{sp} = 1.08 \times 10^7 \times \frac{W^5}{M^5} \approx 10^7 \frac{W^5}{M^5}$$

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We add 25 mL of 1 M AgNO$$_3$$ to 25 mL of 1.05 M KI and need to determine which ions are present in very small quantity in the resulting solution.

The initial moles are: Ag$$^+$$ from AgNO$$_3$$ = $$25 \times 1 = 25$$ mmol, I$$^-$$ from KI = $$25 \times 1.05 = 26.25$$ mmol, K$$^+$$ = 26.25 mmol, and NO$$_3^-$$ = 25 mmol.

The precipitation reaction is

Silver iodide is highly insoluble with $$K_{sp} = 8.5 \times 10^{-17}$$, so the reaction goes essentially to completion.

Since Ag$$^+$$ (25 mmol) and I$$^-$$ (26.25 mmol) react in a 1:1 ratio, Ag$$^+$$ is the limiting reagent. After precipitation, the remaining I$$^-$$ is $$26.25 - 25 = 1.25$$ mmol in 50 mL total, giving a concentration of $$\frac{1.25}{50} \times 1000 = 0.025$$ M.

Now, from the solubility equilibrium of AgI in the presence of excess I$$^-$$,

Comparing the concentrations of all ions in the 50 mL solution: K$$^+$$ = 0.525 M, NO$$_3^-$$ = 0.5 M, I$$^-$$ = 0.025 M, and Ag$$^+$$ = $$3.4 \times 10^{-15}$$ M. Both Ag$$^+$$ and I$$^-$$ are present in very small quantity relative to the spectator ions.

Hence, the correct answer is Option 4: Ag$$^+$$ and I$$^-$$ both.

For the weak electrolyte A$$_2$$B$$_3$$ dissociating as

$$\text{A}_2\text{B}_3 \rightleftharpoons 2\text{A} + 3\text{B}$$

If the initial concentration is $$C$$ and degree of dissociation is $$\alpha$$, then at equilibrium:

$$[\text{A}_2\text{B}_3] = C(1 - \alpha) \approx C$$ (since $$\alpha$$ is small for a concentrated solution), $$[\text{A}] = 2C\alpha$$, and $$[\text{B}] = 3C\alpha$$.

The equilibrium constant is

$$K_{eq} = \frac{[\text{A}]^2[\text{B}]^3}{[\text{A}_2\text{B}_3]} = \frac{(2C\alpha)^2(3C\alpha)^3}{C}$$

$$K_{eq} = \frac{4C^2\alpha^2 \times 27C^3\alpha^3}{C} = 108C^4\alpha^5$$

Solving for $$\alpha$$,

$$\alpha^5 = \frac{K_{eq}}{108C^4}$$

$$\alpha = \left(\frac{K_{eq}}{108C^4}\right)^{1/5}$$

Hence, the correct answer is $$\alpha = \left(\dfrac{K_{eq}}{108c^4}\right)^{1/5}$$.

The pH of a solution is related to the hydrogen ion concentration by the expression $$pH = -\log_{10}[H^+]$$.

Let the initial hydrogen ion concentration be $$[H^+]_1$$. Then, the initial pH is $$pH_1 = -\log[H^+]_1$$.

If the hydrogen ion concentration increases by a factor of 1000, then the new concentration becomes $$[H^+]_2 = 1000[H^+]_1$$.

The new pH is

$$pH_2 = -\log(1000[H^+]_1) = -(\log1000 + \log[H^+]_1) = -(3 + \log[H^+]_1) = pH_1 - 3.$$

Hence, when the hydrogen ion concentration increases by a factor of 1000, the pH decreases by 3 units.

Similarly, if the hydrogen ion concentration decreases by a factor of 1000, then $$[H^+]_2 = \frac{[H^+]_1}{1000}$$.

The new pH becomes

$$pH_2 = -\log\left(\frac{[H^+]_1}{1000}\right) = -(\log[H^+]_1 - 3) = pH_1 + 3.$$

Therefore, a thousandfold decrease in hydrogen ion concentration increases the pH by 3 units.

Hence, for a thousandfold increase in hydrogen ion concentration, the correct statement is that the pH decreases by 3 units. Therefore, the correct answer is $$\boxed{\text{B}}$$.

The equilibrium reaction is: $$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$$

When helium gas is added at constant volume:

1. The total pressure increases, but the partial pressures of the reacting gases remain unchanged.

2. Since the equilibrium constant $$K_p$$ depends on the partial pressures of the reacting gases (not the total pressure), and the partial pressures haven't changed, the system remains at equilibrium.

3. Similarly, at constant volume, the concentrations of the reacting species are unchanged, so $$K_c$$ is also unaffected.

4. Helium is an inert gas and does not participate in the reaction.

Therefore, the addition of helium at constant volume will not affect the equilibrium.

Note: If helium were added at constant pressure (instead of constant volume), the volume would increase, effectively decreasing the concentrations, which would shift the equilibrium forward.

Analyzing each statement:

(A) pH of $$1 \times 10^{-8}$$ M HCl is 8: Incorrect. At such low concentration, we must consider the autoionization of water. The pH will be slightly less than 7, not 8 (an acid cannot have pH > 7).

(B) Conjugate base of H₂PO₄⁻ is HPO₄²⁻: Correct. H₂PO₄⁻ donates a proton to become HPO₄²⁻.

(C) K_w increases with increase in temperature: Correct. The dissociation of water is endothermic, so K_w increases with temperature.

(D) At half neutralization point, pH = ½ pKa: Incorrect. At the half neutralization point of a weak acid, pH = pKa (Henderson-Hasselbalch equation: pH = pKa + log(1) = pKa), not ½ pKa.

Correct statements: B and C.

This matches option 1: (B), (C).

We recall that an acid-base indicator is itself a weak acid or a weak base which exists in two tautomeric/conjugate forms having different colours. Methyl orange behaves exactly in this manner.

Case 1: Nature of methyl orange
 • The indicator may be written as $$HIn$$ (acidic form) in equilibrium with its conjugate base $$In^-$$.
 • The dissociation equilibrium is $$HIn \rightleftharpoons H^{+}+In^{-}$$.
 • Since the position of this equilibrium changes with pH and methyl orange shows a pH range of about $$3.1$$ to $$4.4$$, the acid dissociation constant $$K_a$$ is small. Hence methyl orange is a weak acid.
Therefore Statement-I is correct.

Case 2: Colour of the two structural forms
 • In acidic solution the protonated (benzenoid) structure dominates and the colour observed is red.
 • In basic solution the de-protonated (quinonoid) structure dominates and the colour observed is yellow/orange.
 • Red is visually a deeper or more intense colour than yellow. Hence, for methyl orange, the benzenoid form is more intensely coloured than the quinonoid form.
Therefore Statement-II is also correct.

Because both statements are correct, the most appropriate choice is Option C.

Let the sparingly soluble salt be $$MX$$ and its anion be the conjugate base of the weak acid $$HX$$.

Dissolution equilibrium of the salt: $$MX(s) \rightleftharpoons M^{+} + X^{-}$$
Solubility in the given medium = $$S$$ mol L$$^{-1}$$ ⇒ $$[M^{+}] = S$$.

The anion $$X^{-}$$ is protonated by the hydronium ions present in solution:
$$X^{-} + H^{+} \rightleftharpoons HX$$ with $$K_{a} = \dfrac{[H^{+}][X^{-}]}{[HX]}$$.

Out of the total $$S$$ moles of $$X^{-}$$ produced, only a fraction remains as $$X^{-}$$; let that fraction be $$\alpha$$. Then
$$[X^{-}] = \alpha S,\qquad [HX] = (1-\alpha)S$$.

Applying the definition of $$K_{a}$$:
$$K_{a}= \dfrac{[H^{+}]\,\alpha S}{(1-\alpha)S}= [H^{+}]\,\dfrac{\alpha}{1-\alpha} \; -(1)$$
⇒ $$\dfrac{\alpha}{1-\alpha}= \dfrac{K_{a}}{[H^{+}]} \; -(2)$$
⇒ $$\alpha = \dfrac{K_{a}}{K_{a} + [H^{+}]} \; -(3)$$.

Solubility‐product of the salt:
$$K_{sp} = [M^{+}][X^{-}] = S \times (\alpha S)= S^{2}\,\dfrac{K_{a}}{K_{a} + [H^{+}]} \; -(4)$$.

The same $$K_{sp}$$ must hold at every pH. So for two different pH values,

Case 1: pH 7 ⇒ $$[H^{+}]_{1}=10^{-7}$$ mol L$$^{-1}$$, given $$S_{1}=10^{-4}$$ mol L$$^{-1}$$.
$$K_{sp}= S_{1}^{2}\,\dfrac{K_{a}}{K_{a}+10^{-7}} \; -(5)$$.

Case 2: pH 2 ⇒ $$[H^{+}]_{2}=10^{-2}$$ mol L$$^{-1}$$, given $$S_{2}=10^{-3}$$ mol L$$^{-1}$$.
$$K_{sp}= S_{2}^{2}\,\dfrac{K_{a}}{K_{a}+10^{-2}} \; -(6)$$.

Equate $$(5)$$ and $$(6)$$ because $$K_{sp}$$ is constant:

$$\dfrac{S_{1}^{2}}{K_{a}+10^{-7}} = \dfrac{S_{2}^{2}}{K_{a}+10^{-2}}$$

Insert the numerical solubilities:

$$\dfrac{(10^{-4})^{2}}{K_{a}+10^{-7}} = \dfrac{(10^{-3})^{2}}{K_{a}+10^{-2}}$$

Simplify the solubility ratio:
$$(10^{-4})^{2} = 10^{-8}, \quad (10^{-3})^{2} = 10^{-6}$$
$$\dfrac{10^{-8}}{K_{a}+10^{-7}} = \dfrac{10^{-6}}{K_{a}+10^{-2}}$$

Cross-multiply:

$$10^{-8}(K_{a}+10^{-2}) = 10^{-6}(K_{a}+10^{-7})$$

Divide both sides by $$10^{-8}$$ to clear the powers of 10:

$$(K_{a}+10^{-2}) = 100\,(K_{a}+10^{-7})$$

Expand the right side:

$$K_{a}+10^{-2} = 100K_{a} + 10^{-5}$$

Collect like terms:

$$10^{-2} - 10^{-5} = 100K_{a} - K_{a} = 99K_{a}$$

Evaluate the left side:
$$10^{-2} - 10^{-5} = 0.01 - 0.00001 = 0.00999 = 9.99\times10^{-3}$$

Hence
$$K_{a} = \dfrac{9.99\times10^{-3}}{99} \approx 1.0\times10^{-4}$$

Therefore
$$pK_{a} = -\log_{10}(K_{a}) \approx -\log_{10}(1.0\times10^{-4}) = 4$$

So, the pK$$_a$$ of the weak acid $$HX$$ is 4.

Option B which is: 4

We are given the Freundlich adsorption isotherm plotted as a straight line: $$y = 3x + 2.505$$.

The Freundlich equation is: $$\frac{x}{m} = K \cdot p^{1/n}$$

Taking logarithm:

$$\log\frac{x}{m} = \log K + \frac{1}{n}\log p$$

Comparing with $$y = 3x + 2.505$$:

Here $$y = \log(x/m)$$ and $$x$$-axis = $$\log p$$.

Slope = $$\frac{1}{n} = 3$$

Intercept = $$\log K = 2.505$$

$$\frac{1}{n} = 3$$ and $$\log K = 2.505$$.

The correct answer is Option C: $$3$$ and $$2.505$$.

The flask volume is 1 L, hence the numerical values of moles and molarities are the same.

Initial state (common to both runs):
number of moles of $$A = 6$$,
number of moles of $$P = 0$$.

From the graph (plateau values):
at temperature $$T_1$$, moles of $$P$$ formed $$=4$$ ⇒ moles of $$A$$ left $$=6-4=2$$;
at temperature $$T_2$$, moles of $$P$$ formed $$=2$$ ⇒ moles of $$A$$ left $$=6-2=4$$.

The reaction is $$A(g) \rightleftharpoons P(g)$$, so
$$K = \frac{[P]}{[A]}$$.

Equilibrium constants:
$$K_1 = \frac{4}{2} = 2$$ at $$T_1$$;
$$K_2 = \frac{2}{4} = 0.5$$ at $$T_2$$.

Standard Gibbs free energy relation:
$$\Delta G^\circ = -RT \ln K$$.

Hence
$$\Delta G_1^\circ = -R T_1 \ln K_1$$,   $$\Delta G_2^\circ = -R T_2 \ln K_2$$.

The required difference is
$$\Delta G_2^\circ - \Delta G_1^\circ = -R T_2 \ln K_2 + R T_1 \ln K_1$$.

Given $$T_1 = 2T_2$$, substitute:
$$\Delta G_2^\circ - \Delta G_1^\circ = R T_2 \left(2\ln K_1 - \ln K_2\right)$$.

Combine the logarithms:
$$2\ln K_1 - \ln K_2 = \ln \left(\frac{K_1^2}{K_2}\right)$$.

Therefore
$$\Delta G_2^\circ - \Delta G_1^\circ = R T_2 \ln\!\left(\frac{K_1^2}{K_2}\right).$$

Comparing with $$\Delta G_2^\circ - \Delta G_1^\circ = R T_2 \ln x$$ gives
$$x = \frac{K_1^2}{K_2} = \frac{(2)^2}{0.5} = \frac{4}{0.5} = 8.$$

Hence, the value of $$x$$ is 8.

Find the pH of a buffer solution containing 0.1 mole each of $$NH_3$$ and $$NH_4Cl$$ in 1 L after adding 0.02 mole of HCl.

We first determine the moles after HCl addition. HCl reacts with $$NH_3$$: $$NH_3 + HCl \rightarrow NH_4Cl$$. Moles of $$NH_3$$ remaining = $$0.1 - 0.02 = 0.08$$ mol and moles of $$NH_4^+$$ formed = $$0.1 + 0.02 = 0.12$$ mol.

Next, we apply the Henderson-Hasselbalch equation for a basic buffer: $$pOH = pK_b + \log\frac{[NH_4^+]}{[NH_3]}$$, which yields $$pOH = 4.745 + \log\frac{0.12}{0.08}$$.

Now the logarithm is computed as $$\log\frac{0.12}{0.08} = \log\frac{3}{2} = \log 3 - \log 2 = 0.477 - 0.301 = 0.176$$.

Therefore, $$pOH = 4.745 + 0.176 = 4.921$$ and $$pH = 14 - 4.921 = 9.079$$. Expressing this in the required form gives $$pH = 9.079 = 9079 \times 10^{-3}$$, so the answer is 9079.

We need to find the equilibrium constant for:

$$2\text{NH}_3(g) + \frac{5}{2}\text{O}_2(g) \rightleftharpoons 2\text{NO}(g) + 3\text{H}_2\text{O}(g)$$

Given equilibria:

(1) $$\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$$, $$K_1 = 4 \times 10^5$$

(2) $$\text{N}_2 + \text{O}_2 \rightleftharpoons 2\text{NO}$$, $$K_2 = 1.6 \times 10^{12}$$

(3) $$\text{H}_2 + \frac{1}{2}\text{O}_2 \rightleftharpoons \text{H}_2\text{O}$$, $$K_3 = 1.0 \times 10^{-13}$$

The target reaction can be obtained by combining:

Reverse of (1) + (2) + 3 times (3)

Verification:

Reverse (1): $$2\text{NH}_3 \rightarrow \text{N}_2 + 3\text{H}_2$$

(2): $$\text{N}_2 + \text{O}_2 \rightarrow 2\text{NO}$$

3 × (3): $$3\text{H}_2 + \frac{3}{2}\text{O}_2 \rightarrow 3\text{H}_2\text{O}$$

Adding: $$2\text{NH}_3 + \text{O}_2 + \frac{3}{2}\text{O}_2 \rightarrow 2\text{NO} + 3\text{H}_2\text{O}$$

$$2\text{NH}_3 + \frac{5}{2}\text{O}_2 \rightarrow 2\text{NO} + 3\text{H}_2\text{O}$$ ✓

The equilibrium constant is:

$$ K = \frac{1}{K_1} \times K_2 \times K_3^3 $$

$$ K = \frac{1}{4 \times 10^5} \times 1.6 \times 10^{12} \times (1.0 \times 10^{-13})^3 $$

$$ K = \frac{1.6 \times 10^{12}}{4 \times 10^5} \times 10^{-39} $$

$$ K = 0.4 \times 10^{7} \times 10^{-39} = 4 \times 10^{6} \times 10^{-39} = 4 \times 10^{-33} $$

So the answer is $$4 \times 10^{-33}$$, meaning the coefficient is $$\boxed{4}$$.

