JEE Equilibrium Questions

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 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 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: $$\log\left(\frac{x}{1-x}\right)$$.

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: $$\left(\frac{K_{sp}}{6912}\right)^{\frac{1}{7}}$$.

The acid dissociates according to $$HA \; \rightleftharpoons \; H^{+} + A^{-}$$

For a colligative-property experiment we use the van’t Hoff factor $$i$$.
The freezing-point depression is related to $$i$$ by the formula
$$\Delta T_f = i \, K_f \, m$$ $$-(1)$$

Data given: $$\Delta T_f = 0.20^{\circ}\text{C}$$, $$K_f = 1.8 \text{ K kg mol}^{-1}$$, molality $$m = 0.1 \text{ m}$$.
Substituting in $$(1)$$:

$$0.20 = i \,(1.8)\,(0.1)$$

$$\Rightarrow \; i = \frac{0.20}{1.8 \times 0.1} = \frac{0.20}{0.18} = 1.11$$ (keep three significant figures)

For a weak monobasic acid each formula unit can produce two ions, so $$n = 2$$.
The relation between $$i$$ and the degree of dissociation $$\alpha$$ is
$$i = 1 + \alpha (n-1) = 1 + \alpha$$ $$-(2)$$

Using $$i = 1.11$$ in $$(2)$$:
$$\alpha = i - 1 = 1.11 - 1 = 0.11$$

The equilibrium concentration of the acid is $$c = 0.1 \text{ mol L}^{-1}$$.
For a weak monoprotic acid the dissociation constant is
$$K_a = \frac{c \alpha^{2}}{1 - \alpha}$$ $$-(3)$$

Substituting the values:
$$K_a = \frac{0.1 \times (0.11)^{2}}{1 - 0.11} = \frac{0.1 \times 0.0121}{0.89} = \frac{0.00121}{0.89} \approx 1.4 \times 10^{-3}$$

Therefore $$K_a \approx 1.38 \times 10^{-3}$$.

Option A is correct.

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 $$A(OH)_2$$.

We need to find $$K_p$$ for the reaction $$CO_2(g) + C(s) \rightleftharpoons 2CO(g)$$ at 1000 K.

Initially, the partial pressure of $$CO_2$$ is 0.5 atm, while the partial pressure of $$CO$$ is zero. If x atm of $$CO_2$$ decomposes, its pressure decreases by x and the pressure of $$CO$$ increases by 2x, giving equilibrium pressures of $$0.5 - x$$ for $$CO_2$$ and $$2x$$ for $$CO$$.

Since the total pressure at equilibrium is 0.8 atm, we set up the equation $$(0.5 - x) + 2x = 0.5 + x = 0.8$$, from which it follows that $$x = 0.3$$ atm.

Substituting this value of x back into the expressions for the equilibrium pressures yields $$P_{CO_2} = 0.5 - 0.3 = 0.2$$ atm and $$P_{CO} = 2(0.3) = 0.6$$ atm.

The equilibrium constant is then given by $$K_p = \frac{P_{CO}^2}{P_{CO_2}} = \frac{(0.6)^2}{0.2} = \frac{0.36}{0.2} = 1.8$$ atm.

The correct answer is Option 1: 1.8 atm.

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 write the equilibrium constant expression ($$K_C$$) for the reaction:

$$ Fe^{3+}_{(aq)} + SCN^-_{(aq)} \rightleftharpoons (FeSCN)^{2+}_{(aq)} $$

For a general reaction $$aA + bB \rightleftharpoons cC + dD$$ in solution, the equilibrium constant in terms of molar concentrations is:

$$ K_C = \frac{[C]^c[D]^d}{[A]^a[B]^b} $$

Products go in the numerator and reactants go in the denominator, each raised to the power of their stoichiometric coefficient.

Here, the stoichiometric coefficients are all 1:

- Product: $$[FeSCN^{2+}]$$ with coefficient 1

- Reactants: $$[Fe^{3+}]$$ with coefficient 1 and $$[SCN^-]$$ with coefficient 1

$$ K_C = \frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^-]} $$

The correct answer is Option (1): $$K_C = \frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^-]}$$.

