Chapter 6: Equilibrium

The synthesis of ammonia is described by the equation $$\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$$. The expression for the equilibrium constant in terms of concentrations is

$$K_c = \dfrac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}$$

Substituting the equilibrium concentrations:

$$K_c = \dfrac{(1.2 \times 10^{-2})^2}{(1.5 \times 10^{-2})(3.0 \times 10^{-2})^3}$$

Computing the numerator: $$(1.2 \times 10^{-2})^2 = 1.44 \times 10^{-4}$$.

Computing the denominator: $$(3.0 \times 10^{-2})^3 = 2.7 \times 10^{-5}$$, so $$(1.5 \times 10^{-2})(2.7 \times 10^{-5}) = 4.05 \times 10^{-7}$$.

$$K_c = \dfrac{1.44 \times 10^{-4}}{4.05 \times 10^{-7}} = 3.56 \times 10^{2}\,\mathrm{M^{-2}}$$

For the reaction $$\mathrm{N_2(g) + O_2(g) \rightleftharpoons 2NO(g)}$$, the equilibrium constant is

$$K_c = \dfrac{[\mathrm{NO}]^2}{[\mathrm{N_2}][\mathrm{O_2}]}$$

Substituting the equilibrium concentrations:

$$K_c = \dfrac{(2.8 \times 10^{-3})^2}{(3.0 \times 10^{-3})(4.2 \times 10^{-3})}$$

$$K_c = \dfrac{7.84 \times 10^{-6}}{1.26 \times 10^{-5}} = 0.622$$

Since $$\Delta n = 0$$, $$K_c$$ is dimensionless.

For the dissociation $$\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$$, the equilibrium constant is

$$K_c = \dfrac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]}$$

Substituting the equilibrium concentrations:

$$K_c = \dfrac{(1.59)(1.59)}{1.41} = \dfrac{2.528}{1.41}$$

$$K_c = 1.79\,\mathrm{mol\,L^{-1}}$$

Let $$x$$ moles per litre of $$\mathrm{CO}$$ (and of $$\mathrm{H_2O}$$) react. The ICE table is:

The equilibrium constant expression is

$$K_c = \dfrac{[\mathrm{CO_2}][\mathrm{H_2}]}{[\mathrm{CO}][\mathrm{H_2O}]} = \dfrac{x^2}{(0.10-x)^2} = 4.24$$

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

$$\dfrac{x}{0.10-x} = \sqrt{4.24} = 2.06$$

$$x = 2.06(0.10 - x) = 0.206 - 2.06x$$

$$3.06\,x = 0.206 \;\Rightarrow\; x = 0.0673\,\mathrm{M}$$

Therefore:

$$[\mathrm{CO_2}] = [\mathrm{H_2}] = 0.067\,\mathrm{M}$$
$$[\mathrm{CO}] = [\mathrm{H_2O}] = 0.10 - 0.067 = 0.033\,\mathrm{M}$$

The relation between $$K_p$$ and $$K_c$$ for a gas-phase reaction is

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

where $$\Delta n$$ is the change in number of moles of gas. For $$\mathrm{2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g)}$$, $$\Delta n = (2+1) - 2 = 1$$.

Using $$R = 0.0831\,\mathrm{L\,bar\,K^{-1}\,mol^{-1}}$$ and $$T = 1069\,\mathrm{K}$$:

$$RT = 0.0831 \times 1069 = 88.83\,\mathrm{L\,bar\,mol^{-1}}$$

$$K_p = (3.75 \times 10^{-6}) \times (88.83)^{1} = 3.33 \times 10^{-4}\,\mathrm{bar}$$

Pure solid graphite does not appear in the expression. Let $$x$$ bar of $$\mathrm{CO_2}$$ react at equilibrium. By stoichiometry, $$2x$$ bar of $$\mathrm{CO}$$ is formed.

The equilibrium expression:

$$K_p = \dfrac{P_{\mathrm{CO}}^{2}}{P_{\mathrm{CO_2}}} = \dfrac{(2x)^2}{0.48-x} = 3.0$$

$$4x^2 = 3.0(0.48 - x) = 1.44 - 3x$$

$$4x^2 + 3x - 1.44 = 0$$

Using the quadratic formula:

$$x = \dfrac{-3 + \sqrt{9 + 4(4)(1.44)}}{2 \cdot 4} = \dfrac{-3 + \sqrt{32.04}}{8} = \dfrac{-3 + 5.66}{8} = 0.3325$$

Therefore:

$$P_{\mathrm{CO}} = 2x = 0.665\,\mathrm{bar}$$
$$P_{\mathrm{CO_2}} = 0.48 - 0.3325 = 0.1475 \approx 0.148\,\mathrm{bar}$$

Compute the reaction quotient with the current concentrations:

$$Q_c = \dfrac{[B][C]}{[A]^2} = \dfrac{(3\times 10^{-4})(3 \times 10^{-4})}{(3 \times 10^{-4})^2} = 1$$

Compare with $$K_c = 2 \times 10^{-3}$$:

$$Q_c = 1 \gg K_c = 2 \times 10^{-3}$$

Since $$Q_c > K_c$$, the reaction must proceed in the reverse direction (toward formation of $$A$$) to reach equilibrium.

Step 1: Initial concentration of $$\mathrm{N_2O_4}$$.

Molar mass of $$\mathrm{N_2O_4} = 2(14) + 4(16) = 92\,\mathrm{g\,mol^{-1}}$$.

$$n_0 = \dfrac{13.8}{92} = 0.15\,\mathrm{mol}$$, so initial $$[\mathrm{N_2O_4}]_0 = 0.15\,\mathrm{mol\,L^{-1}}$$.

Step 2: Total moles at equilibrium from total pressure.

From the ideal gas law, $$n_{\text{total}} = \dfrac{PV}{RT}$$. Using $$R = 0.0831\,\mathrm{L\,bar\,K^{-1}\,mol^{-1}}$$, $$V = 1\,\mathrm{L}$$, $$T = 400\,\mathrm{K}$$:

$$n_{\text{total}} = \dfrac{9.15 \times 1}{0.0831 \times 400} = \dfrac{9.15}{33.24} = 0.2752\,\mathrm{mol}$$

Step 3: Set up the ICE table. Let $$x$$ mol of $$\mathrm{N_2O_4}$$ dissociate.

Total moles at equilibrium: $$(0.15 - x) + 2x = 0.15 + x = 0.2752$$. Therefore $$x = 0.1252\,\mathrm{mol}$$.

Step 4: Equilibrium concentrations (V = 1 L).

$$[\mathrm{N_2O_4}] = 0.15 - 0.1252 = 0.0248\,\mathrm{M}$$
$$[\mathrm{NO_2}] = 2(0.1252) = 0.2504\,\mathrm{M}$$

Step 5: $$K_c$$.

$$K_c = \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} = \dfrac{(0.2504)^2}{0.0248} = \dfrac{0.0627}{0.0248} = 2.53\,\mathrm{mol\,L^{-1}}$$

Step 6: Partial pressures. Using $$P_i = c_i\,RT = c_i \times 33.24$$:

$$P_{\mathrm{N_2O_4}} = 0.0248 \times 33.24 = 0.824\,\mathrm{bar}$$
$$P_{\mathrm{NO_2}} = 0.2504 \times 33.24 = 8.32\,\mathrm{bar}$$

Sum $$\approx 9.15\,\mathrm{bar}$$, consistent with the given total pressure.