This matches the recorded answer of 4.

We have the reaction

This is an exothermic reaction with 3 moles of gaseous reactants producing 2 moles of gaseous products. We analyze each factor using Le Chatelier's principle.

Case A: Increasing temperature shifts an exothermic equilibrium backward, decreasing the yield of SO$$_3$$.

Case B: Increasing pressure favours the side with fewer moles of gas (product side with 2 moles vs. 3 moles on the reactant side), so it increases the yield.

Case C: Adding more SO$$_2$$ shifts the equilibrium forward, increasing the yield of SO$$_3$$.

Case D: Adding more O$$_2$$ also shifts the equilibrium forward, increasing the yield.

Case E: Adding a catalyst increases the rate of both forward and backward reactions equally, so it does not affect the equilibrium position or yield.

The factors that increase the yield of SO$$_3$$ are B, C, and D. Hence, the number of factors is $$3$$. So, the answer is $$3$$.

We need to find the solubility of BaSO$$_4$$ in 0.1 M K$$_2$$SO$$_4$$ solution, given $$K_{sp}(\text{BaSO}_4) = 1 \times 10^{-10}$$ at 298 K.

The dissolution equilibrium is:

$$\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)$$

Now, K$$_2$$SO$$_4$$ is a strong electrolyte that completely dissociates, giving an initial SO$$_4^{2-}$$ concentration of 0.1 M (common ion effect). Let the molar solubility of BaSO$$_4$$ be $$s$$ mol/L.

At equilibrium, $$[\text{Ba}^{2+}] = s$$ and $$[\text{SO}_4^{2-}] = 0.1 + s \approx 0.1$$ (since $$s \ll 0.1$$).

Applying the solubility product expression:

$$K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]$$

$$1 \times 10^{-10} = s \times 0.1$$

$$s = \frac{1 \times 10^{-10}}{0.1} = 1 \times 10^{-9}$$ mol/L

Converting to g/L using the molar mass of BaSO$$_4$$ = 233 g/mol:

$$\text{Solubility in g/L} = 1 \times 10^{-9} \times 233 = 233 \times 10^{-9}$$ g/L

So, the answer is $$233$$.

20 mL of 0.1 M NaOH + 50 mL of 0.1 M CH₃COOH.

Moles of NaOH = 0.002 mol

Moles of CH₃COOH = 0.005 mol

After reaction: NaOH fully consumed, remaining CH₃COOH = 0.003 mol, CH₃COONa formed = 0.002 mol.

This is a buffer solution. Using Henderson-Hasselbalch equation:

$$pH = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]} = 4.76 + \log\frac{0.002}{0.003}$$

$$= 4.76 + \log\frac{2}{3} = 4.76 + \log 2 - \log 3$$

$$= 4.76 + 0.30 - 0.48 = 4.58$$

In the required form: $$458 \times 10^{-2}$$

This matches the answer key value of $$\mathbf{458}$$.

600 mL of 0.01 M HCl and 400 mL of 0.01 M H$$_2$$SO$$_4$$ are mixed. The moles of H$$^+$$ from HCl = 0.6 × 0.01 = 0.006 mol, and the moles of H$$^+$$ from H$$_2$$SO$$_4$$ = 0.4 × 0.01 × 2 = 0.008 mol (diprotic acid), giving a total H$$^+$$ = 0.006 + 0.008 = 0.014 mol.

The total volume is 600 + 400 = 1000 mL = 1 L, so [H$$^+$$] = 0.014 M.

$$pH = -\log(0.014) = -\log(14 \times 10^{-3}) = 3 - \log 14$$

$$= 3 - \log 2 - \log 7 = 3 - 0.30 - 0.84 = 1.86$$

pH = 1.86 = $$186 \times 10^{-2}$$, so the answer is $$\boxed{186}$$.

The reaction is: $$\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)$$

Initial moles: CO = 1 mol, H$$_2$$O = 1 mol, CO$$_2$$ = 0, H$$_2$$ = 0

At equilibrium: 40% of water reacts, so 0.4 mol of H$$_2$$O reacts.

$$\text{CO} = 1 - 0.4 = 0.6 \text{ mol}$$

$$\text{H}_2\text{O} = 1 - 0.4 = 0.6 \text{ mol}$$

$$\text{CO}_2 = 0.4 \text{ mol}$$

$$\text{H}_2 = 0.4 \text{ mol}$$

Concentrations (Volume = 10 L):

$$[\text{CO}] = [\text{H}_2\text{O}] = \frac{0.6}{10} = 0.06 \text{ M}$$

$$[\text{CO}_2] = [\text{H}_2] = \frac{0.4}{10} = 0.04 \text{ M}$$

Equilibrium constant:

$$K_C = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} = \frac{0.04 \times 0.04}{0.06 \times 0.06} = \frac{0.0016}{0.0036} = \frac{4}{9}$$

$$K_C \times 10^2 = \frac{4}{9} \times 100 = 44.44 \approx 44$$

When 25 mL of 0.2 M $$CH_3COONa$$ is mixed with 25 mL of 0.02 M $$CH_3COOH$$, the total volume is 50 mL, and the concentrations become $$[CH_3COONa] = \frac{25 \times 0.2}{50} = 0.1$$ M and $$[CH_3COOH] = \frac{25 \times 0.02}{50} = 0.01$$ M.

This is a buffer solution, so using the Henderson-Hasselbalch equation, $$\text{pH} = \text{p}K_a + \log\frac{[\text{Salt}]}{[\text{Acid}]} = \text{p}K_a + \log\frac{0.1}{0.01} = \text{p}K_a + 1$$.

Given pH = 5, it follows that $$\text{p}K_a = 4$$, hence $$K_a = 10^{-4} = 10 \times 10^{-5}$$ and therefore $$x = \boxed{10}$$.

After mixing: $$[CH_3COONa] = \frac{0.2 \times 25}{50} = 0.1$$ M, $$[CH_3COOH] = \frac{0.02 \times 25}{50} = 0.01$$ M.

Henderson-Hasselbalch: $$pH = pK_a + \log\frac{[salt]}{[acid]} = pK_a + \log\frac{0.1}{0.01} = pK_a + 1$$

$$5 = pK_a + 1 \Rightarrow pK_a = 4 \Rightarrow K_a = 10^{-4} = 10 \times 10^{-5}$$

$$x = 10$$.

25.0 mL of 0.050 M Ba(NO$$_3$$)$$_2$$ is mixed with 25.0 mL of 0.020 M NaF. $$K_{sp}$$ of BaF$$_2 = 0.5 \times 10^{-6}$$.

Total volume = 25.0 + 25.0 = 50.0 mL.

$$[\text{Ba}^{2+}] = \frac{0.050 \times 25.0}{50.0} = 0.025 \text{ M}$$

$$[\text{F}^-] = \frac{0.020 \times 25.0}{50.0} = 0.010 \text{ M}$$

$$Q = [\text{Ba}^{2+}][\text{F}^-]^2 = 0.025 \times (0.010)^2$$

$$Q = 0.025 \times 10^{-4} = 2.5 \times 10^{-6}$$

$$\frac{Q}{K_{sp}} = \frac{2.5 \times 10^{-6}}{0.5 \times 10^{-6}} = 5$$

The ratio of $$[\text{Ba}^{2+}][\text{F}^-]^2$$ to $$K_{sp}$$ is $$\mathbf{5}$$.

We are given the reaction $$A(g) \rightleftharpoons 2B(g) + C(g)$$ with initial pressure 450 mmHg and total pressure at time $$t$$ equal to 720 mmHg. We need to find the fraction of A decomposed.

Let $$x$$ be the fraction of A that has decomposed. Initially, only A is present at 450 mmHg.

The stoichiometry tells us: for every mole of A that decomposes, 2 moles of B and 1 mole of C are formed.

The fraction of A decomposed is $$\frac{3}{10}$$. Expressed as a ratio, the numerator is 3.

For the reaction: $$SO_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons SO_3(g)$$

The change in moles of gas: $$\Delta n = 1 - 1 - \frac{1}{2} = -\frac{1}{2}$$

The relation between $$K_p$$ and $$K_c$$:

$$ K_p = K_c(RT)^{\Delta n} $$

$$ K_c = \frac{K_p}{(RT)^{\Delta n}} = K_p \times (RT)^{1/2} $$

At $$T = 27°C = 300$$ K:

$$ RT = 0.082 \times 300 = 24.6 $$

$$ K_c = 2 \times 10^{12} \times \sqrt{24.6} = 2 \times 10^{12} \times 4.96 \approx 9.92 \times 10^{12} \approx 1 \times 10^{13} $$

Therefore, $$K_c \approx 1 \times 10^{13}$$, and the value of $$x = 1$$.

We are given two equilibrium reactions with the same degree of dissociation $$\alpha$$ and same initial concentration:

(i) $$X(g) \rightleftharpoons Y(g) + Z(g)$$, $$K_{p1} = 3$$

(ii) $$A(g) \rightleftharpoons 2B(g)$$, $$K_{p2} = 1$$

For reaction (i), starting with 1 mole of X at pressure $$P_1$$, at equilibrium: moles of X = $$1 - \alpha$$, moles of Y = $$\alpha$$, moles of Z = $$\alpha$$, total moles = $$1 + \alpha$$. The mole fractions give:

$$K_{p1} = \frac{\left(\frac{\alpha}{1+\alpha}\right)^2}{\frac{1-\alpha}{1+\alpha}} \cdot P_1 = \frac{\alpha^2 P_1}{1-\alpha^2} = 3 \quad \cdots (1)$$

For reaction (ii), starting with 1 mole of A at pressure $$P_2$$, at equilibrium: moles of A = $$1 - \alpha$$, moles of B = $$2\alpha$$, total moles = $$1 + \alpha$$. Similarly:

$$K_{p2} = \frac{\left(\frac{2\alpha}{1+\alpha}\right)^2}{\frac{1-\alpha}{1+\alpha}} \cdot P_2 = \frac{4\alpha^2 P_2}{1-\alpha^2} = 1 \quad \cdots (2)$$

Now, dividing equation (1) by equation (2), the $$\alpha^2$$ and $$(1-\alpha^2)$$ terms cancel completely because $$\alpha$$ is the same for both reactions:

$$\frac{K_{p1}}{K_{p2}} = \frac{P_1}{4P_2}$$

Substituting the given values:

$$\frac{3}{1} = \frac{P_1}{4P_2}$$

$$P_1 = 12 P_2$$

Hence, $$\frac{P_1}{P_2} = 12$$, so the answer is $$x = 12$$.

Calcium lactate, $$\text{Ca(C}_3\text{H}_5\text{O}_3)_2$$, is a salt formed from a weak acid (lactic acid) and a strong base ($$\text{Ca(OH)}_2$$).

When dissolved in water, it dissociates completely:

$$\text{Ca(Lactate)}_2 \rightarrow \text{Ca}^{2+} + 2\,\text{Lactate}^-$$

Determine the Concentration of the Lactate Ion

Since each mole of calcium lactate produces $$2$$ moles of lactate ions ($$\text{Lactate}^-$$), the total concentration of the basic anion ($$C$$) is:

$$C = 2 \times 0.005\text{ M} = 0.01\text{ M} = 10^{-2}\text{ M}$$

Calculate the pH

The pH of a salt of a weak acid and a strong base undergoing anionic hydrolysis is calculated using the formula:

$$\text{pH} = 7 + \frac{1}{2}(\text{p}K_a + \log C)$$

Given:

• $$\text{p}K_a = 5$$
• $$\log C = \log(10^{-2}) = -2$$

Substitute these values into the equation:

$$\text{pH} = 7 + \frac{1}{2}(5 - 2)$$
$$\text{pH} = 7 + \frac{1}{2}(3)$$
$$\text{pH} = 7 + 1.5 = 8.5$$

$$\text{pH} = \text{Value} \times 10^{-1}$$

Setting $$8.5 = X \times 10^{-1}$$, we get:

$$X = 85$$

Given the equilibrium reaction:

$$PCl_3 + Cl_2 \rightleftharpoons PCl_5$$

Initial equilibrium concentrations: $$[PCl_3] = 0.2$$ mol/L, $$[Cl_2] = 0.1$$ mol/L, $$[PCl_5] = 0.40$$ mol/L, and $$K_c = 20$$.

After adding 0.2 mol of $$Cl_2$$ (assuming 1 L volume), the new initial concentrations before re-equilibrium are:

$$[PCl_3] = 0.2$$ mol/L, $$[Cl_2] = 0.1 + 0.2 = 0.3$$ mol/L, $$[PCl_5] = 0.40$$ mol/L

Let $$x$$ mol/L shift to the right to restore equilibrium:

$$[PCl_3] = 0.2 - x$$, $$[Cl_2] = 0.3 - x$$, $$[PCl_5] = 0.40 + x$$

Applying the equilibrium expression:

$$K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]} = \frac{0.40 + x}{(0.2 - x)(0.3 - x)} = 20$$

Expanding:

$$0.40 + x = 20(0.06 - 0.5x + x^2) = 1.2 - 10x + 20x^2$$

$$20x^2 - 11x + 0.8 = 0$$

Using the quadratic formula:

$$x = \frac{11 \pm \sqrt{121 - 64}}{40} = \frac{11 \pm \sqrt{57}}{40}$$

Since $$x < 0.2$$, we take the smaller root:

$$x = \frac{11 - \sqrt{57}}{40} = \frac{11 - 7.55}{40} \approx 0.0863$$

The equilibrium concentration of $$PCl_5$$:

$$[PCl_5] = 0.40 + 0.0863 = 0.4863 \text{ mol/L} \approx 49 \times 10^{-2} \text{ mol/L}$$

Therefore, the answer is $$\mathbf{49}$$.

The decomposition reaction is:

$$H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2}O_2(g)$$

with $$K_p = 2 \times 10^{-3}$$ at 2300 K and total pressure = 1 bar.

Let $$\alpha$$ be the fraction of water that decomposes. Starting with 1 mole of H$$_2$$O:

At equilibrium: $$(1-\alpha)$$ mol H$$_2$$O, $$\alpha$$ mol H$$_2$$, $$\frac{\alpha}{2}$$ mol O$$_2$$.

Total moles = $$1 - \alpha + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2}$$.

Partial pressures (with $$P_{total} = 1$$ bar):

$$P_{H_2} = \frac{\alpha}{1 + \alpha/2}, \quad P_{O_2} = \frac{\alpha/2}{1 + \alpha/2}, \quad P_{H_2O} = \frac{1-\alpha}{1 + \alpha/2}$$

The equilibrium expression is:

$$K_p = \frac{P_{H_2} \cdot P_{O_2}^{1/2}}{P_{H_2O}} = \frac{\alpha \cdot (\alpha/2)^{1/2}}{(1-\alpha)(1+\alpha/2)^{1/2}}$$

Since $$K_p = 2 \times 10^{-3}$$ is very small, $$\alpha \ll 1$$. Using the approximation $$1 - \alpha \approx 1$$ and $$1 + \alpha/2 \approx 1$$:

$$K_p \approx \alpha \cdot \left(\frac{\alpha}{2}\right)^{1/2} = \frac{\alpha^{3/2}}{\sqrt{2}}$$

$$\alpha^{3/2} = K_p \sqrt{2} = 2 \times 10^{-3} \times 1.414 = 2.828 \times 10^{-3}$$

$$\alpha = (2.828 \times 10^{-3})^{2/3}$$

Computing: $$(2.828 \times 10^{-3})^{2/3} = (2.828)^{2/3} \times 10^{-2}$$. Since $$(2.828)^{2/3} \approx 2.0$$:

$$\alpha \approx 0.02 = 2\%$$

The percentage of water decomposing is $$\boxed{2}$$ (nearest integer).

$$\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$$

Initial: 4.5 mol, 4.5 mol, 0 mol in 10 L vessel.