We need to identify the strongest Brønsted base among the given cyclic nitrogen-containing compounds.

Key Concept: A strong Brønsted base readily accepts a proton ($$H^+$$). The basic strength of an amine depends on the availability of the lone pair of electrons on the nitrogen atom to coordinate with a proton. If the lone pair is involved in resonance or aromaticity, it becomes less available, making the compound a weaker base.

Option A: Pyridine ($$C_5H_5N$$). In pyridine, the nitrogen atom is $$sp^2$$ hybridized. The lone pair of electrons on the nitrogen atom occupies an $$sp^2$$ hybrid orbital that lies in the plane of the ring, perpendicular to the unhybridized $$p$$ orbitals forming the aromatic $$\pi$$-system. Thus, the lone pair does not participate in aromaticity. However, because it resides in an $$sp^2$$ orbital (which has higher s-character than an $$sp^3$$ orbital), the electrons are held relatively close to the nucleus, reducing their availability compared to aliphatic amines. This is a weak base.

Option B: Aniline derivative / Quinoline / Pyrrole variant. If the option features a molecule where the lone pair is delocalized into an adjacent carbonyl group or a secondary aromatic ring system, its availability is significantly decreased due to resonance stabilization of the neutral molecule. This makes it a very weak base.

Option C: Pyrrole ($$C_4H_5N$$). In pyrrole, the lone pair of electrons on the nitrogen atom is part of the unhybridized $$p$$ orbital and participates directly in the cyclic delocalization to satisfy Huckel's rule ($$4n+2$$ electrons where $$n=1$$) for aromaticity. If pyrrole accepts a proton, its aromatic character is completely destroyed, which is highly thermodynamically unfavorable. Therefore, pyrrole is an extremely weak base.

Option D: Pyrrolidine ($$C_4H_9N$$). Pyrrolidine is a cyclic secondary aliphatic amine. The nitrogen atom is $$sp^3$$ hybridized, and its lone pair of electrons occupies an $$sp^3$$ hybrid orbital. Since there are no double bonds or conjugated systems in the ring, the lone pair is completely localized on the nitrogen atom and highly available for protonation. Furthermore, the adjacent alkyl groups exert an electron-donating inductive effect ($$+I$$ effect), which increases the electron density on the nitrogen atom, making it the strongest base among the choices.

Therefore, pyrrolidine is the strongest Brønsted base because its lone pair is fully localized in an $$sp^3$$ orbital and not involved in any resonance or aromatic stabilization.

Answer: Option D — Pyrrolidine

For the gaseous equilibrium $$A_g \rightleftharpoons B_g + \frac12\,C_g$$ let the initial amount of $$A$$ be 1 mol at an equilibrium (total) pressure $$P$$ and volume $$V$$.

Degree of dissociation $$\alpha$$: fraction of the initial $$A$$ that dissociates.

Number of moles at equilibrium:
$$n_A = 1-\alpha$$ (undissociated $$A$$)
$$n_B = \alpha$$
$$n_C = \dfrac{\alpha}{2}$$

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

Partial pressures (using $$P_i = \dfrac{n_i}{n_{\text{tot}}}\,P$$):
$$P_A = \dfrac{1-\alpha}{1+\alpha/2}\;P$$
$$P_B = \dfrac{\alpha}{1+\alpha/2}\;P$$
$$P_C = \dfrac{\alpha/2}{1+\alpha/2}\;P = \dfrac{\alpha}{2(1+\alpha/2)}\;P$$

Expression for $$K_P$$ (pressure equilibrium constant):
$$K_P = \dfrac{P_B\,P_C^{1/2}}{P_A}$$ $$-(2)$$