Step 7: $$K_p$$. With $$\Delta n = 1$$:

$$K_p = K_c (RT)^{\Delta n} = 2.53 \times 33.24 = 84.1\,\mathrm{bar}$$

Let $$x\,\mathrm{mol\,L^{-1}}$$ of $$\mathrm{PCl_5}$$ dissociate at equilibrium.

$$K_c = \dfrac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]} = \dfrac{x^2}{3.00 - x} = 1.80$$

$$x^2 + 1.80\,x - 5.40 = 0$$

Using the quadratic formula:

$$x = \dfrac{-1.80 + \sqrt{(1.80)^2 + 4(5.40)}}{2} = \dfrac{-1.80 + \sqrt{3.24 + 21.6}}{2} = \dfrac{-1.80 + 4.984}{2} = 1.592$$

Therefore the equilibrium composition is:

$$[\mathrm{PCl_5}] = 3.00 - 1.592 = 1.41\,\mathrm{M}$$
$$[\mathrm{PCl_3}] = [\mathrm{Cl_2}] = 1.59\,\mathrm{M}$$

Use the thermodynamic relation between standard Gibbs free energy and the equilibrium constant:

$$\Delta G^\ominus = -RT \ln K_c = -2.303\,RT\,\log K_c$$

Solving for $$\log K_c$$:

$$\log K_c = -\dfrac{\Delta G^\ominus}{2.303\,RT}$$

With $$\Delta G^\ominus = 13.8\,\mathrm{kJ\,mol^{-1}} = 13800\,\mathrm{J\,mol^{-1}}$$, $$R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$$, $$T = 298\,\mathrm{K}$$:

$$\log K_c = -\dfrac{13800}{2.303 \times 8.314 \times 298} = -\dfrac{13800}{5706} = -2.4189$$

$$K_c = 10^{-2.4189} = 3.81 \times 10^{-3}$$

Use $$\Delta G^\ominus = -2.303\,RT\,\log K_c$$ with $$R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$$ and $$T = 300\,\mathrm{K}$$.

$$\log K_c = \log(2 \times 10^{13}) = \log 2 + 13 = 0.301 + 13 = 13.301$$

$$\Delta G^\ominus = -2.303 \times 8.314 \times 300 \times 13.301$$

$$\Delta G^\ominus = -(5744.14)(13.301)\,\mathrm{J\,mol^{-1}}$$

$$\Delta G^\ominus \approx -76{,}400\,\mathrm{J\,mol^{-1}} = -76.4\,\mathrm{kJ\,mol^{-1}}$$

A Brönsted acid donates a proton; what remains after donation is its conjugate base. Subtract one $$\mathrm{H^+}$$ from each species:

• $$\mathrm{HF} \;\longrightarrow\; \mathrm{F^-}$$
• $$\mathrm{H_2SO_4} \;\longrightarrow\; \mathrm{HSO_4^-}$$
• $$\mathrm{HCO_3^-} \;\longrightarrow\; \mathrm{CO_3^{2-}}$$

A Brönsted base accepts a proton; the species formed after gaining $$\mathrm{H^+}$$ is its conjugate acid. Add one $$\mathrm{H^+}$$ to each:

• $$\mathrm{NH_2^-} \;\longrightarrow\; \mathrm{NH_3}$$
• $$\mathrm{NH_3} \;\longrightarrow\; \mathrm{NH_4^+}$$
• $$\mathrm{HCOO^-} \;\longrightarrow\; \mathrm{HCOOH}$$

A species that can act as both a Brönsted acid and a Brönsted base is called amphoteric (amphiprotic). The conjugate acid is obtained by adding $$\mathrm{H^+}$$ (species behaving as a base), and the conjugate base is obtained by removing $$\mathrm{H^+}$$ (species behaving as an acid).

By definition, $$\mathrm{pH} = -\log_{10}[\mathrm{H^+}]$$.

$$\mathrm{pH} = -\log(3.8 \times 10^{-3}) = -[\log 3.8 + \log 10^{-3}]$$

$$= -[0.5798 - 3] = 3 - 0.5798$$

$$\mathrm{pH} \approx 2.42$$

The soft drink is therefore acidic.

At this very low acid concentration the autoionization of water cannot be neglected, because $$[\mathrm{H^+}]$$ contributed by water ($$\approx 10^{-7}\,\mathrm{M}$$) is larger than the HCl concentration.

Let $$x$$ be the $$[\mathrm{H^+}]$$ contributed by water (then $$[\mathrm{OH^-}] = x$$ also). Including the HCl contribution, the total $$[\mathrm{H^+}] = (10^{-8} + x)$$.

Apply $$K_w = [\mathrm{H^+}][\mathrm{OH^-}] = 10^{-14}$$:

$$(10^{-8} + x)\,x = 10^{-14}$$

$$x^2 + 10^{-8}\,x - 10^{-14} = 0$$

Solving the quadratic:

$$x = \dfrac{-10^{-8} + \sqrt{10^{-16} + 4 \times 10^{-14}}}{2} = \dfrac{-10^{-8} + 2.0025 \times 10^{-7}}{2}$$

$$x = 9.51 \times 10^{-8}\,\mathrm{M}$$

Total $$[\mathrm{H^+}] = 10^{-8} + 9.51 \times 10^{-8} = 1.051 \times 10^{-7}\,\mathrm{M}$$.

$$\mathrm{pH} = -\log(1.051 \times 10^{-7}) = 7 - 0.0216 \approx 6.98$$

For the ionization $$\mathrm{HF + H_2O \rightleftharpoons H_3O^+ + F^-}$$, let $$\alpha$$ be the degree of dissociation in $$c = 0.02\,\mathrm{M}$$ HF.

$$K_a = \dfrac{c\alpha^2}{1-\alpha} = 3.2 \times 10^{-4}$$

$$0.02\,\alpha^2 = 3.2 \times 10^{-4}(1-\alpha)$$

$$\alpha^2 + 1.6 \times 10^{-2}\alpha - 1.6 \times 10^{-2} = 0$$

Quadratic formula:

$$\alpha = \dfrac{-1.6 \times 10^{-2} + \sqrt{(1.6 \times 10^{-2})^2 + 4(1.6 \times 10^{-2})}}{2}$$

$$= \dfrac{-1.6 \times 10^{-2} + \sqrt{6.426 \times 10^{-2}}}{2} = \dfrac{-0.016 + 0.2535}{2} = 0.119$$

So $$\alpha \approx 0.12$$ (about 12%).

Concentrations at equilibrium:

$$[\mathrm{H_3O^+}] = [\mathrm{F^-}] = 0.02 \times 0.12 = 2.4 \times 10^{-3}\,\mathrm{M}$$
$$[\mathrm{HF}] = 0.02(1 - 0.12) = 0.02 \times 0.88 = 1.76 \times 10^{-2}\,\mathrm{M}$$

pH:

$$\mathrm{pH} = -\log(2.4 \times 10^{-3}) = 3 - \log 2.4 = 3 - 0.380 = 2.62$$

For a monobasic acid $$\mathrm{HA \rightleftharpoons H^+ + A^-}$$, the hydrogen-ion concentration is obtained from the pH:

$$[\mathrm{H^+}] = 10^{-\mathrm{pH}} = 10^{-4.50} = 3.162 \times 10^{-5}\,\mathrm{M}$$

By stoichiometry, $$[\mathrm{A^-}] = [\mathrm{H^+}] = 3.162 \times 10^{-5}\,\mathrm{M}$$ (acid water-ionization contribution is negligible at this pH).

Since the dissociated amount is negligible compared with the initial concentration,

$$[\mathrm{HA}] \approx 0.1 - 3.162 \times 10^{-5} \approx 0.1\,\mathrm{M}$$

$$K_a = \dfrac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]} = \dfrac{(3.162 \times 10^{-5})^2}{0.1} = \dfrac{1.0 \times 10^{-9}}{0.1} = 1.0 \times 10^{-8}$$

$$\mathrm{p}K_a = -\log K_a = -\log(10^{-8}) = 8.0$$

For $$\mathrm{HOCl \rightleftharpoons H^+ + OCl^-}$$ with $$K_a = 2.5 \times 10^{-5}$$ and $$c = 0.08\,\mathrm{M}$$.