At equilibrium: 3 mol HI formed → 1.5 mol each of H₂ and I₂ consumed.

$$[\text{H}_2] = \frac{3}{10}$$, $$[\text{I}_2] = \frac{3}{10}$$, $$[\text{HI}] = \frac{3}{10}$$

$$K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(0.3)^2}{(0.3)(0.3)} = 1$$

We need to find the millimoles of Ca(OH)$$_2$$ required to prepare 100 mL of an aqueous solution with pH = 12.

Find [OH$$^-$$] from pH.

$$\text{pH} = 12 \Rightarrow \text{pOH} = 14 - 12 = 2$$

$$[\text{OH}^-] = 10^{-2} = 0.01 \text{ M}$$

Relate [OH$$^-$$] to [Ca(OH)$$_2$$].

Ca(OH)$$_2$$ dissociates completely:

$$\text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\text{OH}^-$$

So $$[\text{OH}^-] = 2 \times [\text{Ca(OH)}_2]$$:

$$[\text{Ca(OH)}_2] = \frac{0.01}{2} = 0.005 \text{ M}$$

Calculate millimoles in 100 mL.

$$\text{Moles} = 0.005 \times 0.1 = 5 \times 10^{-4} \text{ mol} = 0.5 \text{ mmol}$$

Given that the answer is $$x \times 10^{-1}$$ millimoles:

$$0.5 = x \times 10^{-1} \Rightarrow x = 5$$

The value of $$x$$ is $$\boxed{5}$$.

Let us verify each statement about equilibria in physical processes:

(A) Equilibrium is possible only in a closed system at a given temperature. Correct - a closed system is needed to prevent exchange of matter.

(B) Both the opposing processes occur at the same rate. Correct - this is the definition of dynamic equilibrium.

(C) When equilibrium is attained, the value of all its parameters became equal. Incorrect - the parameters (like concentration, pressure) become constant, not necessarily equal.

(D) For dissolution of solids in liquids, the solubility is constant at a given temperature. Correct - at equilibrium, the solubility (amount dissolved) is fixed at a given temperature.

The number of correct statements is A, B, D = 3.

Given: Solid Pb(NO₃)₂ dissolved in 1 litre of water boils at 100.15 °C. Then 0.2 mol NaCl is added, and the solution freezes at −0.8 °C.

$$K_b = 0.5$$ K kg mol$$^{-1}$$, $$K_f = 1.8$$ K kg mol$$^{-1}$$. Molality ≈ Molarity.

Pb(NO₃)₂ → Pb²⁺ + 2NO₃⁻ (van't Hoff factor $$i = 3$$)

$$m = 0.1$$ mol/kg (≈ 0.1 M)

Initial moles: Pb²⁺ = 0.1, NO₃⁻ = 0.2

NaCl → Na⁺ + Cl⁻ (complete dissociation)

Initial species: Pb²⁺ = 0.1, NO₃⁻ = 0.2, Na⁺ = 0.2, Cl⁻ = 0.2

Reaction: Pb²⁺ + 2Cl⁻ → PbCl₂↓

Let $$x$$ mol PbCl₂ precipitate. After equilibrium:

Pb²⁺ = $$(0.1 - x)$$, Cl⁻ = $$(0.2 - 2x)$$, Na⁺ = 0.2, NO₃⁻ = 0.2

Total particle molality = $$0.2 + 0.2 + (0.1-x) + (0.2-2x) = 0.7 - 3x$$

To the nearest integer: the value is $$\mathbf{13} \times 10^{-6}$$.

Let us examine each statement about adsorption:

Statement A: "Water vapours are adsorbed by anhydrous calcium chloride." This is incorrect. Anhydrous $$CaCl_2$$ actually absorbs water vapour (it takes water into its bulk to form hydrated salts like $$CaCl_2 \cdot 6H_2O$$), rather than adsorbing it on the surface. Adsorption is a surface phenomenon, while what $$CaCl_2$$ does is absorption — a bulk phenomenon.

Statement B: "There is a decrease in surface energy during adsorption." This is correct. Adsorption occurs because surface molecules have unbalanced forces. When adsorbate molecules attach to the surface, these residual forces are satisfied, lowering the surface energy.

Statement C: "As the adsorption proceeds, $$\Delta H$$ becomes more and more negative." This is correct. Adsorption is exothermic, and as the process continues, the cumulative enthalpy change becomes increasingly negative.

Statement D: "Adsorption is accompanied by decrease in entropy of the system." This is correct. Gas molecules lose their freedom of movement when they become bound to a surface, resulting in decreased randomness and hence decreased entropy.

Only Statement A is incorrect. The number of incorrect statements is $$\boxed{1}$$.

For the equilibrium reaction: $$A(g) \rightleftharpoons B(g) + \frac{1}{2}C(g)$$ we start with 1 mole of A. At equilibrium, the moles of A, B, and C are $$1-\alpha$$, $$\alpha$$, and $$\frac{\alpha}{2}$$ respectively, giving a total of $$n_{\text{total}} = (1 - \alpha) + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2}$$.

The mole fractions yield the partial pressures (total pressure $$p$$) as $$p_A = \frac{1 - \alpha}{1 + \frac{\alpha}{2}} \cdot p$$, $$p_B = \frac{\alpha}{1 + \frac{\alpha}{2}} \cdot p$$, and $$p_C = \frac{\frac{\alpha}{2}}{1 + \frac{\alpha}{2}} \cdot p = \frac{\alpha}{2\left(1 + \frac{\alpha}{2}\right)} \cdot p$$.

$$K = \frac{p_B \cdot p_C^{1/2}}{p_A}$$

$$K = \frac{\frac{\alpha p}{1 + \frac{\alpha}{2}} \cdot \left(\frac{\alpha p}{2(1 + \frac{\alpha}{2})}\right)^{1/2}}{\frac{(1 - \alpha)p}{1 + \frac{\alpha}{2}}}$$

$$K = \frac{\alpha p}{\left(1 + \frac{\alpha}{2}\right)} \cdot \frac{\alpha^{1/2} p^{1/2}}{2^{1/2}\left(1 + \frac{\alpha}{2}\right)^{1/2}} \cdot \frac{\left(1 + \frac{\alpha}{2}\right)}{(1 - \alpha)p} = \frac{\alpha^{3/2} \cdot p^{1/2}}{2^{1/2} \cdot \left(1 + \frac{\alpha}{2}\right)^{1/2} \cdot (1 - \alpha)}$$

Noting that $$2^{1/2} \cdot \left(1 + \frac{\alpha}{2}\right)^{1/2} = \left(2\left(1 + \frac{\alpha}{2}\right)\right)^{1/2} = (2 + \alpha)^{1/2}$$, we have

$$K = \frac{\alpha^{3/2} \cdot p^{1/2}}{(2 + \alpha)^{1/2} \cdot (1 - \alpha)}$$

The answer is Option A: $$K = \frac{\alpha^{3/2} p^{1/2}}{(2+\alpha)^{1/2}(1-\alpha)}$$.

The three equilibrium reactions are:

(a) $$H_2C_2O_4 \rightleftharpoons H^+ + HC_2O_4^-$$, with equilibrium constant $$K_{a1}$$

(b) $$HC_2O_4^- \rightleftharpoons H^+ + C_2O_4^{2-}$$, with equilibrium constant $$K_{a2}$$

(c) $$H_2C_2O_4 \rightleftharpoons 2H^+ + C_2O_4^{2-}$$, with equilibrium constant $$K_{a3}$$

Reaction (c) is obtained by adding reactions (a) and (b):

$$H_2C_2O_4 \rightleftharpoons H^+ + HC_2O_4^-$$

$$HC_2O_4^- \rightleftharpoons H^+ + C_2O_4^{2-}$$

Adding: $$H_2C_2O_4 \rightleftharpoons 2H^+ + C_2O_4^{2-}$$

When two equilibrium reactions are added, the equilibrium constant of the resultant reaction is the product of the individual equilibrium constants:

$$K_{a3} = K_{a1} \times K_{a2}$$

Therefore, the correct answer is Option D.

Given: 20 mL of 0.1 M $$NH_4OH$$ and 40 mL of 0.05 M $$HCl$$ with $$K_b(NH_4OH) = 1 \times 10^{-5}$$.

The moles of $$NH_4OH$$ equal 20 × 0.1 = 2 mmol and the moles of $$HCl$$ equal 40 × 0.05 = 2 mmol.

They react according to $$NH_4OH + HCl \rightarrow NH_4Cl + H_2O$$ and, since the moles are equal, complete neutralization occurs, forming 2 mmol of $$NH_4Cl$$ in 60 mL of solution.

The $$NH_4^+$$ ion undergoes hydrolysis according to $$NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$$.

The concentration of $$NH_4^+$$ is $$C = \dfrac{2}{60} = \dfrac{1}{30} \text{ M}$$.

The acid dissociation constant of $$NH_4^+$$ is $$K_a = \dfrac{K_w}{K_b} = \dfrac{10^{-14}}{10^{-5}} = 10^{-9}$$.

The hydrogen ion concentration is $$[H^+] = \sqrt{K_a \times C} = \sqrt{10^{-9} \times \dfrac{1}{30}} = \sqrt{\dfrac{10^{-9}}{30}}$$.

Taking the negative logarithm gives $$pH = -\log[H^+] = -\dfrac{1}{2}\log(3.33 \times 10^{-11}) = \dfrac{1}{2}[11 - \log 3.33]$$.

Since $$\log 3.33 = \log\dfrac{10}{3} = \log 10 - \log 3 = 1 - 0.48 = 0.52$$, it follows that $$pH = \dfrac{10.48}{2} = 5.24$$.

Therefore, the correct answer is Option C: $$5.2$$.

We have 200 mL of 0.01 M HCl mixed with 400 mL of 0.01 M $$H_2SO_4$$. We need to find the pH of the mixture.

HCl is a strong monoprotic acid, so the moles of $$H^+$$ from HCl = $$0.01 \times 0.2 = 0.002$$ mol = 2 mmol.

$$H_2SO_4$$ is a strong diprotic acid, contributing 2 moles of $$H^+$$ per mole. So the moles of $$H^+$$ from $$H_2SO_4$$ = $$2 \times 0.01 \times 0.4 = 0.008$$ mol = 8 mmol.

Total moles of $$H^+$$ = $$0.002 + 0.008 = 0.010$$ mol. The total volume is $$200 + 400 = 600$$ mL = 0.6 L.

The concentration of $$H^+$$ in the mixture is: $$[H^+] = \frac{0.010}{0.6} = \frac{1}{60}$$ M.

The pH is: $$\text{pH} = -\log\left(\frac{1}{60}\right) = \log 60 = \log(6 \times 10) = \log 6 + 1 = 0.778 + 1 = 1.778 \approx 1.78$$

Hence, the correct answer is Option B.

We have a buffer solution of pH = 8.26 prepared from ammonia of concentration 0.2 M, given that $$pK_b(NH_3) = 4.74$$ and the molar mass of $$NH_4Cl$$ is $$53.5\text{ g/mol}$$.

Since $$pOH = 14 - pH$$, we find $$pOH = 14 - 8.26 = 5.74$$.

Applying the Henderson‐Hasselbalch equation for a basic buffer, $$pOH = pK_b + \log\frac{[\text{salt}]}{[\text{base}]}$$, gives $$5.74 = 4.74 + \log\frac{[NH_4Cl]}{0.2}$$.

Solving for the ratio yields $$\log\frac{[NH_4Cl]}{0.2} = 1$$, so $$\frac{[NH_4Cl]}{0.2} = 10$$ and therefore $$[NH_4Cl] = 2\text{ M}$$.

Since the required volume is 1 L, the mass of $$NH_4Cl$$ needed is $$\text{Mass} = [NH_4Cl] \times V \times M = 2 \times 1 \times 53.5 = 107\text{ g}$$.

Therefore, the amount of ammonium chloride required is $$107\text{ g}$$. The correct answer is Option C.

Methyl orange is an azo dye indicator used in acid-base titrations. It exists in two structural forms depending on the pH:

In acidic medium (pH < 3.1): Methyl orange exists in the quinonoid form, which is red/pink in colour.

In basic medium (pH > 4.4): Methyl orange exists in the benzenoid (azo) form, which is yellow in colour.

In a base vs. acid titration (i.e., titrating a base with an acid), at the end point, the solution becomes slightly acidic. Methyl orange changes from yellow to orange/pink at the end point.

At the end point, methyl orange is present in the quinonoid form.

Therefore, the correct answer is Option A.

$$K_{sp}$$ of $$Bi_2S_3 = 1.08 \times 10^{-73}$$

$$Bi_2S_3(s) \rightleftharpoons 2Bi^{3+}(aq) + 3S^{2-}(aq)$$

Let the solubility of $$Bi_2S_3$$ be $$s$$ mol/L.

Then: $$[Bi^{3+}] = 2s$$ and $$[S^{2-}] = 3s$$

$$K_{sp} = [Bi^{3+}]^2 [S^{2-}]^3 = (2s)^2(3s)^3$$

$$= 4s^2 \times 27s^3 = 108s^5$$

$$108s^5 = 1.08 \times 10^{-73}$$

$$s^5 = \frac{1.08 \times 10^{-73}}{108} = 1.0 \times 10^{-75}$$

$$s = (10^{-75})^{1/5} = 10^{-15}$$

$$s = 1.0 \times 10^{-15} \text{ mol/L}$$

Hence, the correct answer is Option A.

In a pH-metric titration of a weak base $$\mathrm{(NH_4OH)}$$ against a strong acid $$\mathrm{(HCl)}$$, the titration curve begins at a basic pH.

Since $$\mathrm{NH_4OH}$$ is a weak base, the initial pH is moderately basic and not extremely high.

$$\mathrm{Initial\ pH \approx 11}$$

As $$\mathrm{HCl}$$ is gradually added, the weak base gets neutralised:

$$\mathrm{NH_4OH + HCl \longrightarrow NH_4Cl + H_2O}$$

A buffer solution containing:

$$\mathrm{NH_4OH \ and\ NH_4^+}$$

is formed.

Because buffer solutions resist sudden pH changes, the curve shows a gradual downward slope in this region.

At the equivalence point:

$$\mathrm{NH_4Cl}$$

is present in solution.

Since this salt is formed from a strong acid and a weak base, it undergoes cationic hydrolysis:

$$\mathrm{NH_4^+ + H_2O \rightleftharpoons NH_4OH + H^+}$$

This produces excess $$\mathrm{H^+}$$ ions, making the solution acidic.

Therefore, the equivalence point lies below $$\mathrm{pH = 7}$$.

$$\mathrm{Equivalence\ pH \approx 5\ to\ 5.5}$$

Beyond the equivalence point, excess $$\mathrm{HCl}$$ dominates the solution.

Hence, the pH decreases sharply and finally becomes strongly acidic.

$$\mathrm{Final\ pH \approx 1\ to\ 2}$$

This corresponds to option A. Thus, A is the right option.

We are told that compound X is a weak acid that exhibits a colour change at a pH close to the equivalence point during the neutralisation of NaOH (a strong base) with $$CH_3COOH$$ (a weak acid). It also exists in ionised form in a basic medium.