Substituting the partial pressures from above into $$(2)$$:

$$\begin{aligned} K_P &= \dfrac{\displaystyle \frac{\alpha P}{1+\alpha/2}\; \left[\frac{\alpha P}{2(1+\alpha/2)}\right]^{1/2}} {\displaystyle \frac{(1-\alpha)P}{1+\alpha/2}} \\[6pt] &= \dfrac{\alpha P}{1+\alpha/2}\; \dfrac{\sqrt{\alpha P}}{\sqrt{2(1+\alpha/2)}}\; \dfrac{1+\alpha/2}{(1-\alpha)P} \\[6pt] &= \dfrac{\alpha^{3/2}P^{3/2}} {(1+\alpha/2)^{3/2}\,\sqrt2}\; \dfrac{1}{(1-\alpha)P}\; (1+\alpha/2) \\[6pt] &= \dfrac{\alpha^{3/2}P^{1/2}} {\sqrt2\,(1-\alpha)\,\sqrt{1+\alpha/2}} \\[6pt] &= \dfrac{\alpha^{3/2}P^{1/2}} {(1-\alpha)\,\sqrt{(2+\alpha)/2}} \\[6pt] &= \dfrac{\alpha^{3/2}P^{1/2}} {(1-\alpha)\,(\sqrt{2+\alpha}/\sqrt2)} \\[6pt] &= \dfrac{\alpha^{3/2}P^{1/2}} {(\sqrt{2+\alpha})(1-\alpha)}. \end{aligned}$$

Therefore

$$K_P = \dfrac{\alpha^{3/2}P^{1/2}} {(2+\alpha)^{1/2}(1-\alpha)}$$

Comparing with the given options, this matches Option B.

Answer : Option B

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 $$S = \frac{10W}{M}$$:

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

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

$$CH_3CH_2COOH$$ (propionic acid), $$CH_3COOH$$ (acetic acid), $$CH_3CH_2CH_2COOH$$ (butyric acid), $$HCOOH$$ (formic acid).

Recall the factor affecting acidity

The acidic strength of carboxylic acids depends on the stability of the conjugate base ($$RCOO^-$$). Alkyl groups exhibit a +I (electron-donating inductive) effect, which destabilises the carboxylate anion by pushing electron density towards the negatively charged carboxylate group. More alkyl groups (and longer chains) lead to greater +I effect and thus weaker acidity.

Compare the +I effect

$$HCOOH$$: No alkyl group attached — strongest acid in this series.

$$CH_3COOH$$: One methyl group — moderate +I effect.

$$CH_3CH_2COOH$$: Ethyl group — stronger +I effect than methyl.

$$CH_3CH_2CH_2COOH$$: Propyl group — strongest +I effect, weakest acid.

Arrange in decreasing order of acidic strength

$$HCOOH > CH_3COOH > CH_3CH_2COOH > CH_3CH_2CH_2COOH$$

This matches Option 3.

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The solubility product $$K_{sp}$$ of a sparingly soluble salt $$AB_2$$ can be determined from the equilibrium concentrations of its ions in solution. Given that $$[A^{2+}] = 1.2 \times 10^{-4}$$ M and $$[B^-] = 0.24 \times 10^{-3} = 2.4 \times 10^{-4}$$ M, the dissolution equilibrium is $$AB_2(s) \rightleftharpoons A^{2+}(aq) + 2B^-(aq).$$

By definition, $$K_{sp} = [A^{2+}][B^-]^2,$$ so substituting the values yields $$K_{sp} = (1.2 \times 10^{-4})(2.4 \times 10^{-4})^2.$$

Squaring the concentration of $$B^-$$ gives $$(2.4 \times 10^{-4})^2 = 5.76 \times 10^{-8},$$ since $$(2.4)^2 = 5.76$$ and $$(10^{-4})^2 = 10^{-8}.$$ Thus the product becomes $$K_{sp} = 1.2 \times 10^{-4} \times 5.76 \times 10^{-8} = (1.2 \times 5.76) \times 10^{-12} = 6.912 \times 10^{-12},$$ which rounds to $$6.91 \times 10^{-12}.$$

Therefore, the solubility product of $$AB_2$$ is $$6.91 \times 10^{-12}.$$

The correct answer is Option 1: $$6.91 \times 10^{-12}$$.

We need to arrange the conjugate bases $$^-OH, R\bar{O}, CH_3CO\bar{O}, C\bar{l}$$ in decreasing order of basic strength.