Since $$K_a$$ is small, the approximation $$1-\alpha \approx 1$$ is valid:

$$\alpha = \sqrt{\dfrac{K_a}{c}} = \sqrt{\dfrac{2.5 \times 10^{-5}}{0.08}} = \sqrt{3.125 \times 10^{-4}}$$

$$\alpha = 1.77 \times 10^{-2} \;\;\Rightarrow\;\; 1.77\%$$

Hydrogen-ion concentration:

$$[\mathrm{H^+}] = c\alpha = 0.08 \times 1.77 \times 10^{-2} = 1.414 \times 10^{-3}\,\mathrm{M}$$

$$\mathrm{pH} = -\log(1.414 \times 10^{-3}) = 3 - 0.150 = 2.85$$

Hydrazine ionizes as $$\mathrm{N_2H_4 + H_2O \rightleftharpoons N_2H_5^+ + OH^-}$$.

$$\mathrm{pOH} = 14 - \mathrm{pH} = 14 - 9.7 = 4.3$$

$$[\mathrm{OH^-}] = 10^{-4.3} = 5.01 \times 10^{-5}\,\mathrm{M}$$

By stoichiometry, $$[\mathrm{N_2H_5^+}] = [\mathrm{OH^-}] = 5.01 \times 10^{-5}\,\mathrm{M}$$ and

$$[\mathrm{N_2H_4}] \approx 0.004 - 5.01 \times 10^{-5} \approx 0.004\,\mathrm{M}$$

$$K_b = \dfrac{[\mathrm{N_2H_5^+}][\mathrm{OH^-}]}{[\mathrm{N_2H_4}]} = \dfrac{(5.01 \times 10^{-5})^2}{0.004} = \dfrac{2.51 \times 10^{-9}}{4 \times 10^{-3}}$$

$$K_b = 6.27 \times 10^{-7}$$

$$\mathrm{p}K_b = -\log(6.27 \times 10^{-7}) = 7 - 0.797 = 6.20$$

This is a basic buffer containing the weak base $$\mathrm{NH_3}$$ and its conjugate acid (salt) $$\mathrm{NH_4Cl}$$. Use the Henderson–Hasselbalch type equation for bases:

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

$$\mathrm{pOH} = 4.75 + \log \dfrac{0.2}{0.1} = 4.75 + \log 2 = 4.75 + 0.301 = 5.05$$

$$\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 5.05 = 8.95$$

From Table 6.7, $$K_b(\mathrm{NH_3}) = 1.77 \times 10^{-5}$$.

For $$\mathrm{NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-}$$ in $$c = 0.05\,\mathrm{M}$$:

$$\alpha = \sqrt{\dfrac{K_b}{c}} = \sqrt{\dfrac{1.77 \times 10^{-5}}{0.05}} = \sqrt{3.54 \times 10^{-4}}$$

$$\alpha \approx 1.88 \times 10^{-2}$$ (about 1.88%)

$$[\mathrm{OH^-}] = c\alpha = 0.05 \times 1.88 \times 10^{-2} = 9.4 \times 10^{-4}\,\mathrm{M}$$

$$\mathrm{pOH} = -\log(9.4 \times 10^{-4}) = 3.03$$

$$\mathrm{pH} = 14 - 3.03 = 10.97$$

Conjugate acid of $$\mathrm{NH_3}$$ is $$\mathrm{NH_4^+}$$. Using $$K_a \times K_b = K_w$$:

$$K_a(\mathrm{NH_4^+}) = \dfrac{K_w}{K_b} = \dfrac{1.0 \times 10^{-14}}{1.77 \times 10^{-5}} = 5.65 \times 10^{-10}$$

Part 1: pH of 0.10 M $$\mathrm{NH_3}$$.

$$[\mathrm{OH^-}] = \sqrt{K_b \cdot c} = \sqrt{(1.77 \times 10^{-5})(0.10)} = \sqrt{1.77 \times 10^{-6}} = 1.33 \times 10^{-3}\,\mathrm{M}$$

$$\mathrm{pOH} = -\log(1.33 \times 10^{-3}) = 2.876$$

$$\mathrm{pH} = 14 - 2.876 = 11.12$$

Part 2: After mixing 50.0 mL of 0.10 M $$\mathrm{NH_3}$$ with 25.0 mL of 0.10 M HCl.

Moles of $$\mathrm{NH_3}$$ = $$0.0500 \times 0.10 = 5.0 \times 10^{-3}$$ mol.

Moles of $$\mathrm{HCl}$$ = $$0.0250 \times 0.10 = 2.5 \times 10^{-3}$$ mol.

The acid neutralizes part of the base:

$$\mathrm{NH_3 + HCl \longrightarrow NH_4Cl}$$

Moles of $$\mathrm{NH_3}$$ remaining $$= 5.0 - 2.5 = 2.5 \times 10^{-3}$$ mol.
Moles of $$\mathrm{NH_4^+}$$ formed $$= 2.5 \times 10^{-3}$$ mol.

Total volume = 75.0 mL, so $$[\mathrm{NH_3}] = [\mathrm{NH_4^+}] = 2.5 \times 10^{-3}/0.075 = 0.0333\,\mathrm{M}$$.

This is a basic buffer with equal concentrations of base and salt. Using $$\mathrm{pOH} = \mathrm{p}K_b + \log([\text{salt}]/[\text{base}])$$:

$$\mathrm{p}K_b = -\log(1.77 \times 10^{-5}) = 4.752$$

$$\mathrm{pOH} = 4.752 + \log 1 = 4.752$$

$$\mathrm{pH} = 14 - 4.75 = 9.25$$

Ammonium acetate is the salt of a weak acid (acetic acid, $$K_a$$) and a weak base ($$\mathrm{NH_4OH}$$, $$K_b$$). The standard expression for the pH of such a salt solution is

$$\mathrm{pH} = 7 + \tfrac{1}{2}\,(\mathrm{p}K_a - \mathrm{p}K_b)$$

Substituting the values:

$$\mathrm{pH} = 7 + \tfrac{1}{2}(4.76 - 4.75) = 7 + 0.005 \approx 7.0$$

The solution is essentially neutral because $$\mathrm{p}K_a \approx \mathrm{p}K_b$$ (i.e. the acid and the base ionize to comparable extents and their hydrolyses cancel).

The salt dissociates as $$\mathrm{A_2X_3(s) \rightleftharpoons 2A^{3+}(aq) + 3X^{2-}(aq)}$$. If $$s$$ is the molar solubility:

$$[\mathrm{A^{3+}}] = 2s, \quad [\mathrm{X^{2-}}] = 3s$$

$$K_{sp} = [\mathrm{A^{3+}}]^2[\mathrm{X^{2-}}]^3 = (2s)^2(3s)^3 = 4s^2 \cdot 27s^5 = 108\,s^5$$

Wait — correctly $$(2s)^2 \cdot (3s)^3 = 4s^2 \cdot 27s^3 = 108\,s^5$$.

$$108\,s^5 = 1.1 \times 10^{-23}$$

$$s^5 = \dfrac{1.1 \times 10^{-23}}{108} = 1.019 \times 10^{-25}$$

Taking the fifth root:

$$\log s = \tfrac{1}{5}\log(1.019 \times 10^{-25}) = \tfrac{1}{5}(-24.992) = -4.998$$

$$s = 10^{-4.998} \approx 1.0 \times 10^{-5}\,\mathrm{mol\,L^{-1}}$$

Because the two salts have different stoichiometries, we cannot compare them directly by their $$K_{sp}$$ values; we must first calculate the molar solubility $$s$$ of each.

$$\mathrm{Ni(OH)_2 \rightleftharpoons Ni^{2+} + 2OH^-}$$

$$K_{sp} = [\mathrm{Ni^{2+}}][\mathrm{OH^-}]^2 = s(2s)^2 = 4s^3$$

$$4s^3 = 2.0 \times 10^{-15} \;\Rightarrow\; s^3 = 5.0 \times 10^{-16}$$

$$s = (5.0 \times 10^{-16})^{1/3} \approx 7.94 \times 10^{-6}\,\mathrm{M}$$

$$\mathrm{AgCN \rightleftharpoons Ag^+ + CN^-}$$

$$K_{sp} = s \cdot s = s^2 = 6 \times 10^{-17}$$

$$s = \sqrt{6 \times 10^{-17}} \approx 7.75 \times 10^{-9}\,\mathrm{M}$$

Comparing: $$s(\mathrm{Ni(OH)_2}) \gg s(\mathrm{AgCN})$$. Therefore $$\mathrm{Ni(OH)_2}$$ is the more soluble salt, even though its $$K_{sp}$$ is the larger of the two.

In 0.10 M NaOH the hydroxide ion concentration is essentially fixed by NaOH (common-ion effect); the contribution from $$\mathrm{Ni(OH)_2}$$ is negligible.

$$[\mathrm{OH^-}] \approx 0.10\,\mathrm{M}$$

If the molar solubility of $$\mathrm{Ni(OH)_2}$$ is $$s$$, then $$[\mathrm{Ni^{2+}}] = s$$. Apply the solubility-product expression:

$$K_{sp} = [\mathrm{Ni^{2+}}][\mathrm{OH^-}]^2$$

$$2.0 \times 10^{-15} = s \times (0.10)^2 = 0.01\,s$$

$$s = \dfrac{2.0 \times 10^{-15}}{1.0 \times 10^{-2}} = 2.0 \times 10^{-13}\,\mathrm{mol\,L^{-1}}$$

Solubility is greatly suppressed compared with that in pure water (about $$8 \times 10^{-6}\,\mathrm{M}$$).