The equivalence point of a weak acid-strong base titration lies in the basic pH range (typically pH 8-10), because the salt formed (sodium acetate) undergoes hydrolysis to give a basic solution.

Among the given indicators, phenolphthalein has a pH range of approximately 8.2 to 10.0, which matches the equivalence point of this titration. Phenolphthalein is itself a weak acid — in acidic solution it is colourless (unionised form), while in basic solution it ionises and turns pink. This is consistent with the statement that compound X exists in ionised form in basic medium.

Methyl orange (pH range 3.1-4.4) and methyl red (pH range 4.4-6.2) change colour in acidic pH ranges, which do not match the equivalence point of a weak acid-strong base titration. Eriochrome Black T is used in complexometric titrations, not acid-base titrations.

Hence, the correct answer is Option C.

Required pH of the buffer is 4 and the $$K_a$$ of propanoic acid ($$CH_3CH_2COOH$$) is $$1.3 \times 10^{-5}$$. Using the Henderson-Hasselbalch equation:

$$\text{pH} = \text{p}K_a + \log\frac{[\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]}$$

We calculate $$\text{p}K_a$$ as follows: $$\text{p}K_a = -\log(1.3 \times 10^{-5}) = -\log(1.3) - \log(10^{-5}) = -0.114 + 5 = 4.886$$

Substituting into the Henderson-Hasselbalch equation gives $$4 = 4.886 + \log\frac{[\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]}$$, so $$\log\frac{[\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]} = 4 - 4.886 = -0.886$$.

Thus, $$\frac{[\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]} = 10^{-0.886} = \frac{1}{10^{0.886}}$$, and since $$10^{0.886} \approx 7.7$$, we obtain $$\frac{[\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]} \approx \frac{1}{7.7} \approx 0.13$$.

The correct answer is Option B (0.13).

Analyzing the Assertion:

"The amphoteric behaviour of water is explained by Lewis acid base theory"

The amphoteric nature of water means it can act as both an acid and a base. This behaviour is best explained by the Bronsted-Lowry theory, not the Lewis acid-base theory. According to Bronsted-Lowry theory, water can donate a proton (acting as acid) or accept a proton (acting as base). Lewis theory defines acids as electron pair acceptors and bases as electron pair donors, which is not the primary framework used to explain the amphoteric behaviour of water.

Therefore, the Assertion is false.

Analyzing the Reason:

"Water acts as acid with $$NH_3$$ and base with $$H_2S$$"

With $$NH_3$$: $$H_2O + NH_3 \rightleftharpoons NH_4^+ + OH^-$$. Here, water donates a proton to $$NH_3$$, acting as an acid. This is correct.

With $$H_2S$$: $$H_2O + H_2S \rightleftharpoons H_3O^+ + HS^-$$. Here, water accepts a proton from $$H_2S$$, acting as a base. This is correct.

Therefore, the Reason is true.

Hence, the correct answer is Option D: Assertion is false but Reason is true.

Phenolphthalein is indeed a pH-dependent indicator, remaining colourless in acidic solutions (pH < 8.2) and turning pink in basic solutions (pH 8.2 to 12). Therefore, this statement is true.

Regarding the reason, it is correct that phenolphthalein is a weak acid, but the claim that it does not dissociate in basic medium is false. In basic medium, phenolphthalein actually does dissociate. The undissociated form (HIn) is colourless, and when it dissociates in basic medium to form $$In^-$$ (the conjugate base), it turns pink. The basic medium promotes the dissociation of this weak acid, shifting the equilibrium toward the ionized form.

$$HIn \rightleftharpoons H^+ + In^-$$

(colourless) $$ $$ (pink)

Consequently, the reason that it "doesn't dissociate in basic medium" is incorrect.

Since Assertion A is true but Reason R is false, the correct answer is Option C.

AgCl is a sparingly soluble salt that dissociates as:

$$\text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$$

$$K_{sp} = [\text{Ag}^+][\text{Cl}^-]$$

Analysis of each option:

Option A: 0.01 M HCl

HCl provides Cl$$^-$$ ions (common ion effect). This will suppress the dissolution of AgCl, reducing its solubility.

Option B: 0.01 M KCl

KCl also provides Cl$$^-$$ ions (common ion effect). This will similarly reduce the solubility of AgCl.

Option C: Deionised water

There are no common ions present. The solubility of AgCl will be determined solely by its $$K_{sp}$$ without any suppression.

Option D: 0.01 M AgNO$$_3$$

AgNO$$_3$$ provides Ag$$^+$$ ions (common ion effect). This will also suppress the dissolution of AgCl.

Conclusion:

Since options A, B, and D all introduce a common ion (either Cl$$^-$$ or Ag$$^+$$), the solubility of AgCl is reduced in all of them. Only in deionised water is there no common ion effect, so AgCl has maximum solubility there.

Hence, the correct answer is Option C (Deionised water).

We need to find K_p for the decomposition of PCl₅ at equilibrium.

The equilibrium reaction is $$\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$$.

Using the ideal gas law with a total pressure of 6.0 atm, a volume of 100 L, a temperature of 610 K, and R = 0.082 L atm K⁻¹ mol⁻¹, the total number of moles at equilibrium is calculated as $$n_{\text{total}} = \frac{PV}{RT} = \frac{6.0 \times 100}{0.082 \times 610} = \frac{600}{50.02} = 11.99 \approx 12 \text{ moles}$$.

Argon is inert and remains unchanged at 4.0 moles, so the moles of reacting gases at equilibrium are 12 − 4 = 8 moles. Let x moles of PCl₅ dissociate: PCl₅: $$5 - x$$, PCl₃: $$x$$, Cl₂: $$x$$. The total reacting moles are $$(5 - x) + x + x = 5 + x = 8$$, which gives $$x = 3$$.

The total moles remain 12, so the partial pressures are calculated as follows: $$p_{\text{PCl}_5} = \frac{5-3}{12} \times 6 = \frac{2}{12} \times 6 = 1$$ atm; $$p_{\text{PCl}_3} = \frac{3}{12} \times 6 = 1.5$$ atm; and $$p_{\text{Cl}_2} = \frac{3}{12} \times 6 = 1.5$$ atm.

Substituting these values into the expression for K_p yields $$K_p = \frac{p_{\text{PCl}_3} \times p_{\text{Cl}_2}}{p_{\text{PCl}_5}} = \frac{1.5 \times 1.5}{1} = 2.25$$.

The answer is Option A: 2.25.

First note the equilibria already present in the solution.

1. Complete dissociation of sodium sulphate:
$$Na_2SO_4 \rightarrow 2\,Na^+ + SO_4^{2-}$$
Initial $$[SO_4^{2-}] = 1.8 \times 10^{-2}\,{\rm M}$$.

2. First dissociation of sulphuric acid is virtually complete:
$$H_2SO_4 \rightarrow H^+ + HSO_4^-$$
Hence after this step
$$[H^+] = 1\;{\rm M},\qquad [HSO_4^-] = 1\;{\rm M}$$.

3. The second dissociation of $$H_2SO_4$$ is governed by $$K_{a2}=1.2\times10^{-2}:$$
$$HSO_4^- \rightleftharpoons H^+ + SO_4^{2-}$$.

Let a change of $$y$$ M occur in the above equilibrium, counted positive when the reaction proceeds to the right and negative when it goes to the left.

At equilibrium:
$$[H^+] = 1 + y$$
$$[HSO_4^-] = 1 - y$$
$$[SO_4^{2-}] = 1.8\times10^{-2} + y$$.

Apply the expression for $$K_{a2}:$$
$$K_{a2} = \frac{[H^+][SO_4^{2-}]}{[HSO_4^-]} = 1.2\times10^{-2}$$.

Substituting the concentrations gives
$$\frac{(1+y)\,(1.8\times10^{-2}+y)}{1-y}=1.2\times10^{-2}\;-(1)$$.

Because $$y$$ is expected to be small compared with 1 we solve (1) exactly (quadratic in $$y$$) to avoid error:
$$(1+y)(0.018+y)=0.012(1-y)$$
$$0.018 + 1.018y + y^2 = 0.012 - 0.012y$$
$$y^2 + 1.03y + 0.006 = 0$$.

The positive root is not physically allowed; the negative root is
$$y = -5.85\times10^{-3}\;{\rm M}$$ (still within the limit $$y \gt -0.018$$, so valid).

Therefore the equilibrium sulphate ion concentration is
$$[SO_4^{2-}] = 0.018 + y = 0.018 - 0.00585 \approx 0.012\,{\rm M}$$.

Now let the molar solubility of $$PbSO_4$$ in this medium be $$s$$ M:
$$PbSO_4(s) \rightleftharpoons Pb^{2+} + SO_4^{2-}$$
At equilibrium:
$$[Pb^{2+}] = s,\qquad [SO_4^{2-}] = 0.012 + s \approx 0.012$$ (because $$s$$ will be very small).

Using $$K_{sp}(PbSO_4)=1.6\times10^{-8}$$:
$$K_{sp} = [Pb^{2+}][SO_4^{2-}] \approx s \times 0.012$$
$$\therefore s = \frac{1.6\times10^{-8}}{0.012} = 1.33\times10^{-6}\,{\rm M}$$.

Written in scientific notation, the solubility is $$1.33 \times 10^{-6}$$ M, i.e. $$X\times10^{-Y}$$ with $$Y = 6$$.

Hence, the required value of $$Y$$ is 6.

Initially, each solute furnishes $$0.01\text{ mol}$$ in a total volume of $$100\text{ mL}=0.1\text{ L}$$, so their starting concentrations are $$0.1\text{ M}$$ each:

$$H_2CO_3\;(0.1\text{ M}),\;HCO_3^- \;(0.1\text{ M}),\;CO_3^{2-}\;(0.1\text{ M}),\;OH^-\;(0.1\text{ M})$$

Step 1: Immediate neutralisation by the strong base
The strong base $$OH^-$$ will first de-protonate the strongest acid present, $$H_2CO_3$$:

$$H_2CO_3 + OH^- \;\longrightarrow\; HCO_3^- + H_2O$$

Mole balance (in 0.1 L): both $$H_2CO_3$$ and $$OH^-$$ start with $$0.01\text{ mol}$$, so they completely consume each other.

After this reaction:

$$H_2CO_3 : 0$$    $$OH^- : 0$$    additional $$HCO_3^- : +0.01\text{ mol}$$

Step 2: Concentrations after neutralisation
Total moles now present:

$$HCO_3^- : 0.01 + 0.01 = 0.02\text{ mol}$$
$$CO_3^{2-} : 0.01\text{ mol}$$

Convert to concentrations (divide by $$0.1\text{ L}$$):

$$[HCO_3^-] = \frac{0.02}{0.1} = 0.20\text{ M}$$
$$[CO_3^{2-}] = \frac{0.01}{0.1} = 0.10\text{ M}$$

Step 3: Identify the buffer pair
The remaining species constitute the conjugate acid-base pair of the second dissociation of carbonic acid:

$$HCO_3^- \;\rightleftharpoons\; H^+ + CO_3^{2-} \quad\quad (pK_{a2} = 10.32)$$

Step 4: Apply the Henderson-Hasselbalch equation

$$\text{pH} = pK_{a2} + \log\!\left(\frac{[CO_3^{2-}]}{[HCO_3^-]}\right)$$

Substitute the concentrations:

$$\text{pH} = 10.32 + \log\!\left(\frac{0.10}{0.20}\right) = 10.32 + \log(0.50)$$

Using $$\log 2 = 0.30 \;,$$ so $$\log(0.50) = -\log 2 = -0.30$$, we get

$$\text{pH} = 10.32 - 0.30 = 10.02$$

Hence, the pH of the resulting solution is 10.02.

We need to find the standard free energy change for the reaction: $$2O_3(g) \rightleftharpoons 3O_2(g)$$ at 300 K where ozone is 50% dissociated.

Let us start with 2 moles of $$O_3$$ to match the stoichiometric coefficient. At 50% dissociation, the degree of dissociation $$\alpha = 0.50$$. Starting with 2 moles of $$O_3$$, 50% dissociation means 1 mole of $$O_3$$ reacts (since 50% of 2 = 1 mole reacts according to the forward reaction with coefficient 2, i.e., the reaction proceeds by $$x = 0.5$$ times). Using extent of reaction $$\xi$$: if $$\alpha = 0.5$$, then from 2 mol of $$O_3$$, 50% dissociates = 1 mol $$O_3$$ reacts. The initial amounts are $$O_3 = 2$$ mol and $$O_2 = 0$$ mol; the change is $$O_3: -1$$ mol and $$O_2: +1.5$$ mol (by stoichiometry: for every 2 mol $$O_3$$ that react, 3 mol $$O_2$$ form, so 1 mol of $$O_3$$ reacting gives 1.5 mol of $$O_2$$); at equilibrium $$O_3 = 1$$ mol and $$O_2 = 1.5$$ mol.

From the equilibrium amounts the total moles are $$n_{total} = 1 + 1.5 = 2.5$$ mol, giving mole fractions $$x_{O_3} = \frac{1}{2.5} = 0.4$$ and $$x_{O_2} = \frac{1.5}{2.5} = 0.6$$.

At total pressure P = 1 atm the partial pressures are $$p_{O_3} = 0.4 \times 1 = 0.4$$ atm and $$p_{O_2} = 0.6 \times 1 = 0.6$$ atm.

The equilibrium constant in terms of partial pressures is $$K_p = \frac{(p_{O_2})^3}{(p_{O_3})^2} = \frac{(0.6)^3}{(0.4)^2}$$ and $$K_p = \frac{0.216}{0.16} = 1.35$$.

Using the formula for standard free energy change gives $$\Delta G° = -RT \ln K_p$$, then $$\Delta G° = -(8.3)(300) \ln(1.35)$$, followed by $$\Delta G° = -(8.3)(300)(0.3)$$, and thus $$\Delta G° = -747 \text{ J mol}^{-1}$$.

The magnitude of the standard free energy change is 747 J mol$$^{-1}$$.

We need to find the volume at which all liquid water evaporates while maintaining equilibrium vapour pressure. The mass of water is 0.90 g and its molar mass is 18 g/mol, so $$n = \frac{0.90}{18} = 0.05 \text{ mol}$$. The vapour pressure is 32.0 Torr, which is $$P = 32.0 \text{ Torr} = \frac{32.0}{760} \text{ atm} = 0.04211 \text{ atm}$$.

When all liquid water evaporates, all 0.05 mol exists as vapour at the equilibrium pressure. By the ideal gas law, $$PV = nRT$$, so $$V = \frac{nRT}{P} = \frac{0.05 \times 0.082 \times 300}{32/760}$$, which gives $$= \frac{0.05 \times 0.082 \times 300}{0.04211}$$ and then $$= \frac{1.23}{0.04211} = 29.2 \approx 29 \text{ litres}$$.

The answer is 29 litres.

Given reaction: $$2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g)$$

Initial moles of NOCl = 2.0 mol in a 1 L flask, so initial concentration = 2.0 M

At equilibrium, concentration of NO = 0.4 mol/L

Setting up the ICE table:

$$2NOCl \rightleftharpoons 2NO + Cl_2$$

Initial: 2.0, 0, 0

Change: $$-2x$$, $$+2x$$, $$+x$$

Equilibrium: $$2.0 - 2x$$, $$2x$$, $$x$$

Since $$[NO] = 2x = 0.4$$ mol/L:

$$x = 0.2$$ mol/L

At equilibrium:

$$[NOCl] = 2.0 - 2(0.2) = 1.6$$ mol/L

$$[NO] = 0.4$$ mol/L

$$[Cl_2] = 0.2$$ mol/L

The equilibrium constant expression is:

$$K_c = \frac{[NO]^2[Cl_2]}{[NOCl]^2}$$

Substituting the values:

$$K_c = \frac{(0.4)^2 \times (0.2)}{(1.6)^2}$$

$$K_c = \frac{0.16 \times 0.2}{2.56}$$

$$K_c = \frac{0.032}{2.56}$$

$$K_c = 0.0125 = 125 \times 10^{-4}$$

Hence, the answer is 125.