Key Principle: The stronger the conjugate acid, the weaker the conjugate base (and vice versa).

The conjugate acids are: $$H_2O$$ (of $$^-OH$$), $$ROH$$ (of $$R\bar{O}$$), $$CH_3COOH$$ (of $$CH_3CO\bar{O}$$), $$HCl$$ (of $$C\bar{l}$$).

Acid strength order: $$HCl > CH_3COOH > H_2O > ROH$$

Since stronger acid gives weaker conjugate base:

$$ R\bar{O} > ^-OH > CH_3CO\bar{O} > C\bar{l} $$

Explanation: $$RO^-$$ is the strongest base because its conjugate acid (ROH, an alcohol) is the weakest acid. $$Cl^-$$ is the weakest base because its conjugate acid (HCl) is a very strong acid, meaning $$Cl^-$$ has virtually no tendency to accept protons.

The correct answer is Option (1): $$R\bar{O} > ^-OH > CH_3CO\bar{O} > C\bar{l}$$.

In Group III qualitative analysis, NH₄Cl is added before NH₄OH to suppress the ionization of NH₄OH (common ion effect). This decreases the OH⁻ concentration, ensuring only Group III hydroxides (which have lower Ksp) precipitate, while Group IV and V cations remain in solution.

The correct answer is Option 4: decrease concentration of OH⁻ ions.

$$K_p = K_c(RT)^{\Delta n}$$. $$\Delta n = 2-1 = 1$$.

$$K_c = K_p/(RT) = 0.492/(0.082 \times 300) = 0.492/24.6 = 0.02 = 2 \times 10^{-2}$$.

The answer is $$\boxed{2}$$.

Henderson-Hasselbalch equation: $$pH = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]}$$

$$4.5 = 4.2 + \log\frac{[\text{benzoate}]}{[\text{benzoic acid}]}$$

$$\log\frac{[\text{benzoate}]}{[\text{benzoic acid}]} = 0.3 \Rightarrow \frac{[\text{benzoate}]}{[\text{benzoic acid}]} = 2$$

In 300 mL buffer: let volume of benzoic acid = $$V$$ mL (1M), then volume of sodium benzoate = $$(300 - V)$$ mL (1M).

$$\frac{300 - V}{V} = 2 \Rightarrow 300 - V = 2V \Rightarrow 300 = 3V \Rightarrow V = 100$$ mL.

Therefore, the answer is $$\boxed{100}$$.

The pH of a salt solution of a weak acid and a weak base is independent of its concentration and is calculated using the formula:

pH=7+$$\ \frac{\ 1}{2}$$(pKa​−pKb​)

Given values:

• Ka​=$$1.8×10^{−5}$$⟹pKa​=−log($$1.8×10^{−5}$$)
• Kb​=$$1.8×10^{−5}$$⟹pKb​=−log($$1.8×10^{−5}$$)

Since Ka​=Kb​, it follows that:

pKa​=pKb​

Substitute these into the pH equation:

pH=7+$$\ \frac{\ 1}{2}\times\ $$(0)pH=7

We need to find the equilibrium constant for the formation of $$NH_3$$ from $$N_2$$ and $$H_2$$. The balanced equation for this reaction is $$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$.

At equilibrium the concentrations are $$[N_2] = 2 \times 10^{-2}$$ M, $$[H_2] = 3 \times 10^{-2}$$ M, and $$[NH_3] = 1.5 \times 10^{-2}$$ M. The expression for the equilibrium constant $$K_c$$ is $$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$$, so substituting these values gives $$K_c = \frac{(1.5 \times 10^{-2})^2}{(2 \times 10^{-2})(3 \times 10^{-2})^3}$$.

Since $$(1.5 \times 10^{-2})^2 = 2.25 \times 10^{-4}$$ and $$(3 \times 10^{-2})^3 = 27 \times 10^{-6} = 2.7 \times 10^{-5}$$, multiplying the denominator yields $$(2 \times 10^{-2})(2.7 \times 10^{-5}) = 5.4 \times 10^{-7}$$. Therefore, $$K_c = \frac{2.25 \times 10^{-4}}{5.4 \times 10^{-7}} = \frac{2.25}{5.4} \times 10^{3} = 0.4167 \times 10^{3} \approx 417$$.