For the reaction $$\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$$, the equilibrium constant is

$$K_c = \dfrac{[\mathrm{SO_3}]^2}{[\mathrm{SO_2}]^2\,[\mathrm{O_2}]}$$

Substitute the equilibrium concentrations:

$$K_c = \dfrac{(1.90)^2}{(0.60)^2(0.82)} = \dfrac{3.61}{0.36 \times 0.82} = \dfrac{3.61}{0.2952} \approx 12.23$$

The change in the number of moles of gas is $$\Delta n_g = 2 - 3 = -1$$, so $$K_c$$ carries units of $$(\text{concentration})^{-1}$$, i.e. $$\mathrm{L\,mol^{-1}}$$ (equivalently $$\mathrm{M^{-1}}$$).

$$K_c \approx 12.23\,\mathrm{L\,mol^{-1}}$$

For a gas mixture, percentage by volume equals mole-fraction percentage (Avogadro's law). So:

$$x_{\mathrm{I}} = 0.40, \quad x_{\mathrm{I_2}} = 0.60$$

Partial pressures (Dalton's law):

$$P_{\mathrm{I}} = x_{\mathrm{I}} \cdot P_{\text{total}} = 0.40 \times 10^5 = 4 \times 10^4\,\mathrm{Pa}$$
$$P_{\mathrm{I_2}} = 0.60 \times 10^5 = 6 \times 10^4\,\mathrm{Pa}$$

The equilibrium constant:

$$K_p = \dfrac{P_{\mathrm{I}}^2}{P_{\mathrm{I_2}}} = \dfrac{(4 \times 10^4)^2}{6 \times 10^4} = \dfrac{1.6 \times 10^9}{6 \times 10^4}$$

$$K_p \approx 2.67 \times 10^4\,\mathrm{Pa}$$

For any equilibrium, the equilibrium constant of the reverse reaction is the reciprocal of that of the forward reaction:

$$K_c^{\text{(rev)}} = \dfrac{1}{K_c^{\text{(fwd)}}}$$

$$K_c^{\text{(rev)}} = \dfrac{1}{6.3 \times 10^{14}} = 1.59 \times 10^{-15}$$

The activity of any species in the equilibrium expression is its concentration relative to a chosen standard state. For a pure liquid or solid, the standard state is the pure substance itself; its concentration (mass per unit volume divided by molar mass, i.e. density/M) depends only on the substance and the temperature.

For example, the concentration of pure water is approximately $$\dfrac{1000\,\mathrm{g\,L^{-1}}}{18\,\mathrm{g\,mol^{-1}}} \approx 55.5\,\mathrm{mol\,L^{-1}}$$, which is constant in any dilute aqueous solution. Similarly, the molar concentration of a pure solid is a fixed property of its density and molar mass and does not change as long as some solid is present.

Because these concentrations are constants, they are absorbed into the equilibrium constant itself. In the standard convention their activities are set equal to unity, so pure solids and pure liquids simply do not appear in the equilibrium expression.

Initial concentrations (volume 10 L):

$$[\mathrm{N_2}]_0 = \dfrac{0.482}{10} = 0.0482\,\mathrm{M}, \quad [\mathrm{O_2}]_0 = \dfrac{0.933}{10} = 0.0933\,\mathrm{M}$$

Let $$2x\,\mathrm{mol\,L^{-1}}$$ of $$\mathrm{N_2O}$$ be formed at equilibrium. Then:

Because $$K_c$$ is extraordinarily small ($$\sim 10^{-37}$$), the reaction proceeds only to a negligibly tiny extent. We may safely approximate $$0.0482 - 2x \approx 0.0482$$ and $$0.0933 - x \approx 0.0933$$.

$$K_c = \dfrac{[\mathrm{N_2O}]^2}{[\mathrm{N_2}]^2\,[\mathrm{O_2}]} = \dfrac{(2x)^2}{(0.0482)^2(0.0933)} = 2.0 \times 10^{-37}$$

$$(2x)^2 = 2.0 \times 10^{-37} \times (0.0482)^2 \times 0.0933$$

$$(2x)^2 = 2.0 \times 10^{-37} \times 2.324 \times 10^{-3} \times 0.0933 = 4.337 \times 10^{-41}$$

$$2x = \sqrt{4.337 \times 10^{-41}} \approx 6.59 \times 10^{-21}\,\mathrm{M}$$

In a 10 L vessel:

$$n(\mathrm{N_2O}) = (6.59 \times 10^{-21}) \times 10 = 6.6 \times 10^{-20}\,\mathrm{mol}$$

The amounts of $$\mathrm{N_2}$$ and $$\mathrm{O_2}$$ are essentially unchanged (0.482 mol and 0.933 mol respectively).

Stoichiometry of $$\mathrm{2NO + Br_2 \rightarrow 2NOBr}$$ tells us that producing 2 mol $$\mathrm{NOBr}$$ consumes 2 mol $$\mathrm{NO}$$ and 1 mol $$\mathrm{Br_2}$$.

Moles of $$\mathrm{NOBr}$$ at equilibrium = 0.0518.

$$\Rightarrow$$ Moles of $$\mathrm{NO}$$ consumed = 0.0518 (1:1 with $$\mathrm{NOBr}$$).
$$\Rightarrow$$ Moles of $$\mathrm{Br_2}$$ consumed = $$\tfrac{1}{2}(0.0518) = 0.0259$$.

Equilibrium amounts:

$$n(\mathrm{NO}) = 0.087 - 0.0518 = 0.0352\,\mathrm{mol}$$
$$n(\mathrm{Br_2}) = 0.0437 - 0.0259 = 0.0178\,\mathrm{mol}$$

For the reaction $$\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$$, $$\Delta n_{\text{gas}} = 2 - 3 = -1$$. Use:

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

$$K_c = K_p (RT)^{-\Delta n} = K_p \cdot (RT)^{+1}$$

With $$R = 0.0831\,\mathrm{L\,bar\,K^{-1}\,mol^{-1}}$$ and $$T = 450\,\mathrm{K}$$:

$$RT = 0.0831 \times 450 = 37.395\,\mathrm{L\,bar\,mol^{-1}}$$

$$K_c = (2.0 \times 10^{10}\,\mathrm{bar^{-1}}) \times (37.395\,\mathrm{L\,bar\,mol^{-1}}) = 7.48 \times 10^{11}\,\mathrm{L\,mol^{-1}}$$

HI consumed:

$$\Delta P_{\mathrm{HI}} = 0.2 - 0.04 = 0.16\,\mathrm{atm}$$

By stoichiometry ($$\mathrm{2HI \rightarrow H_2 + I_2}$$), the pressures of $$\mathrm{H_2}$$ and $$\mathrm{I_2}$$ formed are each half of the HI consumed:

$$P_{\mathrm{H_2}} = P_{\mathrm{I_2}} = \dfrac{0.16}{2} = 0.08\,\mathrm{atm}$$

$$K_p = \dfrac{P_{\mathrm{H_2}}\,P_{\mathrm{I_2}}}{P_{\mathrm{HI}}^2} = \dfrac{(0.08)(0.08)}{(0.04)^2} = \dfrac{0.0064}{0.0016} = 4.0$$

$$\Delta n = 0$$, so $$K_p$$ is dimensionless.

Compute concentrations (V = 20 L):

$$[\mathrm{N_2}] = 1.57/20 = 0.0785\,\mathrm{M}$$
$$[\mathrm{H_2}] = 1.92/20 = 0.0960\,\mathrm{M}$$
$$[\mathrm{NH_3}] = 8.13/20 = 0.4065\,\mathrm{M}$$

Reaction quotient:

$$Q_c = \dfrac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3} = \dfrac{(0.4065)^2}{(0.0785)(0.0960)^3}$$

$$(0.0960)^3 = 8.85 \times 10^{-4}$$
Denominator: $$0.0785 \times 8.85 \times 10^{-4} = 6.95 \times 10^{-5}$$
Numerator: $$(0.4065)^2 = 0.1652$$

$$Q_c = \dfrac{0.1652}{6.95 \times 10^{-5}} \approx 2.38 \times 10^3$$

Compare with $$K_c = 1.7 \times 10^2$$: $$Q_c \gg K_c$$. The mixture is not at equilibrium; to reduce $$Q_c$$, products must decrease, so the reaction proceeds in the reverse direction (toward formation of $$\mathrm{N_2}$$ and $$\mathrm{H_2}$$).