5 moles of $$PCl_5$$ and 2 moles of $$N_2$$ in a 200 L vessel at 600 K and an equilibrium pressure of 2.46 atm are considered for the dissociation reaction $$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$$. Let $$x$$ moles of $$PCl_5$$ dissociate at equilibrium, giving $$PCl_5 = 5 - x$$ moles, $$PCl_3 = x$$ moles, $$Cl_2 = x$$ moles, while $$N_2$$ remains unchanged at 2 moles (inert gas). The total moles at equilibrium are $$(5 - x) + x + x + 2 = 7 + x$$.

Using the ideal gas law $$P_{total} = \frac{n_{total} \cdot R \cdot T}{V}$$, we write $$2.46 = \frac{(7 + x) \times 0.082 \times 600}{200}$$ which simplifies to $$2.46 = \frac{(7 + x) \times 49.2}{200}$$ and then to $$2.46 = (7 + x) \times 0.246$$, so $$7 + x = \frac{2.46}{0.246} = 10$$ and hence $$x = 3$$.

With a total of 10 moles at equilibrium, the partial pressures using mole fractions and total pressure are $$p_{PCl_5} = \frac{5 - 3}{10} \times 2.46 = \frac{2}{10} \times 2.46 = 0.492 \text{ atm}$$, $$p_{PCl_3} = \frac{3}{10} \times 2.46 = 0.738 \text{ atm}$$, and $$p_{Cl_2} = \frac{3}{10} \times 2.46 = 0.738 \text{ atm}$$.

Finally, $$K_p = \frac{p_{PCl_3} \times p_{Cl_2}}{p_{PCl_5}} = \frac{0.738 \times 0.738}{0.492} = \frac{0.544644}{0.492} = 1.107 \text{ atm}$$ and $$K_p = 1107 \times 10^{-3}$$. The answer is 1107.

We need to find the pH when 25 mL of 0.1 M NaOH is added to 50 mL of 0.1 M $$CH_3COOH$$.

Calculate moles of acid and base

Moles of $$CH_3COOH$$ = 50 × 0.1 = 5 mmol

Moles of NaOH = 25 × 0.1 = 2.5 mmol

After reaction

$$CH_3COOH + NaOH \to CH_3COONa + H_2O$$

NaOH is the limiting reagent. After the reaction:

- $$CH_3COOH$$ remaining = 5 - 2.5 = 2.5 mmol

- $$CH_3COONa$$ formed = 2.5 mmol

This is a buffer solution (half-equivalence point)

Since moles of acid = moles of conjugate base (salt), this is the half-equivalence point.

Using the Henderson-Hasselbalch equation:

$$pH = pK_a + \log\frac{[CH_3COO^-]}{[CH_3COOH]}$$

$$pH = pK_a + \log\frac{2.5}{2.5}$$

$$pH = pK_a + \log(1)$$

$$pH = pK_a + 0$$

$$pH = 4.76$$

Express the answer in the required format

The pH = 4.76 = $$476 \times 10^{-2}$$

The correct answer is 476.

We are given the solubility product of PbS as $$K_{sp} = 8 \times 10^{-28}$$ and need to find its solubility in pure water.

PbS dissociates in water as:

$$PbS \rightleftharpoons Pb^{2+} + S^{2-}$$

Let the solubility be $$s$$ mol L$$^{-1}$$. Then $$[Pb^{2+}] = s$$ and $$[S^{2-}] = s$$.

The solubility product expression is:

$$K_{sp} = [Pb^{2+}][S^{2-}] = s \times s = s^2$$

$$s^2 = 8 \times 10^{-28}$$

$$s = \sqrt{8 \times 10^{-28}} = \sqrt{8} \times 10^{-14}$$

Now, $$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} = 2 \times 1.41 = 2.82$$.

$$s = 2.82 \times 10^{-14} \text{ mol L}^{-1}$$

We need to express this as $$x \times 10^{-16}$$ mol L$$^{-1}$$:

$$s = 2.82 \times 10^{-14} = 282 \times 10^{-16} \text{ mol L}^{-1}$$

Hence, the correct answer is 282.

We need to find the solubility product $$K_{sp}$$ of $$CaF_2$$.

Since the solubility of $$CaF_2$$ at 310 K is $$2.34 \times 10^{-3} \text{ g/100 mL}$$ and the molar mass of $$CaF_2$$ is $$78 \text{ g mol}^{-1}$$, we first convert the solubility to mol/L.

Because the solubility is given as $$2.34 \times 10^{-3} \text{ g per 100 mL}$$, converting to grams per liter involves multiplying by 10, which yields $$2.34 \times 10^{-2} \text{ g/L}$$. Substituting this into the molar solubility formula gives $$s = \frac{2.34 \times 10^{-2}}{78} = 3.0 \times 10^{-4} \text{ mol/L}$$.

Next, we write the dissociation equation: $$CaF_2 \rightleftharpoons Ca^{2+} + 2F^-$$. If $$s$$ is the molar solubility, then $$[Ca^{2+}] = s = 3.0 \times 10^{-4} \text{ M}$$ and $$[F^-] = 2s = 6.0 \times 10^{-4} \text{ M}$$.

To calculate $$K_{sp}$$, we use the expression $$K_{sp} = [Ca^{2+}][F^-]^2 = s \times (2s)^2 = 4s^3$$. Substituting the value of $$s$$ gives $$K_{sp} = 4 \times (3.0 \times 10^{-4})^3$$, which simplifies to $$K_{sp} = 4 \times 2.7 \times 10^{-11}$$, then $$K_{sp} = 10.8 \times 10^{-11}$$, and finally $$K_{sp} = 1.08 \times 10^{-10}$$.

Expressing this in the required form yields $$K_{sp} = 1.08 \times 10^{-10} = 0.0108 \times 10^{-8} \approx 0 \times 10^{-8}$$. Rounding to the nearest integer in units of $$\times 10^{-8} (\text{mol/L})^3$$ gives $$K_{sp} = 0.0108 \times 10^{-8}$$.

The nearest integer is 0, and therefore the correct answer is 0.

We are given the reaction $$2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)$$ at 600 K. Initially, 2 mol of NO and 1 mol of $$O_2$$ are mixed, and the total pressure at equilibrium is 1 atm. At equilibrium, 0.6 mol of $$O_2$$ is present. We need to find the equilibrium constant $$K_p$$.

Let us set up an ICE table. Let $$x$$ be the moles of $$O_2$$ consumed.

Since 1 mol of $$O_2$$ was initially present and 0.6 mol remains at equilibrium, $$x = 1 - 0.6 = 0.4$$ mol of $$O_2$$ has been consumed.

From the stoichiometry, 2 mol of NO are consumed for every 1 mol of $$O_2$$, so moles of NO consumed = $$2 \times 0.4 = 0.8$$ mol. Also, 2 mol of $$NO_2$$ are formed for every 1 mol of $$O_2$$ consumed, so moles of $$NO_2$$ formed = $$2 \times 0.4 = 0.8$$ mol.

At equilibrium:

$$n_{NO} = 2 - 0.8 = 1.2 \text{ mol}$$

$$n_{O_2} = 1 - 0.4 = 0.6 \text{ mol}$$

$$n_{NO_2} = 0 + 0.8 = 0.8 \text{ mol}$$

Total moles at equilibrium: $$n_{total} = 1.2 + 0.6 + 0.8 = 2.6 \text{ mol}$$

Now we calculate the mole fractions and partial pressures (using $$P_{total} = 1$$ atm):

$$P_{NO} = \frac{1.2}{2.6} \times 1 = \frac{1.2}{2.6} \text{ atm}$$

$$P_{O_2} = \frac{0.6}{2.6} \times 1 = \frac{0.6}{2.6} \text{ atm}$$

$$P_{NO_2} = \frac{0.8}{2.6} \times 1 = \frac{0.8}{2.6} \text{ atm}$$

The equilibrium constant expression is:

$$K_p = \frac{(P_{NO_2})^2}{(P_{NO})^2 \cdot P_{O_2}}$$

Substituting:

$$K_p = \frac{\left(\frac{0.8}{2.6}\right)^2}{\left(\frac{1.2}{2.6}\right)^2 \times \frac{0.6}{2.6}}$$

$$= \frac{\frac{0.64}{6.76}}{\frac{1.44}{6.76} \times \frac{0.6}{2.6}}$$

$$= \frac{0.64}{1.44 \times \frac{0.6}{2.6}}$$

$$= \frac{0.64 \times 2.6}{1.44 \times 0.6}$$

$$= \frac{1.664}{0.864}$$

$$\approx 1.926$$

Rounding to the nearest integer, $$K_p \approx 2$$.

Hence, the correct answer is 2.

We are given that the acid dissociation constant $$K_a$$ for butyric acid ($$C_3H_7COOH$$) is $$2 \times 10^{-5}$$, the concentration $$C$$ is $$0.2$$ M, and $$\log 2 = 0.30$$.

Since butyric acid is a weak acid, the hydrogen ion concentration can be determined using the relation $$[H^+] = \sqrt{K_a \times C}$$. Substituting the known values into this expression yields $$[H^+] = \sqrt{2 \times 10^{-5} \times 0.2} = \sqrt{4 \times 10^{-6}} = 2 \times 10^{-3} \text{ M}$$.

From the above hydrogen ion concentration, we calculate the pH as follows: $$pH = -\log[H^+] = -\log(2 \times 10^{-3}) = -[\log 2 + \log 10^{-3}] = -[0.30 + (-3)] = -[0.30 - 3] = -(-2.70) = 2.70$$.

Expressing this pH in the required form gives $$pH = 2.70 = 27 \times 10^{-1}$$. Therefore, the value is 27.

We need to find the pH of a 0.001 M NaOH solution. NaOH is a strong base that completely dissociates in water: $$NaOH \rightarrow Na^+ + OH^-$$.

Since NaOH dissociates completely: $$[OH^-] = 0.001 \text{ M} = 10^{-3} \text{ M}$$.

$$pOH = -\log[OH^-] = -\log(10^{-3}) = 3$$.

At 25°C, using the relation $$pH + pOH = 14$$: $$pH = 14 - pOH = 14 - 3 = 11$$.

The pH of the solution is 11.

To determine the basicity of acid 'A', we apply the law of equivalence, which states that in a complete neutralization reaction, the milliequivalents (mEq) of the acid must equal the milliequivalents of the base. The formula for calculating milliequivalents is Molarity × Volume × n-factor. First, we calculate the milliequivalents for the base, $$M(OH)_2$$. Its volume is 30 mL, its molarity is 0.05 M, and its acidity (n-factor) is 2 because it can donate two hydroxide ions. Multiplying these together gives $$30 \times 0.05 \times 2 = 3$$ mEq. Next, we set up the expression for acid 'A', which has a volume of 10 mL, a molarity of 0.1 M, and an unknown basicity (n-factor) that we will call '$$x$$'. This gives us $$10 \times 0.1 \times x = 1x$$ mEq. By setting the milliequivalents equal to each other ($$1x = 3$$), we find that $x$ equals 3. Therefore, the basicity of the acid is 3.

We are given that the equilibrium constant for the reaction:

$$Cu(s) + 2Ag^+(aq) \rightleftharpoons Cu^{2+}(aq) + 2Ag(s)$$

is $$K_1 = 2 \times 10^{15}$$ at 298 K.

We need to find the equilibrium constant for:

$$\frac{1}{2}Cu^{2+}(aq) + Ag(s) \rightleftharpoons \frac{1}{2}Cu(s) + Ag^+(aq)$$

The second reaction is the reverse of the first reaction, divided by 2.

When a reaction is reversed, the new equilibrium constant is the reciprocal: $$K_{reverse} = \frac{1}{K_1}$$. When a reaction is divided by 2, the new equilibrium constant is the square root: $$K_2 = \sqrt{K_{reverse}}$$.

Therefore:

$$K_2 = \sqrt{\frac{1}{K_1}} = \frac{1}{\sqrt{K_1}} = \frac{1}{\sqrt{2 \times 10^{15}}}$$

Calculating the value gives:

$$K_2 = \frac{1}{\sqrt{2 \times 10^{15}}} = \frac{1}{\sqrt{2} \times 10^{7.5}} = \frac{1}{1.414 \times 3.162 \times 10^7} = \frac{1}{4.472 \times 10^7}$$

$$K_2 = 2.236 \times 10^{-8} \approx 2 \times 10^{-8}$$

Since $$K_2 = x \times 10^{-8}$$, we have $$x = 2$$.

Therefore, the correct answer is 2.

We need to find the standard free energy change for 50% dissociation of $$N_2O_4$$ into $$NO_2$$. Setting up the dissociation equilibrium $$N_2O_4 \rightleftharpoons 2NO_2$$ and starting with 1 mol of $$N_2O_4$$, 50% dissociation corresponds to 0.5 mol dissociating, leaving $$1 - 0.5 = 0.5$$ mol of $$N_2O_4$$ and producing $$2 \times 0.5 = 1$$ mol of $$NO_2$$, which makes a total of $$0.5 + 1 = 1.5$$ mol.

At a total pressure of 1 atm, the partial pressures are given by the mole fractions, namely $$p_{N_2O_4} = \frac{0.5}{1.5} \times 1 = \frac{1}{3}$$ atm and $$p_{NO_2} = \frac{1}{1.5} \times 1 = \frac{2}{3}$$ atm, from which we calculate:

$$K_p = \frac{(p_{NO_2})^2}{p_{N_2O_4}} = \frac{(2/3)^2}{1/3} = \frac{4/9}{1/3} = \frac{4}{3} \approx 1.33$$

To find the standard free energy change, we use the relation $$\Delta G° = -RT \ln K_p = -RT \times 2.303 \times \log K_p$$ at $$T = 27°C = 300$$ K. Thus,

$$\Delta G° = -8.31 \times 300 \times 2.3 \times \log(1.33)$$
$$= -8.31 \times 300 \times 2.3 \times 0.1239$$
$$= -8.31 \times 300 \times 0.28497$$
$$= -8.31 \times 85.491$$
$$= -710.43 \text{ J mol}^{-1}$$

Hence, since $$\Delta G° = -x$$ J mol$$^{-1}$$, we find $$x \approx 710$$ (nearest integer).

The correct answer is $$\mathbf{710}$$.

We begin by recalling that in any acid-base titration the chosen indicator must change colour as close as possible to the pH of the equivalence point. This is because, at that pH, the steep vertical rise in the titration curve occurs, and even a single extra drop of titrant causes a sudden jump in pH. If the indicator’s transition range lies completely inside that vertical portion, the end-point and the equivalence point coincide, giving minimum error.

Now, let us examine Statement I: “In the titration between a strong acid and a weak base methyl orange is suitable as an indicator.” A strong acid-weak base titration curve starts at a very low pH (because of the strong acid) and ends at an acidic equivalence point. In fact, the pH at equivalence is usually around $$\text{pH}\approx 3\text{ to }5$$, because the conjugate acid of the weak base hydrolyses and releases $$\mathrm{H^{+}}$$ ions. Hence the vertical segment of the curve lies roughly between $$\text{pH}=3$$ and $$\text{pH}=6$$.

Methyl orange has a colour change interval $$\text{pH}=3.1\;{\text{to}}\;4.4$$. Clearly this entire interval is inside the steep section of the strong-acid-weak-base titration curve. Therefore methyl orange will flip colour exactly where the equivalence occurs, giving an accurate end-point. So Statement I is true.

Next, we analyse Statement II: “For titration of acetic acid with NaOH phenolphthalein is not a suitable indicator.” Acetic acid is a weak acid while sodium hydroxide is a strong base. In a weak acid-strong base titration the pH at equivalence is greater than 7, typically around $$\text{pH}\approx 8.7$$. The vertical portion of this curve generally spans roughly $$\text{pH}=7.5$$ to $$\text{pH}=10$$.