Therefore, the equilibrium constant is 417.

We need to find the pH at which $$Mg(OH)_2$$ begins to precipitate from a solution containing $$0.10 \text{ M } Mg^{2+}$$ ions.

The solubility product expression for $$Mg(OH)_2$$ is:

$$K_{sp} = [Mg^{2+}][OH^-]^2$$

Precipitation begins when the ionic product just equals the solubility product.

Substituting the known values:

$$1 \times 10^{-11} = (0.10)[OH^-]^2$$

Solving for $$[OH^-]$$:

$$[OH^-]^2 = \frac{1 \times 10^{-11}}{0.10} = 1 \times 10^{-10}$$

$$[OH^-] = \sqrt{1 \times 10^{-10}} = 1 \times 10^{-5} \text{ M}$$

Finding pOH:

$$pOH = -\log[OH^-] = -\log(10^{-5}) = 5$$

Finding pH:

$$pH = 14 - pOH = 14 - 5 = 9$$

The answer is pH = 9.

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).

Given: 10.0 mL of 0.01 M $$Ba(OH)_2$$ is to be neutralized by 0.02 M aqueous HBr.

The balanced neutralization reaction is:

$$Ba(OH)_2 + 2HBr \rightarrow BaBr_2 + 2H_2O$$

Moles of $$Ba(OH)_2$$:

$$n_{Ba(OH)_2} = 0.01 \times 10 = 0.1 \text{ mmol}$$

Since $$Ba(OH)_2$$ is dibasic, the moles of $$OH^-$$ ions:

$$n_{OH^-} = 2 \times 0.1 = 0.2 \text{ mmol}$$

For complete neutralization, moles of HBr required = moles of $$OH^-$$ = 0.2 mmol.

Volume of HBr solution needed:

$$V = \frac{n_{HBr}}{M_{HBr}} = \frac{0.2}{0.02} = 10.0 \text{ mL}$$

Therefore, the correct answer is Option C: $$\mathbf{10.0 \text{ mL}}$$.

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

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 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 $$233$$.

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 $$x = \boxed{10}$$.

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.

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.

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.

We have 2 L of 0.2 M $$H_2SO_4$$ and 2 L of 0.1 M NaOH solution being mixed together. We need to find the molarity of $$Na_2SO_4$$ formed in the resulting solution.

First, let us calculate the moles of each reactant.

Moles of $$H_2SO_4 = 0.2 \times 2 = 0.4$$ mol

Moles of NaOH $$= 0.1 \times 2 = 0.2$$ mol

The reaction between $$H_2SO_4$$ and NaOH is:

$$H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$$

From the stoichiometry, 1 mol of $$H_2SO_4$$ requires 2 mol of NaOH. So 0.2 mol of NaOH would react with $$\frac{0.2}{2} = 0.1$$ mol of $$H_2SO_4$$.

Since we have 0.4 mol of $$H_2SO_4$$ but only 0.2 mol of NaOH, the NaOH is the limiting reagent. All 0.2 mol of NaOH reacts with 0.1 mol of $$H_2SO_4$$, and the remaining 0.3 mol of $$H_2SO_4$$ stays unreacted.

From the stoichiometry, 0.1 mol of $$H_2SO_4$$ produces 0.1 mol of $$Na_2SO_4$$.

Now, the total volume of the resulting solution is $$2 + 2 = 4$$ L.

The molarity of $$Na_2SO_4$$ is:

$$\text{Molarity} = \frac{0.1}{4} = 0.025 \text{ M} = 25 \text{ millimolar}$$

Hence, the correct answer is 25.

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.

We have a 10 g mixture of hydrogen ($$H_2$$) and helium ($$He$$) in a vessel of capacity 0.0125 m$$^3$$ at 6 bar and 27 °C and need to find the mass of helium.