Species in the numerator of $$K_c$$ are products; species in the denominator are reactants. The exponents are the stoichiometric coefficients in the balanced equation.

Products: $$\mathrm{NH_3}$$ (coeff. 4) and $$\mathrm{O_2}$$ (coeff. 5).
Reactants: $$\mathrm{NO}$$ (coeff. 4) and $$\mathrm{H_2O}$$ (coeff. 6).

The balanced equation is therefore

$$\mathrm{4NO(g) + 6H_2O(g) \rightleftharpoons 4NH_3(g) + 5O_2(g)}$$

Check atom balance: N: 4 = 4 ✓; O: 4 + 6 = 5×2 = 10 ✓; H: 12 = 4×3 = 12 ✓.

Initial moles: $$n_0(\mathrm{H_2O}) = n_0(\mathrm{CO}) = 1$$.

40% of water reacts, i.e. 0.4 mol $$\mathrm{H_2O}$$ is consumed. By stoichiometry the same number of moles of $$\mathrm{CO}$$ react and 0.4 mol each of $$\mathrm{H_2}$$ and $$\mathrm{CO_2}$$ are formed.

Since $$\Delta n_{\text{gas}} = 0$$, the volume cancels in the $$K_c$$ expression and we can use mole ratios directly:

$$K_c = \dfrac{[\mathrm{H_2}][\mathrm{CO_2}]}{[\mathrm{H_2O}][\mathrm{CO}]} = \dfrac{(0.4)(0.4)}{(0.6)(0.6)} = \dfrac{0.16}{0.36}$$

$$K_c = 0.444 \approx 4/9$$

For $$\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$$, $$K_c = 54.8 = \dfrac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}$$.

Starting from pure HI, by stoichiometry the system produces equal amounts of $$\mathrm{H_2}$$ and $$\mathrm{I_2}$$. Let $$[\mathrm{H_2}] = [\mathrm{I_2}] = y$$ at equilibrium.

$$54.8 = \dfrac{(0.5)^2}{y^2} = \dfrac{0.25}{y^2}$$

$$y^2 = \dfrac{0.25}{54.8} = 4.56 \times 10^{-3}$$

$$y = 6.75 \times 10^{-2}\,\mathrm{mol\,L^{-1}}$$

Hence $$[\mathrm{H_2}] = [\mathrm{I_2}] \approx 6.75 \times 10^{-2}\,\mathrm{M}$$.

Let $$x\,\mathrm{mol\,L^{-1}}$$ of $$\mathrm{I_2}$$ (and of $$\mathrm{Cl_2}$$) form at equilibrium. Then $$2x\,\mathrm{mol\,L^{-1}}$$ of $$\mathrm{ICl}$$ is consumed.

$$K_c = \dfrac{x^2}{(0.78 - 2x)^2} = 0.14$$

Take square roots:

$$\dfrac{x}{0.78 - 2x} = \sqrt{0.14} = 0.3742$$

$$x = 0.3742(0.78 - 2x) = 0.2919 - 0.7483\,x$$

$$1.7483\,x = 0.2919 \;\Rightarrow\; x = 0.167$$

Equilibrium concentrations:

$$[\mathrm{I_2}] = [\mathrm{Cl_2}] = 0.167\,\mathrm{M}$$
$$[\mathrm{ICl}] = 0.78 - 2(0.167) = 0.446\,\mathrm{M}$$

Let $$p$$ atm of $$\mathrm{C_2H_6}$$ dissociate. Then $$P_{\mathrm{C_2H_4}} = P_{\mathrm{H_2}} = p$$ and $$P_{\mathrm{C_2H_6}} = 4.0 - p$$.

$$K_p = \dfrac{P_{\mathrm{C_2H_4}}\,P_{\mathrm{H_2}}}{P_{\mathrm{C_2H_6}}} = \dfrac{p^2}{4.0 - p} = 0.04$$

$$p^2 + 0.04\,p - 0.16 = 0$$

$$p = \dfrac{-0.04 + \sqrt{(0.04)^2 + 4(0.16)}}{2} = \dfrac{-0.04 + \sqrt{0.6416}}{2} = \dfrac{-0.04 + 0.801}{2} = 0.381\,\mathrm{atm}$$

$$P_{\mathrm{C_2H_6}} = 4.0 - 0.381 = 3.62\,\mathrm{atm}$$

Convert to molar concentration using $$c = P/RT$$, with $$R = 0.0821\,\mathrm{L\,atm\,K^{-1}\,mol^{-1}}$$, $$T = 899\,\mathrm{K}$$:

$$RT = 0.0821 \times 899 = 73.81\,\mathrm{L\,atm\,mol^{-1}}$$

$$[\mathrm{C_2H_6}] = \dfrac{3.62}{73.81} \approx 4.9 \times 10^{-2}\,\mathrm{mol\,L^{-1}}$$

Because the vessel started with pure $$\mathrm{PCl_5}$$, by stoichiometry $$[\mathrm{PCl_3}] = [\mathrm{Cl_2}] = x$$ at equilibrium.

$$K_c = \dfrac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]} = \dfrac{x^2}{0.05}$$

$$x^2 = K_c \times [\mathrm{PCl_5}] = (8.3 \times 10^{-3})(0.05) = 4.15 \times 10^{-4}$$

$$x = \sqrt{4.15 \times 10^{-4}} = 2.04 \times 10^{-2}\,\mathrm{mol\,L^{-1}}$$

Therefore $$[\mathrm{PCl_3}] = [\mathrm{Cl_2}] \approx 2.04 \times 10^{-2}\,\mathrm{M}$$.

Solids drop out, so $$K_p = \dfrac{P_{\mathrm{CO_2}}}{P_{\mathrm{CO}}}$$ (note: this is actually dimensionless because $$\Delta n_{\text{gas}} = 0$$).

Compute the initial quotient:

$$Q_p = \dfrac{0.80}{1.4} \approx 0.571$$

Since $$Q_p (0.571) > K_p (0.265)$$, the reaction shifts in the reverse direction (CO_2 → CO). Let the partial pressure of $$\mathrm{CO_2}$$ decrease by $$p$$ atm:

$$K_p = \dfrac{0.80 - p}{1.4 + p} = 0.265$$

$$0.80 - p = 0.265(1.4 + p) = 0.371 + 0.265\,p$$

$$0.429 = 1.265\,p \;\Rightarrow\; p = 0.339\,\mathrm{atm}$$

Equilibrium pressures:

$$P_{\mathrm{CO}} = 1.4 + 0.339 = 1.739\,\mathrm{atm}$$
$$P_{\mathrm{CO_2}} = 0.80 - 0.339 = 0.461\,\mathrm{atm}$$

Compute the reaction quotient:

$$Q_c = \dfrac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3} = \dfrac{(0.5)^2}{(3.0)(2.0)^3} = \dfrac{0.25}{(3.0)(8.0)} = \dfrac{0.25}{24} \approx 1.04 \times 10^{-2}$$

Compare with $$K_c = 0.061$$.

$$Q_c (1.04 \times 10^{-2}) < K_c (0.061)$$, so the system is not at equilibrium. To increase $$Q_c$$, the products must form, so the reaction proceeds in the forward direction (toward $$\mathrm{NH_3}$$).