Phenolphthalein changes colour in the interval $$\text{pH}=8.3\;{\text{to}}\;10.0$$. We observe that this entire range lies wholly within the steep rise of the weak-acid-strong-base titration curve, embracing the equivalence pH of $$\approx 8.7$$. Hence phenolphthalein will signal the end-point accurately and is, in fact, the preferred indicator for this titration.

Therefore Statement II is false.

Combining our results, Statement I is true while Statement II is false. The option matching this combination is Option D.

Hence, the correct answer is Option D.

For $$Ca(OH)_2$$, the dissociation in water is: $$Ca(OH)_2 \rightleftharpoons Ca^{2+} + 2OH^-$$. If the solubility is $$s$$, then $$[Ca^{2+}] = s$$ and $$[OH^-] = 2s$$.

The solubility product is $$K_{sp} = [Ca^{2+}][OH^-]^2 = s \cdot (2s)^2 = 4s^3$$.

Given $$K_{sp} = 5.5 \times 10^{-6}$$, we have $$4s^3 = 5.5 \times 10^{-6}$$, so $$s^3 = \frac{5.5 \times 10^{-6}}{4} = 1.375 \times 10^{-6}$$.

Taking the cube root: $$s = (1.375 \times 10^{-6})^{1/3} = (1.375)^{1/3} \times 10^{-2}$$. Now $$(1.375)^{1/3} \approx 1.112$$.

So $$s \approx 1.11 \times 10^{-2}$$ mol/L.

Therefore, the solubility of $$Ca(OH)_2$$ is $$1.11 \times 10^{-2}$$, which is option (3).

To determine which precipitate forms first, we need to find the minimum concentration of $$Ag^+$$ required to start precipitation of each ion.

For AgCl: $$K_{sp}(\text{AgCl}) = [\text{Ag}^+][\text{Cl}^-] = 1.7 \times 10^{-10}$$

With $$[\text{Cl}^-] = 0.1$$ M, precipitation begins when:

$$[\text{Ag}^+] = \frac{K_{sp}}{[\text{Cl}^-]} = \frac{1.7 \times 10^{-10}}{0.1} = 1.7 \times 10^{-9} \text{ M}$$

For $$Ag_2CrO_4$$: $$K_{sp}(\text{Ag}_2\text{CrO}_4) = [\text{Ag}^+]^2[\text{CrO}_4^{2-}] = 1.9 \times 10^{-12}$$

With $$[\text{CrO}_4^{2-}] = 0.001$$ M, precipitation begins when:

$$[\text{Ag}^+]^2 = \frac{K_{sp}}{[\text{CrO}_4^{2-}]} = \frac{1.9 \times 10^{-12}}{10^{-3}} = 1.9 \times 10^{-9}$$

$$[\text{Ag}^+] = \sqrt{1.9 \times 10^{-9}} \approx 4.4 \times 10^{-5} \text{ M}$$

Since $$1.7 \times 10^{-9}$$ M $$\ll$$ $$4.4 \times 10^{-5}$$ M, a much smaller amount of $$Ag^+$$ is needed to precipitate $$Cl^-$$ as AgCl. Therefore, AgCl precipitates first because the amount of $$Ag^+$$ needed is lower.

We need to find the solubility of AgCN in a buffer solution of pH = 3. Given: $$K_{sp}(\text{AgCN}) = 2.2 \times 10^{-16}$$ and $$K_a(\text{HCN}) = 6.2 \times 10^{-10}$$.

Let the solubility of AgCN be $$s$$. When AgCN dissolves: $$\text{AgCN} \rightleftharpoons \text{Ag}^+ + \text{CN}^-$$. So $$[\text{Ag}^+] = s$$.

In the acidic buffer (pH = 3), the $$\text{CN}^-$$ ions react with $$\text{H}^+$$: $$\text{CN}^- + \text{H}^+ \rightleftharpoons \text{HCN}$$. Since $$[\text{H}^+] = 10^{-3}\,\text{M}$$ is much larger than $$K_a = 6.2 \times 10^{-10}$$, most of the $$\text{CN}^-$$ is converted to HCN.

The total cyanide concentration is $$[\text{CN}^-] + [\text{HCN}] = s$$. From the acid dissociation: $$K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]}$$, so $$[\text{CN}^-] = \frac{K_a \cdot [\text{HCN}]}{[\text{H}^+]}$$. Since $$[\text{H}^+] \gg K_a$$, essentially all cyanide is as HCN, so $$[\text{HCN}] \approx s$$ and $$[\text{CN}^-] = \frac{K_a \cdot s}{[\text{H}^+]} = \frac{6.2 \times 10^{-10} \times s}{10^{-3}} = 6.2 \times 10^{-7} \cdot s$$.

Applying the solubility product: $$K_{sp} = [\text{Ag}^+][\text{CN}^-] = s \times 6.2 \times 10^{-7} \cdot s = 6.2 \times 10^{-7} \cdot s^2$$.

Solving: $$s^2 = \frac{2.2 \times 10^{-16}}{6.2 \times 10^{-7}} = \frac{2.2}{6.2} \times 10^{-9} \approx 0.3548 \times 10^{-9} = 3.548 \times 10^{-10}$$.

$$s = \sqrt{3.548 \times 10^{-10}} \approx 1.88 \times 10^{-5} \approx 1.9 \times 10^{-5}$$.

Therefore, the solubility of AgCN is $$1.9 \times 10^{-5}$$, which corresponds to option (1).

Statement I states that the pH of rain water is normally ~5.6. This is true because atmospheric $$CO_2$$ dissolves in rainwater to form carbonic acid: $$CO_2 + H_2O \rightarrow H_2CO_3$$. This weak acid lowers the pH of pure rainwater from 7.0 to approximately 5.6.

Statement II states that if the pH of rain water drops below 5.6, it is called acid rain. This is also true. Acid rain is defined as rain with a pH below the normal value of 5.6, typically caused by pollutants such as $$SO_2$$ and $$NO_x$$ that form sulphuric acid and nitric acid in the atmosphere.

Since both statements are true, the correct answer is option (1): Both Statement I and Statement II are true.

When the sparingly soluble salt $$A_3B_2$$ dissolves, it dissociates into its constituent ions based on its stoichiometry:

$$A_3B_2(s) \rightleftharpoons 3A(aq) + 2B(aq)$$

If the molar solubility of the entire salt is $$s$$ (which is equal to $$\frac{x}{M}$$), then at equilibrium, the concentrations of the individual ions will be proportional to their coefficients:

• Concentration of $$A$$ ions: $$[A] = 3s$$
• Concentration of $$B$$ ions: $$[B] = 2s$$

he $$K_{sp}$$ expression is the product of the ion concentrations, each raised to the power of their stoichiometric coefficients from the balanced equation:

$$K_{sp} = [A]^3[B]^2$$ = $$108 \times s^5$$ = $$108 \times \frac{x}{M}$$.

For the reaction $$\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$$, the change in moles of gas is $$\Delta n = 2 - 1 = 1$$.

The relationship between $$K_P$$ and $$K_C$$ is given by $$K_P = K_C \cdot (RT)^{\Delta n}$$. Substituting $$\Delta n = 1$$:

$$K_P = K_C \cdot RT$$

Solving for $$T$$:

$$T = \frac{K_P}{K_C \cdot R} = \frac{600.1}{20.4 \times 0.0831}$$

Computing the denominator: $$20.4 \times 0.0831 = 1.69524$$

$$T = \frac{600.1}{1.69524} = 354.0 \text{ K}$$

Rounding to the nearest integer, the temperature is 354 K.

The answer is 354.

We have the homogeneous gaseous reaction

$$\mathrm{A + B \rightleftharpoons C + D}$$

The equilibrium constant at 298 K is given as

$$K_c = 100.$$

Initially all four species are present in equal molar concentration, viz.

$$[\mathrm A]_0 = [\mathrm B]_0 = [\mathrm C]_0 = [\mathrm D]_0 = 1\ \text{M}.$$

The reaction quotient at the start is therefore

$$Q_0 = \frac{[\mathrm C]_0[\mathrm D]_0}{[\mathrm A]_0[\mathrm B]_0} = \frac{1 \times 1}{1 \times 1} = 1.$$

Because $$K_c = 100 > Q_0 = 1,$$ the reaction must proceed in the forward direction so that more products (C and D) are formed until equilibrium is reached.

Let $$x\ \text{M}$$ of A and the same amount of B be consumed. By stoichiometry the same $$x\ \text{M}$$ of C and of D will be produced. Hence the concentrations at equilibrium become

$$[\mathrm A]_{eq} = 1 - x,$$

$$[\mathrm B]_{eq} = 1 - x,$$

$$[\mathrm C]_{eq} = 1 + x,$$

$$[\mathrm D]_{eq} = 1 + x.$$

We now impose the definition of the equilibrium constant:

$$K_c = \frac{[\mathrm C]_{eq}[\mathrm D]_{eq}} {[\mathrm A]_{eq}[\mathrm B]_{eq}} = \frac{(1 + x)(1 + x)}{(1 - x)(1 - x)} = \left(\frac{1 + x}{1 - x}\right)^2.$$

Substituting the given value $$K_c = 100$$ we get

$$100 = \left(\frac{1 + x}{1 - x}\right)^2.$$

Taking the positive square root (concentrations must stay positive)

$$\sqrt{100} = 10 = \frac{1 + x}{1 - x}.$$

Now we solve this linear equation for $$x$$:

$$10(1 - x) = 1 + x,$$

$$10 - 10x = 1 + x,$$

$$10 - 1 = x + 10x,$$

$$9 = 11x,$$

$$x = \frac{9}{11}\ \text{M} \approx 0.818181\ \text{M}.$$

Therefore the equilibrium concentration of D is

$$[\mathrm D]_{eq} = 1 + x = 1 + \frac{9}{11} = \frac{20}{11}\ \text{M} \approx 1.818181\ \text{M}.$$

The problem asks for this value in the form $$\;n \times 10^{-2}\ \text{M}.$$

We convert as follows:

$$[\mathrm D]_{eq} = 1.818181\ \text{M} = 1.818181 \times 10^{0}\ \text{M} = 181.8181 \times 10^{-2}\ \text{M}.$$

The nearest integer to $$181.8181$$ is $$182.$$ Hence, the correct answer is Option 182.

We are given $$\Delta G^\circ = +25.2 \text{ kJ mol}^{-1} = 25200 \text{ J mol}^{-1}$$ at $$T = 400 \text{ K}$$, and need to find $$K_C$$. We use the relation $$\Delta G^\circ = -RT \ln K_p$$, where $$R = 8.3 \text{ J mol}^{-1} \text{K}^{-1}$$ (as given in the problem).

Rearranging: $$\ln K_p = -\frac{\Delta G^\circ}{RT} = -\frac{25200}{8.3 \times 400} = -\frac{25200}{3320} = -7.59$$.

Converting to $$\log_{10}$$: since $$\ln x = 2.3 \times \log_{10} x$$, we get $$\log_{10} K_p = \frac{\ln K_p}{2.3} = \frac{-7.59}{2.3} = -3.3$$. Writing this as $$\log_{10} K_p = -3 - 0.3$$, we get $$K_p = 10^{-3} \times 10^{-0.3}$$. Since $$\log_{10} 2 = 0.30$$, we have $$10^{0.3} = 2$$, so $$10^{-0.3} = \frac{1}{2} = 0.5$$. Therefore $$K_p = 0.5 \times 10^{-3} = 5 \times 10^{-4}$$.

Now we convert $$K_p$$ to $$K_C$$ using the relation $$K_p = K_C \times (RT)^{\Delta n}$$. For the reaction $$2A(g) \rightleftharpoons A_2(g)$$, the change in moles of gas is $$\Delta n = 1 - 2 = -1$$. So $$K_p = K_C \times (RT)^{-1}$$, which gives $$K_C = K_p \times (RT)^{1} = K_p \times RT$$.

Here we need $$RT$$ in units of $$\text{L atm mol}^{-1}$$ (or equivalently $$\text{L bar mol}^{-1}$$, since the problem states 1 atm = 1 bar). We use $$R = 0.0831 \text{ L bar mol}^{-1} \text{K}^{-1}$$, so $$RT = 0.0831 \times 400 = 33.24 \text{ L bar mol}^{-1}$$. Note that this is a different numerical value of $$R$$ from the one used earlier because the units are different: $$8.3 \text{ J mol}^{-1}\text{K}^{-1}$$ is used for energy calculations, while $$0.0831 \text{ L bar mol}^{-1}\text{K}^{-1}$$ is used for the $$K_p$$-$$K_C$$ conversion involving volume units.

Therefore $$K_C = 5 \times 10^{-4} \times 33.24 = 1.662 \times 10^{-2} \approx 2 \times 10^{-2}$$.

The answer is $$2$$ (i.e., $$K_C = 2 \times 10^{-2}$$).

We first calculate the number of millimoles present before mixing. For an acid or a base, the relation between molarity, volume and millimoles is given by

$$\text{millimoles} = M \times V(\text{mL})$$

For the strong acid $$\mathrm{HCl}$$ we have

$$50\ \text{mL}\times 1\ \text{M}=50\ \text{millimoles}.$$

For the strong base $$\mathrm{NaOH}$$ we have

$$30\ \text{mL}\times 1\ \text{M}=30\ \text{millimoles}.$$

The neutralisation reaction is

$$\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}$$

Since the reaction proceeds 1 : 1, $$30$$ millimoles of $$\mathrm{HCl}$$ are completely consumed by the $$30$$ millimoles of $$\mathrm{NaOH}$$. The excess unreacted acid is

$$50-30 = 20\ \text{millimoles \ of } \mathrm{HCl}.$$

These $$20$$ millimoles of $$\mathrm{HCl}$$ remain in the final solution. The total volume after mixing becomes

$$50\ \text{mL}+30\ \text{mL}=80\ \text{mL}=0.08\ \text{L}.$$

Hence the molarity of $$\mathrm{H^+}$$ ions in the mixture is

$$[\mathrm H^+] = \frac{20\ \text{mmol}}{0.08\ \text{L}} = \frac{0.020\ \text{mol}}{0.08\ \text{L}} = 0.25\ \text{M} = 2.5\times 10^{-1}\ \text{M}.$$

We now use the definition of pH:

$$\text{pH} = -\log[\mathrm H^+].$$

Substituting $$[\mathrm H^+] = 2.5\times10^{-1}\ \text{M}$$ gives

$$\text{pH} = -\log\!\left(2.5\times10^{-1}\right).$$

Using the logarithmic identity $$\log(ab)=\log a +\log b$$, we write

$$\log\!\left(2.5\times10^{-1}\right)=\log 2.5 +\log10^{-1} =0.3979-1 =-0.6021.$$

Therefore

$$\text{pH}= -(-0.6021)=0.6021.$$

The problem states the pH in the form $$x\times10^{-4}$$. We write

$$0.6021 = 6021\times10^{-4}.$$

Thus $$x = 6021$$ (nearest integer).

So, the answer is $$6021$$.

This is a double indicator acid-base titration of a mixture containing NaOH, $$Na_2CO_3$$, and inert impurities with $$\frac{N}{10}$$ HCl.

In the first step using phenolphthalein indicator, all of the NaOH is neutralised and half of the $$Na_2CO_3$$ is converted to $$NaHCO_3$$. The volume of HCl used is $$V_1 = 17.5$$ mL.

In the second step after adding methyl orange indicator, the remaining half of $$Na_2CO_3$$ (now as $$NaHCO_3$$) is neutralised. The volume of HCl used is $$V_2 = 1.5$$ mL.

The volume of HCl used for the second half of $$Na_2CO_3$$ equals $$V_2 = 1.5$$ mL. So the total volume of HCl for complete neutralisation of $$Na_2CO_3$$ is $$2 \times V_2 = 2 \times 1.5 = 3.0$$ mL.