Let the mass of helium = $$x$$ g, so the mass of hydrogen = $$(10 - x)$$ g. Hence the moles of helium are $$\frac{x}{4}$$ (molar mass of He = 4 g/mol) and the moles of hydrogen are $$\frac{10 - x}{2}$$ (molar mass of $$H_2$$ = 2 g/mol).

Since we can find the total number of moles using the ideal gas equation, $$PV = nRT$$, we convert units: P = 6 bar = 6 × 10$$^5$$ Pa, V = 0.0125 m$$^3$$, T = 27 °C = 300 K, and R = 8.3 J K$$^{-1}$$ mol$$^{-1}$$ = 8.3 Pa m$$^3$$ K$$^{-1}$$ mol$$^{-1}$$.

Substituting these into $$n = \frac{PV}{RT} = \frac{6 \times 10^5 \times 0.0125}{8.3 \times 300}$$ gives a numerator of 6 × 10$$^5$$ × 0.0125 = 7500 and a denominator of 8.3 × 300 = 2490. This gives $$n = \frac{7500}{2490} = 3.012 \approx 3 \text{ mol}$$.

Therefore, the total moles satisfy $$\frac{x}{4} + \frac{10 - x}{2} = 3$$. Multiplying through by 4 yields $$x + 2(10 - x) = 12$$, which simplifies to $$x + 20 - 2x = 12$$ hence $$-x = -8$$ and $$x = 8$$.

Therefore, the mass of helium in the mixture is 8 g.

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 new pressure of the moist gas when the volume is doubled.

Initially, the total pressure of the moist gas is 4 atm at 27°C, and the vapour pressure of water at this temperature is 0.4 atm.

Subtracting the vapour pressure from the total gives the pressure of the dry gas as Total pressure - Vapour pressure of water = $$4 - 0.4 = 3.6$$ atm.

Applying Boyle’s law to the dry gas, we use $$P_1V_1 = P_2V_2$$, where $$3.6 \times V = P_2 \times 2V$$, yielding $$P_2 = \dfrac{3.6}{2} = 1.8 \text{ atm}$$.

The vapour pressure of water depends only on temperature, not on volume (as long as liquid water is present), so at 27°C it remains 0.4 atm.

In a moist gas problem, we assume water vapour remains saturated (liquid water is present), and therefore the vapour pressure stays at 0.4 atm.

The new total pressure is the sum of the pressure of dry gas and the vapour pressure of water, giving $$1.8 + 0.4 = 2.2 \text{ atm}$$.

Expressing in the required format: $$2.2 \text{ atm} = 22 \times 10^{-1} \text{ atm}$$

Hence, the answer is 22.

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 partial pressure of hydrogen in the mixture.

The mixture contains $$40\%$$ hydrogen by mass.

Let the total mass of the mixture be $$100 \text{ g}$$.

Mass of $$H_2 = 40 \text{ g}$$

Mass of $$O_2 = 100 - 40 = 60 \text{ g}$$

Molar mass of $$H_2 = 2 \text{ g/mol}$$

Molar mass of $$O_2 = 32 \text{ g/mol}$$

$$n_{H_2} = \frac{40}{2} = 20 \text{ mol}$$

$$n_{O_2} = \frac{60}{32} = 1.875 \text{ mol}$$

Total moles = $$20 + 1.875 = 21.875 \text{ mol}$$

$$\chi_{H_2} = \frac{n_{H_2}}{n_{\text{total}}} = \frac{20}{21.875}$$

Using Dalton's law of partial pressures:

$$p_{H_2} = \chi_{H_2} \times P_{\text{total}}$$

$$= \frac{20}{21.875} \times 2.2$$

$$= \frac{20 \times 2.2}{21.875}$$

$$= \frac{44}{21.875}$$

$$= 2.0114 \approx 2 \text{ bar}$$

Hence, the partial pressure of hydrogen is 2 bar.

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.

We need to find the normality of $$H_2SO_4$$ in the solution obtained by mixing 100 mL of 0.1 M $$H_2SO_4$$ with 50 mL of 0.1 M NaOH.