Let $$x\,\mathrm{mol\,L^{-1}}$$ of $$\mathrm{Br_2}$$ (and of $$\mathrm{Cl_2}$$) form, with $$2x$$ of $$\mathrm{BrCl}$$ consumed.

$$K_c = \dfrac{x^2}{(3.3 \times 10^{-3} - 2x)^2} = 32$$

Take square roots:

$$\dfrac{x}{3.3 \times 10^{-3} - 2x} = \sqrt{32} = 5.657$$

$$x = 5.657(3.3 \times 10^{-3} - 2x) = 1.867 \times 10^{-2} - 11.31\,x$$

$$12.31\,x = 1.867 \times 10^{-2}$$

$$x = 1.516 \times 10^{-3}\,\mathrm{M}$$

Therefore:

$$[\mathrm{BrCl}] = 3.3 \times 10^{-3} - 2(1.516 \times 10^{-3}) = 0.268 \times 10^{-3} \approx 2.68 \times 10^{-4}\,\mathrm{M}$$

Step 1: Convert mass% to mole fractions. Per 100 g of gas mixture: 90.55 g CO and 9.45 g $$\mathrm{CO_2}$$.

$$n_{\mathrm{CO}} = \dfrac{90.55}{28} = 3.234\,\mathrm{mol}$$
$$n_{\mathrm{CO_2}} = \dfrac{9.45}{44} = 0.2148\,\mathrm{mol}$$
$$n_{\text{total}} = 3.234 + 0.2148 = 3.449\,\mathrm{mol}$$

$$x_{\mathrm{CO}} = \dfrac{3.234}{3.449} = 0.9377, \quad x_{\mathrm{CO_2}} = 0.0623$$

Step 2: Partial pressures (total = 1 atm):

$$P_{\mathrm{CO}} = 0.9377\,\mathrm{atm}, \quad P_{\mathrm{CO_2}} = 0.0623\,\mathrm{atm}$$

Step 3: $$K_p$$. Solid C drops out:

$$K_p = \dfrac{P_{\mathrm{CO}}^2}{P_{\mathrm{CO_2}}} = \dfrac{(0.9377)^2}{0.0623} = \dfrac{0.8793}{0.0623} = 14.11\,\mathrm{atm}$$

Step 4: Convert to $$K_c$$. $$\Delta n_{\text{gas}} = 2 - 1 = 1$$, so $$K_p = K_c (RT)^1$$.

$$RT = 0.0821 \times 1127 = 92.53\,\mathrm{L\,atm\,mol^{-1}}$$

$$K_c = \dfrac{K_p}{RT} = \dfrac{14.11}{92.53} \approx 0.153\,\mathrm{mol\,L^{-1}}$$

(a) Standard Gibbs energy of reaction.

$$\Delta_r G^\ominus = \Delta_f G^\ominus(\mathrm{NO_2}) - \Delta_f G^\ominus(\mathrm{NO}) - \tfrac{1}{2}\Delta_f G^\ominus(\mathrm{O_2})$$

$$\Delta_r G^\ominus = 52.0 - 87.0 - \tfrac{1}{2}(0) = -35.0\,\mathrm{kJ\,mol^{-1}}$$

(b) Equilibrium constant.

$$\Delta_r G^\ominus = -2.303\,RT\,\log K$$

$$\log K = -\dfrac{\Delta_r G^\ominus}{2.303\,RT} = -\dfrac{-35000}{2.303 \times 8.314 \times 298}$$

$$\log K = \dfrac{35000}{5706} = 6.134$$

$$K = 10^{6.134} \approx 1.36 \times 10^{6}$$

Since only HBr is initially present, the reaction must proceed in the reverse direction. Consider the reverse reaction $$\mathrm{2HBr \rightleftharpoons H_2 + Br_2}$$ with equilibrium constant

$$K' = \dfrac{1}{1.6 \times 10^5} = 6.25 \times 10^{-6}$$

Let $$p$$ bar of $$\mathrm{H_2}$$ (and of $$\mathrm{Br_2}$$) form at equilibrium, with $$2p$$ bar of HBr consumed.

$$K' = \dfrac{P_{\mathrm{H_2}}P_{\mathrm{Br_2}}}{P_{\mathrm{HBr}}^2} = \dfrac{p^2}{(10 - 2p)^2} = 6.25 \times 10^{-6}$$

Take square roots:

$$\dfrac{p}{10 - 2p} = 2.5 \times 10^{-3}$$

$$p = (2.5 \times 10^{-3})(10 - 2p) = 0.025 - 0.005\,p$$

$$1.005\,p = 0.025 \;\Rightarrow\; p \approx 0.0249\,\mathrm{bar}$$

Equilibrium pressures:

$$P_{\mathrm{H_2}} = P_{\mathrm{Br_2}} \approx 2.5 \times 10^{-2}\,\mathrm{bar}$$
$$P_{\mathrm{HBr}} = 10 - 2(0.0249) \approx 9.95\,\mathrm{bar}$$

Let $$p$$ bar of $$\mathrm{H_2}$$ (and of $$\mathrm{CO_2}$$) form at equilibrium. By stoichiometry $$p$$ bar of $$\mathrm{CO}$$ and $$p$$ bar of steam are consumed.

$$K_p = \dfrac{P_{\mathrm{CO_2}}P_{\mathrm{H_2}}}{P_{\mathrm{CO}}P_{\mathrm{H_2O}}} = \dfrac{p^2}{(4 - p)^2} = 10.1$$

Take square roots:

$$\dfrac{p}{4 - p} = \sqrt{10.1} = 3.178$$

$$p = 3.178(4 - p) = 12.71 - 3.178\,p$$

$$4.178\,p = 12.71 \;\Rightarrow\; p \approx 3.04\,\mathrm{bar}$$

Therefore $$P_{\mathrm{H_2}} \approx 3.04\,\mathrm{bar}$$.

From the equilibrium expression:

$$K_c = \dfrac{[\mathrm{O_3}]^2}{[\mathrm{O_2}]^3}$$

$$[\mathrm{O_3}]^2 = K_c \times [\mathrm{O_2}]^3 = (2.0 \times 10^{-50})(1.6 \times 10^{-2})^3$$

$$(1.6 \times 10^{-2})^3 = 4.096 \times 10^{-6}$$

$$[\mathrm{O_3}]^2 = (2.0 \times 10^{-50})(4.096 \times 10^{-6}) = 8.192 \times 10^{-56}$$

$$[\mathrm{O_3}] = \sqrt{8.192 \times 10^{-56}} \approx 2.86 \times 10^{-28}\,\mathrm{mol\,L^{-1}}$$

With V = 1 L, the molar concentration equals the number of moles. Thus $$[\mathrm{CO}] = 0.30\,\mathrm{M}$$, $$[\mathrm{H_2}] = 0.10\,\mathrm{M}$$, $$[\mathrm{H_2O}] = 0.02\,\mathrm{M}$$. Let $$[\mathrm{CH_4}] = x$$.

$$K_c = \dfrac{[\mathrm{CH_4}][\mathrm{H_2O}]}{[\mathrm{CO}][\mathrm{H_2}]^3}$$

$$3.90 = \dfrac{x \times 0.02}{0.30 \times (0.10)^3} = \dfrac{0.02\,x}{0.30 \times 0.001} = \dfrac{0.02\,x}{3 \times 10^{-4}}$$

$$3.90 = 66.67\,x \;\Rightarrow\; x = \dfrac{3.90}{66.67} \approx 0.0585\,\mathrm{M}$$

Conjugate acid–base pair: two species that differ from one another by a single proton ($$\mathrm{H^+}$$). The species with the extra proton is the conjugate acid; the species without it is the conjugate base. They are always related by $$\mathrm{HA \rightleftharpoons H^+ + A^-}$$.

For each species below, the conjugate partner is obtained by adding or removing one $$\mathrm{H^+}$$:

A Lewis acid is any species that can accept an electron pair, typically owing to an empty orbital or an incomplete octet.

• $$\mathrm{H_2O}$$ — oxygen has two lone pairs and a filled octet; primarily a Lewis base, not a Lewis acid.
• $$\mathrm{BF_3}$$ — boron has only 6 valence electrons and a vacant 2p orbital. It accepts a lone pair, so it is a Lewis acid.
• $$\mathrm{H^+}$$ — a bare proton with empty 1s orbital; it readily accepts an electron pair. Lewis acid.
• $$\mathrm{NH_4^+}$$ — nitrogen has a complete octet and no lone pair to donate (and cannot easily accept another pair); not normally classified as a Lewis acid in this context.

Therefore, $$\mathrm{BF_3}$$ and $$\mathrm{H^+}$$ are Lewis acids.