The milliequivalents of $$Na_2CO_3$$ are: $$3.0 \times \frac{1}{10} = 0.3$$ meq. Since the equivalent weight of $$Na_2CO_3$$ is $$\frac{106}{2} = 53$$ g/eq, the weight of $$Na_2CO_3$$ is: $$\frac{0.3 \times 53}{1000} = 0.0159$$ g.

The weight percentage of $$Na_2CO_3$$ in the mixture is: $$\frac{0.0159}{0.4} \times 100 = 3.975\% \approx 4\%$$

The weight percentage of $$Na_2CO_3$$ in the mixture is $$\mathbf{4}$$.

For the equilibrium $$2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g})$$, the equilibrium constant $$K_P$$ is defined as:

$$K_P = \frac{(P_{\text{SO}_3})^2}{(P_{\text{SO}_2})^2 \cdot P_{\text{O}_2}}$$

The given partial pressures are: $$P_{\text{SO}_3} = 43 \text{ kPa}$$, $$P_{\text{O}_2} = 530 \text{ Pa} = 0.530 \text{ kPa}$$, and $$P_{\text{SO}_2} = 45 \text{ kPa}$$. All pressures must be in consistent units (kPa).

Substituting:

$$K_P = \frac{(43)^2}{(45)^2 \times 0.530} = \frac{1849}{2025 \times 0.530} = \frac{1849}{1073.25}$$

$$K_P = \frac{1849}{1073.25} \approx 1.7228 \text{ kPa}^{-1}$$

Expressing as $$K_P = p \times 10^{-2}$$: $$p = 1.7228 \times 10^{2} \times 10^{-2} = 172.28$$

Rounding to the nearest integer, $$K_P \approx 172 \times 10^{-2} \text{ kPa}^{-1}$$, so the answer is $$\boxed{172}$$.

We are given the reaction $$A(g) \rightleftharpoons B(g)$$ at 495 K with $$\Delta_r G^\circ = -9.478$$ kJ/mol. We start with 22 millimoles of A in a closed container and need to find the amount of B at equilibrium.

First, we find the equilibrium constant using $$\Delta_r G^\circ = -RT \ln K$$. Rearranging: $$\ln K = \frac{-\Delta_r G^\circ}{RT} = \frac{9478}{8.314 \times 495} = \frac{9478}{4115.43} = 2.3026$$.

Since $$\ln 10 = 2.303$$, we get $$K = 10$$.

For the reaction $$A \rightleftharpoons B$$, if we start with 22 millimoles of A and let $$x$$ millimoles react, at equilibrium we have $$(22 - x)$$ millimoles of A and $$x$$ millimoles of B.

Since both A and B are gases in the same container, the equilibrium constant in terms of mole fractions (which equals $$K_p$$ for this equal-moles reaction) gives us $$K = \frac{x}{22 - x}$$. For a reaction with equal total moles of gas on both sides, $$K_p = K$$ in terms of the ratio of partial pressures, which equals the ratio of moles.

So $$10 = \frac{x}{22 - x}$$. Solving: $$10(22 - x) = x$$, giving $$220 - 10x = x$$, so $$11x = 220$$, hence $$x = 20$$.

The amount of B at equilibrium is $$\mathbf{20}$$ millimoles.

For sulphurous acid ($$H_2SO_3$$) with $$K_{a1} = 1.7 \times 10^{-2}$$ and $$K_{a2} = 6.4 \times 10^{-8}$$, since $$K_{a1} \gg K_{a2}$$, the pH is primarily determined by the first dissociation.

The first dissociation is: $$H_2SO_3 \rightleftharpoons H^+ + HSO_3^-$$. Let the concentration of $$H^+$$ produced be $$x$$. Then: $$K_{a1} = \frac{x^2}{0.588 - x} = 1.7 \times 10^{-2}$$.

Since $$K_{a1}$$ is not negligible compared to the concentration, we cannot ignore $$x$$ in the denominator. Rearranging: $$x^2 + 0.017x - 0.009996 = 0$$.

Using the quadratic formula: $$x = \frac{-0.017 + \sqrt{(0.017)^2 + 4(0.009996)}}{2} = \frac{-0.017 + \sqrt{0.000289 + 0.039984}}{2} = \frac{-0.017 + \sqrt{0.040273}}{2}$$.

$$x = \frac{-0.017 + 0.2007}{2} = \frac{0.1837}{2} = 0.09185$$ M.

$$pH = -\log(0.09185) = -\log(9.185 \times 10^{-2}) = 2 - \log(9.185) = 2 - 0.963 = 1.037$$.

Rounding off to the nearest integer, the pH is $$1$$.

For any equilibrium involving gases, the equilibrium constant in terms of pressure, $$K_p$$, is written by taking the partial pressure of every gaseous species raised to the power of its stoichiometric coefficient and dividing the product of such terms for products by that for reactants.

The balanced reaction supplied is

$$A(s) \;\rightleftharpoons\; M(s) + \tfrac{1}{2}\,O_2(g)$$

Here, both $$A(s)$$ and $$M(s)$$ are solids. According to the law of mass action, the activity (or “effective concentration”) of a pure solid is taken as unity, so solids do not appear explicitly in the expression for $$K_p$$. Only the gaseous component, $$O_2(g)$$, contributes.

Writing the expression, we have

$$K_p \;=\; \bigl(P_{O_2}\bigr)^{\,\tfrac{1}{2}}$$

because the stoichiometric coefficient of $$O_2(g)$$ is $$\tfrac12$$, and the pressure term must be raised to that power. All other species are solids and hence are omitted (equivalently, treated as 1).

The numerical value of the equilibrium constant is given as $$K_p = 4$$. Substituting this value, we get

$$4 \;=\; \bigl(P_{O_2}\bigr)^{\tfrac{1}{2}}$$

To isolate $$P_{O_2}$$, we remove the square root (i.e., we square both sides). Stating the algebraic operation explicitly:

$$\bigl(P_{O_2}\bigr)^{\tfrac{1}{2}} \;=\; 4 \;\;\Longrightarrow\;\; \left[\bigl(P_{O_2}\bigr)^{\tfrac{1}{2}}\right]^{2} \;=\; 4^{2}$$

Since $$\bigl(x^{1/2}\bigr)^{2} = x$$, the left-hand side simplifies directly to $$P_{O_2}$$. Performing the squaring on the right-hand side, we obtain

$$P_{O_2} \;=\; 16$$

The unit carried by a partial pressure in a $$K_p$$ calculation is atmospheres (atm). Therefore, the equilibrium partial pressure of oxygen is

$$P_{O_2} = 16\ \text{atm}$$

Because the question instructs us to round to the nearest integer, and 16 is already an integer, no further adjustment is necessary.

So, the answer is $$16$$.

We begin by noting that the mixture contains a weak base, ammonia NH$$_3$$, and its conjugate acid, the ammonium ion NH$$_4^+$$ (supplied by NH$$_4$$Cl). Such a pair constitutes a buffer system that obeys the equilibrium

$$\mathrm{NH_3\;+\;H_2O \rightleftharpoons NH_4^+ \;+\; OH^-}$$

For this equilibrium, the base-dissociation constant is given as

$$$K_b \;=\;\frac{[\mathrm{NH_4^+}]\,[\mathrm{OH^-}]}{[\mathrm{NH_3}]} \;=\;1.8\times10^{-5}$$$

Our first task is to calculate the molar concentrations of NH$$_3$$ and NH$$_4^+$$ after mixing the two solutions.

Number of moles of NH$$_3$$ present:

$$$n_{\mathrm{NH_3}} \;=\; V \times M \;=\; (2.0\;\text{mL})\;\Big(\frac{1\;\text{L}}{1000\;\text{mL}}\Big)\;\times 0.0210\;\text{M}$$$

$$$=\;0.0020\;\text{L}\times0.0210\;\text{mol L}^{-1} \;=\;4.20\times10^{-5}\;\text{mol}$$$

Number of moles of NH$$_4^+$$ (from NH$$_4$$Cl):

$$$n_{\mathrm{NH_4^+}} \;=\; V \times M \;=\; (5.0\;\text{mL})\;\Big(\frac{1\;\text{L}}{1000\;\text{mL}}\Big)\;\times 0.0504\;\text{M}$$$

$$$=\;0.0050\;\text{L}\times0.0504\;\text{mol L}^{-1} \;=\;2.52\times10^{-4}\;\text{mol}$$$

The total volume after mixing is simply the sum of the two volumes (volume additivity is assumed):

$$$V_{\text{total}} \;=\; 5.0\;\text{mL} \;+\; 2.0\;\text{mL} \;=\; 7.0\;\text{mL} \;=\; 0.0070\;\text{L}$$$

Hence the initial molarities (before any further equilibrium shift) are

$$$[\mathrm{NH_3}]_0 \;=\;\frac{4.20\times10^{-5}\;\text{mol}}{0.0070\;\text{L}} \;=\;6.00\times10^{-3}\;\text{M}$$$

$$$[\mathrm{NH_4^+}]_0 \;=\;\frac{2.52\times10^{-4}\;\text{mol}}{0.0070\;\text{L}} \;=\;3.60\times10^{-2}\;\text{M}$$$

Because NH$$_3$$ is weak and NH$$_4^+$$ is its conjugate acid, only a very small additional amount of NH$$_3$$ will convert to NH$$_4^+$$ and OH$$^-$$. We let the equilibrium concentration of OH$$^-$$ be $$y$$:

$$$\begin{aligned} \mathrm{NH_3} &\;\;+\;\; \mathrm{H_2O} &\rightleftharpoons&\;\; \mathrm{NH_4^+} &+;;& \mathrm{OH^-} \\ \text{Initial (M)} &\;6.00\times10^{-3} &&\;3.60\times10^{-2} && 0 \\ \text{Change (M)} &\;-y && +y && +y\\ \text{Equilibrium (M)} &6.00\times10^{-3}-y && 3.60\times10^{-2}+y && y \end{aligned}$$$

Because the buffer is fairly concentrated and $$y$$ will be very small compared with the starting concentrations, we make the common buffer approximation $$6.00\times10^{-3}-y \approx 6.00\times10^{-3}$$ and $$3.60\times10^{-2}+y \approx 3.60\times10^{-2}$$. Substituting these equilibrium values into the definition of $$K_b$$ gives

$$$K_b \;=\;\frac{(3.60\times10^{-2})\,y}{6.00\times10^{-3}}$$$

Solving for $$y$$ (which equals $$[\mathrm{OH^-}]$$):

$$$y \;=\;\frac{K_b\; \bigl(6.00\times10^{-3}\bigr)}{3.60\times10^{-2}}$$$

$$$=\;\frac{1.8\times10^{-5}\;\times\;6.00\times10^{-3}}{3.60\times10^{-2}}$$$

Now we simplify step by step:

$$$1.8\times10^{-5}\;\times\;6.00\times10^{-3} = 10.8\times10^{-8} = 1.08\times10^{-7}$$$

$$$\frac{1.08\times10^{-7}}{3.60\times10^{-2}} = \frac{1.08}{3.60}\times10^{-5} = 0.30\times10^{-5} = 3.0\times10^{-6}$$$

Thus

$$[\mathrm{OH^-}] \;=\;3.0\times10^{-6}\;\text{M}$$

The question states this concentration in the form $$x\times10^{-6}\;\text{M}$$, so clearly

$$x = 3$$

As 3 is already an integer, taking the nearest integer does not alter the value.

Hence, the correct answer is Option C.

Ammonium phosphate $$(NH_4)_3PO_4$$ is a salt formed from the weak base ammonium hydroxide ($$NH_4OH$$) and the weak acid phosphoric acid ($$H_3PO_4$$).

For a salt of a weak acid and a weak base, the pH of the aqueous solution is given by the formula:

$$pH = 7 + \frac{1}{2}(pK_a - pK_b)$$

Here, $$pK_a$$ refers to the dissociation constant of the weak acid and $$pK_b$$ refers to the dissociation constant of the weak base. We are given $$pK_a = 5.23$$ (for phosphoric acid) and $$pK_b = 4.75$$ (for ammonium hydroxide).

Note that this formula applies when both the cation and anion hydrolyse. The ammonium ion ($$NH_4^+$$) undergoes cationic hydrolysis: $$NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$$. The phosphate ion ($$PO_4^{3-}$$) undergoes anionic hydrolysis: $$PO_4^{3-} + H_2O \rightleftharpoons HPO_4^{2-} + OH^-$$.

Since both ions hydrolyse, the pH depends on the relative strengths of the acid and base. Substituting the given values into the formula:

$$pH = 7 + \frac{1}{2}(5.23 - 4.75) = 7 + \frac{1}{2}(0.48) = 7 + 0.24 = 7.24$$

Rounded to the nearest integer, the pH of the ammonium phosphate solution is $$7$$.

$$\text{CdSO}_4$$ dissociates as $$\text{CdSO}_4 \rightleftharpoons \text{Cd}^{2+} + \text{SO}_4^{2-}$$. In pure water the solubility is $$s = 8.0 \times 10^{-4}$$ mol/L, so the solubility product is $$K_{sp} = s \times s = s^2 = (8.0 \times 10^{-4})^2 = 6.4 \times 10^{-7}$$.

In 0.01 M $$\text{H}_2\text{SO}_4$$ solution, the sulphate ion concentration from the acid is $$[\text{SO}_4^{2-}] = 0.01$$ M (since $$\text{H}_2\text{SO}_4$$ is a strong acid that fully dissociates). Let the solubility of $$\text{CdSO}_4$$ in this solution be $$s'$$. Since $$s' \ll 0.01$$ M (as given), the total sulphate concentration is approximately 0.01 M.

Applying the solubility product expression: $$K_{sp} = s' \times (0.01 + s') \approx s' \times 0.01$$. Therefore $$s' = \frac{K_{sp}}{0.01} = \frac{6.4 \times 10^{-7}}{10^{-2}} = 6.4 \times 10^{-5} = 64 \times 10^{-6}$$ mol/L.

The answer is $$64$$.

For the equilibrium $$\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)}$$, the relationship between $$K_P$$ and $$K_C$$ is given by $$K_P = K_C(RT)^{\Delta n}$$, where $$\Delta n$$ is the change in moles of gaseous species.

Here $$\Delta n = 2 - 1 = 1$$. Substituting the values: $$47.9 = K_C \times (0.083 \times 288)^1 = K_C \times 23.904$$.

Solving for $$K_C$$: $$K_C = \frac{47.9}{23.904} = 2.004$$.

Rounding to the nearest integer, the answer is $$2$$.

First, we recall that barium hydroxide dissociates completely in water. The balanced ionic equation is

$$\text{Ba(OH)}_2 \;\longrightarrow\; \text{Ba}^{2+} + 2\,\text{OH}^-$$

We are told that the formal concentration of the Ba(OH)$$_2$$ solution is $$0.005\ \text{mol L}^{-1}$$. Because one formula unit of Ba(OH)$$_2$$ furnishes two hydroxide ions, the concentration of hydroxide ions produced is obtained by simple stoichiometry:

$$[\text{OH}^-] = 2 \times 0.005\ \text{mol L}^{-1} = 0.010\ \text{mol L}^{-1}$$

Next, we invoke the ionic product of water. At $$298\ \text{K}$$ the experimentally measured constant is

$$K_w = [\text{H}_3\text{O}^+]\, [\text{OH}^-] = 1.0 \times 10^{-14}$$

Solving this relation for the hydronium-ion concentration gives

$$[\text{H}_3\text{O}^+] = \frac{K_w}{[\text{OH}^-]}$$

Now we substitute the numerical values:

$$[\text{H}_3\text{O}^+] = \frac{1.0 \times 10^{-14}}{0.010} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-2}}$$

Dividing the powers of ten, we obtain

$$[\text{H}_3\text{O}^+] = 1.0 \times 10^{-12}\ \text{mol L}^{-1}$$

The question asks for the numerical factor that multiplies $$10^{-12}$$. We have found that factor to be exactly $$1.0$$, which rounds to the nearest integer as $$1$$.

So, the answer is $$1$$.