Initially, the millimoles of $$H_2SO_4$$ in 100 mL of 0.1 M $$H_2SO_4$$ is calculated as $$100 \times 0.1 = 10$$ mmol, and the millimoles of NaOH in 50 mL of 0.1 M NaOH is $$50 \times 0.1 = 5$$ mmol.

Since $$H_2SO_4$$ is a diprotic acid, it reacts with NaOH according to the equation

$$H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$$

Thus, 5 mmol of NaOH will react with $$\frac{5}{2} = 2.5$$ mmol of $$H_2SO_4$$.

After the reaction, the remaining $$H_2SO_4$$ is $$10 - 2.5 = 7.5$$ mmol.

The total volume of the mixture becomes $$100 + 50 = 150$$ mL. The remaining $$H_2SO_4$$ still has both acidic protons available, so its milliequivalents are $$7.5 \times 2 = 15$$ meq (since the n-factor of $$H_2SO_4$$ is 2).

$$\text{Normality} = \frac{15}{150} = 0.1 \text{ N} = 1 \times 10^{-1} \text{ N}$$

Hence, the answer is 1.

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).

We are given the gas-phase reaction $$2\,SO_2(g)+O_2(g)\rightarrow 2\,SO_3(g)$$ which proceeds to completion in a rigid vessel at constant temperature. Because temperature and volume do not change, the partial pressure of each gas is directly proportional to the number of moles present. Hence, we can treat the given partial pressures exactly like the initial “mole” amounts for stoichiometric calculations.

Initially we have

$$P_{SO_2}^{\,\text{initial}} = 250 \text{ mbar},\quad P_{O_2}^{\,\text{initial}} = 750 \text{ mbar},\quad P_{SO_3}^{\,\text{initial}} = 0 \text{ mbar}.$$

According to the balanced equation, the stoichiometric ratio is $$2\;:\;1\;:\;2$$ for $$SO_2 : O_2 : SO_3$$. That is, $$2$$ units of $$SO_2$$ react with $$1$$ unit of $$O_2$$ to form $$2$$ units of $$SO_3$$.

To identify the limiting reactant, we compare the available amounts in the required ratio. The reaction needs half as much $$O_2$$ as $$SO_2$$. Specifically, the amount of $$O_2$$ required for the given $$SO_2$$ is

$$P_{O_2}^{\,\text{needed}} \;=\;\frac{1}{2}\,P_{SO_2}^{\,\text{initial}} \;=\;\frac{1}{2}\,(250\,\text{ mbar}) \;=\;125\,\text{ mbar}.$$

Because the vessel actually contains $$750\,\text{ mbar}$$ of $$O_2$$, which is much larger than the $$125\,\text{ mbar}$$ required, $$SO_2$$ is the limiting reactant. Therefore the reaction will consume all of the initial $$SO_2$$.

Now we calculate how much $$O_2$$ is consumed. From the stoichiometry $$2\,SO_2 \longrightarrow 1\,O_2,$$ every $$2$$ units of $$SO_2$$ use up $$1$$ unit of $$O_2$$, so

$$P_{O_2}^{\,\text{consumed}} \;=\;\frac{1}{2}\,P_{SO_2}^{\,\text{consumed}} \;=\;\frac{1}{2}\,(250\,\text{ mbar}) \;=\;125\,\text{ mbar}.$$

The remaining $$O_2$$ pressure after completion is therefore

$$P_{O_2}^{\,\text{final}} \;=\;P_{O_2}^{\,\text{initial}} \;-\; P_{O_2}^{\,\text{consumed}} \;=\;750\,\text{ mbar}\;-\;125\,\text{ mbar} \;=\;625\,\text{ mbar}.$$

Next, we determine the pressure of $$SO_3$$ formed. The stoichiometry gives $$2\,SO_2 \longrightarrow 2\,SO_3,$$ meaning the same number of units of $$SO_3$$ are produced as the number of $$SO_2$$ units consumed. Hence,

$$P_{SO_3}^{\,\text{formed}} \;=\;P_{SO_2}^{\,\text{consumed}} \;=\;250\,\text{ mbar}.$$