Remove one $$\mathrm{H^+}$$ from each Brönsted acid to get its conjugate base:

• $$\mathrm{HF} \longrightarrow \mathrm{F^-}$$
• $$\mathrm{H_2SO_4} \longrightarrow \mathrm{HSO_4^-}$$
• $$\mathrm{HCO_3^-} \longrightarrow \mathrm{CO_3^{2-}}$$

Add one $$\mathrm{H^+}$$ to each Brönsted base:

• $$\mathrm{NH_2^-} \longrightarrow \mathrm{NH_3}$$
• $$\mathrm{NH_3} \longrightarrow \mathrm{NH_4^+}$$
• $$\mathrm{HCOO^-} \longrightarrow \mathrm{HCOOH}$$

Each species is amphoteric: it can donate or accept a proton. The conjugate acid is obtained by adding $$\mathrm{H^+}$$; the conjugate base by removing $$\mathrm{H^+}$$.

(a) $$\mathrm{OH^-}$$ — three lone pairs on O; donates an electron pair. Lewis base. Example: $$\mathrm{OH^- + H^+ \longrightarrow H_2O}$$.

(b) $$\mathrm{F^-}$$ — four lone pairs on F; donates an electron pair. Lewis base. Example: $$\mathrm{BF_3 + F^- \longrightarrow BF_4^-}$$.

(c) $$\mathrm{H^+}$$ — empty 1s orbital; accepts an electron pair. Lewis acid. Example: $$\mathrm{H^+ + :NH_3 \longrightarrow NH_4^+}$$.

(d) $$\mathrm{BCl_3}$$ — boron has an incomplete octet (vacant 2p orbital); accepts an electron pair. Lewis acid. Example: $$\mathrm{BCl_3 + :NH_3 \longrightarrow Cl_3B\!\leftarrow\!NH_3}$$.

Use the definition of pH:

$$\mathrm{pH} = -\log_{10}[\mathrm{H^+}] = -\log(3.8 \times 10^{-3})$$

$$= -[\log 3.8 + \log 10^{-3}] = -[0.5798 - 3] = 2.42$$

The drink is acidic.

By definition, $$[\mathrm{H^+}] = 10^{-\mathrm{pH}}$$.

$$[\mathrm{H^+}] = 10^{-3.76} = 10^{-4} \times 10^{0.24}$$

$$10^{0.24} \approx 1.74$$

$$[\mathrm{H^+}] \approx 1.74 \times 10^{-4}\,\mathrm{mol\,L^{-1}}$$

For any conjugate acid–base pair $$K_a \times K_b = K_w = 1.0 \times 10^{-14}$$, so $$K_b = K_w/K_a$$.

For $$\mathrm{F^-}$$ (conjugate base of HF):

$$K_b = \dfrac{10^{-14}}{6.8 \times 10^{-4}} \approx 1.47 \times 10^{-11}$$

For $$\mathrm{HCOO^-}$$ (conjugate base of HCOOH):

$$K_b = \dfrac{10^{-14}}{1.8 \times 10^{-4}} \approx 5.56 \times 10^{-11}$$

For $$\mathrm{CN^-}$$ (conjugate base of HCN):

$$K_b = \dfrac{10^{-14}}{4.8 \times 10^{-9}} \approx 2.08 \times 10^{-6}$$

Part 1: 0.05 M phenol (no common ion).

Phenol: $$\mathrm{C_6H_5OH \rightleftharpoons C_6H_5O^- + H^+}$$, $$K_a = 1.0 \times 10^{-10}$$.

If $$x = [\mathrm{C_6H_5O^-}] = [\mathrm{H^+}]$$ (small) and $$[\mathrm{C_6H_5OH}] \approx 0.05$$:

$$K_a = \dfrac{x^2}{0.05} \;\Rightarrow\; x^2 = (1.0 \times 10^{-10})(0.05) = 5 \times 10^{-12}$$

$$x = 2.236 \times 10^{-6}\,\mathrm{M}$$

So $$[\mathrm{C_6H_5O^-}] \approx 2.24 \times 10^{-6}\,\mathrm{M}$$.

Part 2: 0.05 M phenol + 0.01 M sodium phenolate.

Sodium phenolate fully dissociates to give $$[\mathrm{C_6H_5O^-}] = 0.01\,\mathrm{M}$$ (common-ion effect). Let $$\alpha$$ be the degree of ionization of phenol. Then phenol contributes $$0.05\alpha$$ to $$[\mathrm{C_6H_5O^-}]$$ (negligible compared with 0.01) and $$[\mathrm{H^+}] = 0.05\alpha$$.

$$K_a = \dfrac{(0.05\alpha)(0.01)}{0.05(1-\alpha)} \approx 0.01\,\alpha$$

$$\alpha = \dfrac{K_a}{0.01} = \dfrac{1.0 \times 10^{-10}}{1.0 \times 10^{-2}} = 1.0 \times 10^{-8}$$

Two equilibria:

$$\mathrm{H_2S \rightleftharpoons H^+ + HS^-}; \quad K_{a1} = 9.1 \times 10^{-8}$$
$$\mathrm{HS^- \rightleftharpoons H^+ + S^{2-}}; \quad K_{a2} = 1.2 \times 10^{-13}$$

(i) In 0.1 M $$\mathrm{H_2S}$$ alone. Since $$K_{a1} \gg K_{a2}$$, almost all $$\mathrm{H^+}$$ comes from the first ionization. Let $$[\mathrm{H^+}] = [\mathrm{HS^-}] = x$$:

$$K_{a1} = \dfrac{x^2}{0.1} \;\Rightarrow\; x^2 = 9.1 \times 10^{-9}$$

$$x \approx 9.54 \times 10^{-5}\,\mathrm{M}$$

So $$[\mathrm{HS^-}] \approx 9.54 \times 10^{-5}\,\mathrm{M}$$.

The second ionization is so small that $$[\mathrm{S^{2-}}]$$ is found from $$K_{a2}$$. Using $$[\mathrm{H^+}] \approx [\mathrm{HS^-}]$$ in $$K_{a2} = [\mathrm{H^+}][\mathrm{S^{2-}}]/[\mathrm{HS^-}]$$ gives

$$[\mathrm{S^{2-}}] \approx K_{a2} = 1.2 \times 10^{-13}\,\mathrm{M}$$

(ii) In 0.1 M $$\mathrm{H_2S}$$ + 0.1 M HCl (common-ion effect). HCl fixes $$[\mathrm{H^+}] \approx 0.1\,\mathrm{M}$$, suppressing the ionization of $$\mathrm{H_2S}$$.

$$K_{a1} = \dfrac{[\mathrm{H^+}][\mathrm{HS^-}]}{[\mathrm{H_2S}]} \;\Rightarrow\; [\mathrm{HS^-}] = \dfrac{K_{a1}\times[\mathrm{H_2S}]}{[\mathrm{H^+}]} = \dfrac{(9.1 \times 10^{-8})(0.1)}{0.1} = 9.1 \times 10^{-8}\,\mathrm{M}$$

So $$[\mathrm{HS^-}]$$ drops from $$9.5 \times 10^{-5}\,\mathrm{M}$$ to $$9.1 \times 10^{-8}\,\mathrm{M}$$.

$$[\mathrm{S^{2-}}] = \dfrac{K_{a2}[\mathrm{HS^-}]}{[\mathrm{H^+}]} = \dfrac{(1.2 \times 10^{-13})(9.1 \times 10^{-8})}{0.1} \approx 1.09 \times 10^{-19}\,\mathrm{M}$$

For $$\mathrm{CH_3COOH \rightleftharpoons H^+ + CH_3COO^-}$$ with $$K_a = 1.74 \times 10^{-5}$$ and $$c = 0.05\,\mathrm{M}$$, since $$K_a$$ is small the approximation $$1-\alpha \approx 1$$ is good:

$$\alpha = \sqrt{\dfrac{K_a}{c}} = \sqrt{\dfrac{1.74 \times 10^{-5}}{0.05}} = \sqrt{3.48 \times 10^{-4}}$$

$$\alpha \approx 1.87 \times 10^{-2}$$ (about 1.87%)

$$[\mathrm{CH_3COO^-}] = [\mathrm{H^+}] = c\alpha = 0.05 \times 1.87 \times 10^{-2} = 9.33 \times 10^{-4}\,\mathrm{M}$$

$$\mathrm{pH} = -\log(9.33 \times 10^{-4}) = 4 - \log 9.33 = 4 - 0.970 = 3.03$$

$$[\mathrm{H^+}] = 10^{-4.15}$$. Compute:

$$10^{-4.15} = 10^{-5} \times 10^{0.85} \approx 10^{-5} \times 7.08 = 7.08 \times 10^{-5}\,\mathrm{M}$$