We are given that at 1990 K and 1 atm pressure, there are equal number of $$Cl_2$$ molecules and $$Cl$$ atoms in the equilibrium: $$Cl_{2}(g) \rightleftharpoons 2Cl(g)$$.

Since the number of $$Cl_2$$ molecules equals the number of $$Cl$$ atoms, their moles are also equal. Let $$n_{Cl_2} = n_{Cl} = n$$.

The total number of moles is $$n + n = 2n$$.

The mole fraction of $$Cl_2$$ is $$\frac{n}{2n} = \frac{1}{2}$$ and the mole fraction of $$Cl$$ is $$\frac{n}{2n} = \frac{1}{2}$$.

Since the total pressure is 1 atm, the partial pressures are:

Now, $$K_p$$ for the reaction $$Cl_2(g) \rightleftharpoons 2Cl(g)$$ is:

We are told $$K_p = x \times 10^{-1}$$. So:

So, the answer is $$5$$.

For the homogeneous reaction $$\mathrm{A + B \rightleftharpoons 2C}$$ at 298 K the equilibrium constant is given to be $$K_c = 100$$.

Initially all three species have the same concentration, $$[\mathrm A]_0 = [\mathrm B]_0 = [\mathrm C]_0 = 1\;\text{M}$$.

Let the extent of the forward reaction at equilibrium be $$y$$ M. For every mole of A that reacts, one mole of B also reacts and two moles of C are produced. Therefore:

$$ \begin{aligned} [\mathrm A]_{\text{eq}} &= 1 - y,\\[3pt] [\mathrm B]_{\text{eq}} &= 1 - y,\\[3pt] [\mathrm C]_{\text{eq}} &= 1 + 2y. \end{aligned} $$

The equilibrium constant expression for the reaction $$\mathrm{A + B \rightleftharpoons 2C}$$ is

$$ K_c \;=\; \frac{[\mathrm C]_{\text{eq}}^{\,2}}{[\mathrm A]_{\text{eq}}\,[\mathrm B]_{\text{eq}}}. $$

Substituting the equilibrium concentrations gives

$$ 100 \;=\; \frac{(1 + 2y)^{2}}{(1 - y)(1 - y)} \;=\; \left(\frac{1 + 2y}{1 - y}\right)^{2}. $$

Taking the positive square root (all concentrations are positive) we obtain

$$ \frac{1 + 2y}{1 - y} \;=\; 10. $$

Now we solve for $$y$$:

$$ 1 + 2y \;=\; 10(1 - y) \quad\Longrightarrow\quad 1 + 2y \;=\; 10 - 10y. $$

Collecting the $$y$$ terms on one side:

$$ 2y + 10y \;=\; 10 - 1 \quad\Longrightarrow\quad 12y \;=\; 9. $$

Hence

$$ y \;=\; \frac{9}{12} \;=\; \frac34 \;=\; 0.75. $$

Substituting this value back to find the equilibrium concentration of C:

$$ [\mathrm C]_{\text{eq}} \;=\; 1 + 2y \;=\; 1 + 2(0.75) \;=\; 1 + 1.5 \;=\; 2.5\;\text{M}. $$

This concentration is written in the required form as

$$ 2.5\;\text{M} \;=\; 25 \times 10^{-1}\;\text{M}, $$

so the value of $$x$$ is $$25$$.

So, the answer is $$25$$.

We consider the equilibrium reaction

$$PCl_5\; \rightleftharpoons\; PCl_3 + Cl_2$$

with the given equilibrium constant

$$K_c = 1.844$$

Initially, 3.0 moles of $$PCl_5$$ are placed in a 1 L vessel at 380 K. Because the volume is 1 L, the numerical value of the concentration equals the number of moles. Hence the initial concentrations are

$$[PCl_5]_0 = 3.0\;{\rm M}, \qquad [PCl_3]_0 = 0,\qquad [Cl_2]_0 = 0.$$

Let $$x$$ be the number of moles of $$PCl_5$$ that dissociate at equilibrium. The changes and the equilibrium amounts are therefore

$$\begin{aligned} PCl_5 &: \; 3.0 - x,\\ PCl_3 &: \; 0 + x = x,\\ Cl_2 &: \; 0 + x = x. \end{aligned}$$

Again, because the volume is 1 L, these numbers are simultaneously the equilibrium concentrations.

The law of mass action states that for the reaction $$PCl_5 \rightleftharpoons PCl_3 + Cl_2$$ the equilibrium constant is

$$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}.$$

Substituting the equilibrium concentrations we obtain

$$1.844 = \frac{(x)(x)}{3.0 - x} = \frac{x^2}{3.0 - x}.$$

Cross-multiplying gives

$$x^2 = 1.844\,(3.0 - x).$$

Expanding the right-hand side,

$$x^2 = 1.844 \times 3.0 - 1.844\,x = 5.532 - 1.844\,x.$$

Bringing every term to the left produces a quadratic equation,

$$x^2 + 1.844\,x - 5.532 = 0.$$

We now solve this quadratic using the quadratic formula. For an equation $$ax^2 + bx + c = 0,$$ the roots are

$$x = \frac{-b \pm \sqrt{\,b^2 - 4ac\,}}{2a}.$$

Here $$a = 1,\; b = 1.844,\; c = -5.532.$$ Substituting these values,

$$\begin{aligned} x &= \frac{-1.844 \pm \sqrt{(1.844)^2 - 4(1)(-5.532)}}{2(1)}\\[4pt] &= \frac{-1.844 \pm \sqrt{3.400336 + 22.128}}{2}\\[4pt] &= \frac{-1.844 \pm \sqrt{25.528336}}{2}\\[4pt] &= \frac{-1.844 \pm 5.052}{2}. \end{aligned}$$

The negative sign would give a negative value of $$x$$, which is physically impossible, so we take the positive sign:

$$x = \frac{-1.844 + 5.052}{2} = \frac{3.208}{2} = 1.604.$$

Therefore, the equilibrium amount of $$PCl_5$$ is

$$3.0 - x = 3.0 - 1.604 = 1.396\;{\rm moles}.$$

Expressing this in the form “number × 10−3”,

$$1.396\;{\rm moles} = 1.396 \times 10^{0}\;{\rm moles} = 1396 \times 10^{-3}\;{\rm moles}.$$

Rounding 1396 to the nearest integer yields 1400.

Hence, the correct answer is Option 1400.

The dissolution equilibrium for $$PbI_2$$ is:

$$PbI_2 \rightleftharpoons Pb^{2+} + 2I^-$$

The solubility product expression is $$K_{sp} = [Pb^{2+}][I^-]^2 = 8.0 \times 10^{-9}$$.

In a 0.1 M lead nitrate solution, the lead ion concentration from lead nitrate is 0.1 M. Let the solubility of $$PbI_2$$ in this solution be $$s$$ mol/L.

Then $$[Pb^{2+}] = 0.1 + s \approx 0.1$$ M (since $$s$$ is very small compared to 0.1), and $$[I^-] = 2s$$.

Substituting into the $$K_{sp}$$ expression:

$$8.0 \times 10^{-9} = (0.1)(2s)^2 = 0.1 \times 4s^2 = 0.4s^2$$

$$s^2 = \frac{8.0 \times 10^{-9}}{0.4} = 2.0 \times 10^{-8}$$

$$s = \sqrt{2.0 \times 10^{-8}} = \sqrt{2} \times 10^{-4} = 1.41 \times 10^{-4} \text{ mol/L}$$

This equals $$141 \times 10^{-6}$$ mol/L, so the value of $$x$$ is $$\textbf{141}$$.

We are given that salts $$A_2X$$ and $$MX$$ have the same solubility product $$K_{sp} = 4.0 \times 10^{-12}$$. We need to find the ratio of their molar solubilities.

For salt $$A_2X$$, it dissociates as $$A_2X \rightleftharpoons 2A^+ + X^{2-}$$. If the molar solubility is $$S_1$$, then $$[A^+] = 2S_1$$ and $$[X^{2-}] = S_1$$. Therefore, $$K_{sp} = (2S_1)^2 \times S_1 = 4S_1^3$$.

So $$4S_1^3 = 4.0 \times 10^{-12}$$, which gives $$S_1^3 = 1.0 \times 10^{-12}$$, hence $$S_1 = 10^{-4}$$ mol/L.

For salt $$MX$$, it dissociates as $$MX \rightleftharpoons M^+ + X^-$$. If the molar solubility is $$S_2$$, then $$K_{sp} = S_2 \times S_2 = S_2^2$$.

So $$S_2^2 = 4.0 \times 10^{-12}$$, which gives $$S_2 = 2.0 \times 10^{-6}$$ mol/L.

The ratio of molar solubilities is $$\frac{S_1}{S_2} = \frac{10^{-4}}{2.0 \times 10^{-6}} = \frac{10^{-4}}{2 \times 10^{-6}} = 50$$.

The answer is $$\mathbf{50}$$.

The weak acid HA dissociates as $$HA \rightleftharpoons H^+ + A^-$$ with $$K_a = 2.0 \times 10^{-6}$$. When 0.01 mol of HA is dissolved in 1.0 L of 0.1 M HCl, the $$H^+$$ concentration from HCl is 0.1 M, which strongly suppresses the dissociation of HA (common ion effect).

Let $$\alpha$$ be the degree of dissociation of HA. Since $$\alpha \ll 1$$, the concentration of undissociated HA remains approximately $$C = 0.01$$ M. The $$A^-$$ concentration is $$C\alpha = 0.01\alpha$$, and the total $$[H^+] \approx 0.1$$ M (dominated by HCl).

Applying the equilibrium expression: $$K_a = \frac{[H^+][A^-]}{[HA]} = \frac{0.1 \times 0.01\alpha}{0.01} = 0.1\alpha$$. Solving for $$\alpha$$: $$\alpha = \frac{K_a}{0.1} = \frac{2.0 \times 10^{-6}}{0.1} = 2.0 \times 10^{-5}$$.

Expressed as $$\_\_ \times 10^{-5}$$, the degree of dissociation is 2.

The dissolution equilibrium for zinc hydroxide is
$$\text{Zn(OH)}_2(s) \rightleftharpoons \text{Zn}^{2+}(aq) + 2\;\text{OH}^-(aq)$$

For every mole (solubility $$s$$) of $$\text{Zn(OH)}_2$$ that dissolves:
  • $$[\text{Zn}^{2+}] = s$$
  • $$[\text{OH}^-] = 0.10\;\text{M (from NaOH)} + 2s$$

The given solubility product expression is
$$K_{sp} = [\text{Zn}^{2+}]\,[\text{OH}^-]^2$$

Substituting the equilibrium concentrations:
$$2 \times 10^{-20} = s\,(0.10 + 2s)^2 \; -(1)$$

The added hydroxide $$2s$$ is extremely small compared with the fixed $$0.10\;\text{M}$$ (because solubilities of such sparingly soluble hydroxides are usually of the order of $$10^{-17}\;\text{M}$$ or less). Hence,
$$0.10 + 2s \approx 0.10$$

Applying this approximation to $$(1)$$:
$$2 \times 10^{-20} \approx s\,(0.10)^2$$
$$2 \times 10^{-20} = s\,(1 \times 10^{-2})$$

Solving for $$s$$:
$$s = \frac{2 \times 10^{-20}}{1 \times 10^{-2}} = 2 \times 10^{-18}\;\text{M}$$

Thus the molar solubility can be written as $$x \times 10^{-18}\;\text{M}$$ with $$x = 2$$.

Answer (nearest integer): $$2$$.

We consider the equilibrium

$$\text{NH}_4\text{HS(s)} \;\rightleftharpoons\; \text{NH}_3(\text{g}) + \text{H}_2\text{S}(\text{g})$$

The solid does not appear in the expression for $$K_p$$, only the gaseous products do.

First we calculate how many moles of solid are initially taken. The molar mass of NH$$_4$$HS is

$$M = 14 \;(\text{N}) + 5 \;(\text{H}) + 32 \;(\text{S}) = 51 \,\text{g mol}^{-1}$$

Hence, from 5.1 g of the solid the number of moles present initially is

$$n_0 = \frac{5.1\ \text{g}}{51\ \text{g mol}^{-1}} = 0.10\ \text{mol}$$

Only 20 % of this solid decomposes. Therefore the amount that actually undergoes decomposition is

$$n_{\text{decomp}} = 0.20 \times 0.10 = 0.02\ \text{mol}$$

Because the stoichiometric coefficients of NH$$_3$$ and H$$_2$$S are each 1, the decomposition of 0.02 mol of solid produces

$$n_{\text{NH}_3} = 0.02\ \text{mol},\qquad n_{\text{H}_2\text{S}} = 0.02\ \text{mol}$$

The gases are contained in a 2 L flask at 27 °C, i.e. 300 K. From the ideal-gas equation $$P = \dfrac{nRT}{V}$$ their partial pressures are

$$P_{\text{NH}_3} = \frac{(0.02)\,(0.082)\,(300)}{2}$$

First multiply the numerator:

$$0.02 \times 0.082 = 0.00164$$

$$0.00164 \times 300 = 0.492$$

Now divide by 2 L:

$$P_{\text{NH}_3} = \frac{0.492}{2} = 0.246\ \text{atm}$$

The amount of H$$_2$$S is the same, so

$$P_{\text{H}_2\text{S}} = 0.246\ \text{atm}$$

The equilibrium constant in terms of pressure is defined by

$$K_p = P_{\text{NH}_3}\;P_{\text{H}_2\text{S}}$$

Substituting the values just obtained,

$$K_p = (0.246)\,(0.246) = 0.060516 \,\text{atm}^2$$

This can be written in scientific notation as

$$K_p \approx 6.05 \times 10^{-2}$$

The question states $$K_p = x \times 10^{-2}$$, so clearly

$$x \approx 6.05 \; \Longrightarrow \; x = 6\ \text{(nearest integer)}$$

So, the answer is $$6$$.

For adsorption on a solid surface the Freundlich isotherm is used, which in its usual form is written as

$$\frac{x}{m}=kP^{\,1/n}$$

Here $$x/m$$ is the amount adsorbed per gram of adsorbent. In this question the experimentally measured volume $$V$$ of CH$$_4$$ (in mL) plays the role of $$x/m$$, so we may write

$$V=k\,P^{\,1/n}$$

Taking common (base-10) logarithms on both sides, we obtain the linear equation

$$\log V=\log k+\frac{1}{n}\,\log P$$

Two data points are given:

At $$P_1=100\,$$mm Hg, $$V_1=10\,$$mL.
At $$P_2=200\,$$mm Hg, $$V_2=15\,$$mL.

We first evaluate the logarithms needed.

For the first point:

$$\log V_1=\log 10=1$$ $$\log P_1=\log 100=2$$

For the second point:

$$\log V_2=\log 15=\log 3+\log 5 =0.4771+0.6990 =1.1761$$

$$\log P_2=\log 200=\log(2\times100) =\log 2+\log 100 =0.3010+2 =2.3010$$

Substituting each set of values into $$\log V=\log k+\dfrac{1}{n}\log P$$ gives two simultaneous equations:

$$1=\log k+\frac{1}{n}(2)\quad\text{(A)}$$

$$1.1761=\log k+\frac{1}{n}(2.3010)\quad\text{(B)}$$

Now we subtract (A) from (B) to eliminate $$\log k$$:

$$1.1761-1=\frac{1}{n}(2.3010-2)$$ $$0.1761=\frac{1}{n}(0.3010)$$

Hence

$$\frac{1}{n}=\frac{0.1761}{0.3010}=0.585\;(\text{to }3\text{ dp})$$

Substituting this value into equation (A) to find $$\log k$$:

$$1=\log k+2\left(0.585\right)$$ $$1=\log k+1.17$$ $$\log k=1-1.17=-0.17$$

Now we calculate the volume adsorbed at $$P_3=300\,$$mm Hg. First,

$$\log P_3=\log 300=\log(3\times100)=\log 3+\log 100 =0.4771+2 =2.4771$$