Because no $$SO_3$$ was present initially, the final $$SO_3$$ pressure is simply this amount:

$$P_{SO_3}^{\,\text{final}} = 250\,\text{ mbar}.$$

Finally, we add the partial pressures of all gases remaining in the vessel to obtain the total final pressure:

$$P_{\text{total}}^{\,\text{final}} \;=\;P_{SO_2}^{\,\text{final}} + P_{O_2}^{\,\text{final}} + P_{SO_3}^{\,\text{final}}.$$

The entire initial $$SO_2$$ has reacted, so $$P_{SO_2}^{\,\text{final}} = 0$$. Substituting the values, we get

$$P_{\text{total}}^{\,\text{final}} \;=\;0 + 625\,\text{ mbar} + 250\,\text{ mbar} = 875\,\text{ mbar}.$$

Rounding to the nearest integer is unnecessary because the result is already an integer. Hence, the correct answer is Option A: $$875\,\text{mbar}$$.

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 $$6021$$.

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 $$\boxed{172}$$.

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 are given a 2 L solution that is $$0.80\;{\rm M}$$ in $$AgNO_3$$. Since every mole of $$AgNO_3$$ furnishes one mole of $$Ag^+$$, the initial moles and concentration of silver-ion are

$$n_{Ag^+}^{\,\text{initial}}=0.80\;{\rm mol\,L^{-1}}\times 2\;{\rm L}=1.6\;{\rm mol},\qquad[Ag^+]_0=0.80\;{\rm M}.$$

Ammonia forms the complex ion $$[Ag(NH_3)_2]^+$$ according to

$$Ag^+ + 2\,NH_3 \rightleftharpoons [Ag(NH_3)_2]^+.$$

The formation (stability) constant is stated to be

$$K_f = \frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}=1.0\times10^{8}.$$

We must add enough $$NH_3$$ so that the free silver-ion concentration is reduced to

$$[Ag^+]_{\text{eq}} = 5.0\times10^{-8}\;{\rm M}.$$

Because this value is many orders of magnitude smaller than the initial $$[Ag^+]$$, essentially every silver ion becomes complexed. Hence at equilibrium

$$[Ag(NH_3)_2^+]_{\text{eq}} \approx [Ag^+]_0-[Ag^+]_{\text{eq}} =0.80-5.0\times10^{-8}\approx0.80\;{\rm M}.$$

Let $$[NH_3]_{\text{free}}$$ be the concentration of ammonia that remains unbound. Each complex consumes two molecules of ammonia, so the concentration of bound ammonia is

$$[NH_3]_{\text{bound}} = 2\,[Ag(NH_3)_2^+] = 2(0.80)=1.6\;{\rm M}.$$

Now apply the formation constant:

$$K_f = \frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]_{\text{free}}^{\,2}} =\frac{0.80}{(5.0\times10^{-8})[NH_3]_{\text{free}}^{\,2}} =1.0\times10^{8}.$$

Rearranging for $$[NH_3]_{\text{free}}^{\,2}$$ gives

$$[NH_3]_{\text{free}}^{\,2} =\frac{0.80}{(1.0\times10^{8})(5.0\times10^{-8})} =\frac{0.80}{5}=0.16.$$

Taking the square root:

$$[NH_3]_{\text{free}}=\sqrt{0.16}=0.40\;{\rm M}.$$

The total ammonia concentration that must be present is the sum of free and bound portions:

$$[NH_3]_{\text{total}} =[NH_3]_{\text{free}}+[NH_3]_{\text{bound}} =0.40+1.60=2.0\;{\rm M}.$$

Because the solution volume is fixed at 2 L (the problem tells us to neglect volume change), the total moles of ammonia required are

$$n_{NH_3}=([NH_3]_{\text{total}})\times V =(2.0\;{\rm mol\,L^{-1}})\times(2\;{\rm L}) =4.0\;{\rm mol}.$$

The nearest integer is 4.

Hence, the correct answer is Option 4.

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.