By stoichiometry of a monoprotic acid $$\mathrm{HA \to H^+ + A^-}$$, $$[\mathrm{A^-}] = [\mathrm{H^+}] \approx 7.08 \times 10^{-5}\,\mathrm{M}$$.

$$[\mathrm{HA}] \approx 0.01 - 7.08 \times 10^{-5} \approx 0.01\,\mathrm{M}$$.

$$K_a = \dfrac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]} = \dfrac{(7.08 \times 10^{-5})^2}{0.01} = \dfrac{5.01 \times 10^{-9}}{1.0 \times 10^{-2}}$$

$$K_a \approx 5.01 \times 10^{-7}$$

$$\mathrm{p}K_a = -\log(5.01 \times 10^{-7}) = 7 - 0.700 = 6.30$$

$$[\mathrm{H^+}] = c\alpha = 0.1 \times 0.132 = 1.32 \times 10^{-2}\,\mathrm{M}$$

$$\mathrm{pH} = -\log(1.32 \times 10^{-2}) = 2 - \log 1.32 = 2 - 0.121 = 1.88$$

$$K_a = \dfrac{c\alpha^2}{1-\alpha} = \dfrac{0.1 \times (0.132)^2}{1 - 0.132} = \dfrac{0.1 \times 0.01742}{0.868}$$

$$K_a = \dfrac{1.742 \times 10^{-3}}{0.868} \approx 2.01 \times 10^{-3}$$

$$\mathrm{p}K_a = -\log(2.01 \times 10^{-3}) = 3 - 0.303 = 2.70$$

Codeine (denoted B) is a weak base: $$\mathrm{B + H_2O \rightleftharpoons BH^+ + OH^-}$$.

$$\mathrm{pOH} = 14 - 9.95 = 4.05$$

$$[\mathrm{OH^-}] = 10^{-4.05} = 10^{-5} \times 10^{0.95} \approx 8.91 \times 10^{-5}\,\mathrm{M}$$

By stoichiometry, $$[\mathrm{BH^+}] = [\mathrm{OH^-}] = 8.91 \times 10^{-5}\,\mathrm{M}$$ and $$[\mathrm{B}] \approx 0.005\,\mathrm{M}$$.

$$K_b = \dfrac{(8.91 \times 10^{-5})^2}{0.005} = \dfrac{7.94 \times 10^{-9}}{5 \times 10^{-3}} \approx 1.59 \times 10^{-6}$$

$$\mathrm{p}K_b = -\log(1.59 \times 10^{-6}) = 6 - 0.201 = 5.80$$

From Table 6.7, $$K_b(\mathrm{C_6H_5NH_2}) = 4.27 \times 10^{-10}$$. Using $$[\mathrm{OH^-}] = \sqrt{K_b \cdot c}$$:

$$[\mathrm{OH^-}] = \sqrt{(4.27 \times 10^{-10})(10^{-3})} = \sqrt{4.27 \times 10^{-13}}$$

$$[\mathrm{OH^-}] \approx 6.53 \times 10^{-7}\,\mathrm{M}$$

$$\mathrm{pOH} = -\log(6.53 \times 10^{-7}) \approx 6.19$$

$$\mathrm{pH} = 14 - 6.19 = 7.81$$

Degree of ionization:

$$\alpha = \dfrac{[\mathrm{OH^-}]}{c} = \dfrac{6.53 \times 10^{-7}}{10^{-3}} = 6.53 \times 10^{-4}$$

For the conjugate acid (anilinium ion, $$\mathrm{C_6H_5NH_3^+}$$):

$$K_a = \dfrac{K_w}{K_b} = \dfrac{10^{-14}}{4.27 \times 10^{-10}} \approx 2.34 \times 10^{-5}$$

$$K_a = 10^{-4.74} = 1.82 \times 10^{-5}$$.

Acetic acid alone (0.05 M). Assuming $$1-\alpha \approx 1$$:

$$\alpha = \sqrt{\dfrac{K_a}{c}} = \sqrt{\dfrac{1.82 \times 10^{-5}}{0.05}} = \sqrt{3.64 \times 10^{-4}} \approx 1.91 \times 10^{-2}$$

With 0.01 M HCl. The strong acid fixes $$[\mathrm{H^+}] \approx 0.01\,\mathrm{M}$$. Using $$K_a = [\mathrm{H^+}][\mathrm{Ac^-}]/[\mathrm{HAc}]$$ and $$[\mathrm{Ac^-}] = c\alpha'$$, $$[\mathrm{HAc}] \approx c$$:

$$K_a = \dfrac{(0.01)(c\alpha')}{c} = 0.01\,\alpha'$$

$$\alpha' = \dfrac{K_a}{0.01} = \dfrac{1.82 \times 10^{-5}}{0.01} = 1.82 \times 10^{-3}$$

With 0.1 M HCl. $$[\mathrm{H^+}] \approx 0.1\,\mathrm{M}$$:

$$\alpha'' = \dfrac{K_a}{0.1} = \dfrac{1.82 \times 10^{-5}}{0.1} = 1.82 \times 10^{-4}$$

Hence the common-ion effect of HCl suppresses the ionization of acetic acid: from $$\sim 1.9\%$$ down to $$0.18\%$$ (0.01 M HCl) and $$0.018\%$$ (0.1 M HCl).

In 0.02 M dimethylamine alone.

$$\alpha = \sqrt{\dfrac{K_b}{c}} = \sqrt{\dfrac{5.4 \times 10^{-4}}{0.02}} = \sqrt{0.027} \approx 0.164$$

So the degree of ionization is $$\approx 0.16$$ (about 16%).

With 0.1 M NaOH (common-ion effect). NaOH fixes $$[\mathrm{OH^-}] \approx 0.1\,\mathrm{M}$$. Let $$\alpha'$$ be the new degree of ionization of dimethylamine.

$$K_b = \dfrac{[\mathrm{(CH_3)_2NH_2^+}][\mathrm{OH^-}]}{[\mathrm{(CH_3)_2NH}]} = \dfrac{(c\alpha')(0.1)}{c(1-\alpha')} \approx 0.1\,\alpha'$$

$$\alpha' = \dfrac{K_b}{0.1} = \dfrac{5.4 \times 10^{-4}}{0.1} = 5.4 \times 10^{-3}$$

Percentage ionization $$= 5.4 \times 10^{-3} \times 100\% \approx 0.54\%$$.

Use $$[\mathrm{H^+}] = 10^{-\mathrm{pH}}$$ for each fluid.

Molar mass of KOH = 39 + 16 + 1 = 56 g mol⁻¹.

$$n(\mathrm{KOH}) = \dfrac{0.561}{56} \approx 1.00 \times 10^{-2}\,\mathrm{mol}$$

$$[\mathrm{KOH}] = \dfrac{1.00 \times 10^{-2}}{0.200} = 5.00 \times 10^{-2}\,\mathrm{M}$$

KOH is a strong base and dissociates completely:

$$[\mathrm{K^+}] = [\mathrm{OH^-}] = 5.00 \times 10^{-2}\,\mathrm{M}$$

$$[\mathrm{H^+}] = \dfrac{K_w}{[\mathrm{OH^-}]} = \dfrac{1.0 \times 10^{-14}}{5.0 \times 10^{-2}} = 2.0 \times 10^{-13}\,\mathrm{M}$$

$$\mathrm{pH} = -\log(2.0 \times 10^{-13}) = 13 - 0.301 = 12.70$$

Molar mass of $$\mathrm{Sr(OH)_2} = 87.6 + 2(17) = 121.6\,\mathrm{g\,mol^{-1}}$$.

Molar solubility:

$$s = \dfrac{19.23}{121.6} \approx 0.1581\,\mathrm{mol\,L^{-1}}$$

Strong base, fully dissociated:

$$[\mathrm{Sr^{2+}}] = s = 0.158\,\mathrm{M}$$
$$[\mathrm{OH^-}] = 2s = 0.316\,\mathrm{M}$$

$$\mathrm{pOH} = -\log(0.316) = -\log(3.16 \times 10^{-1}) = 1 - 0.500 = 0.500$$

$$\mathrm{pH} = 14 - 0.50 = 13.50$$

Propanoic acid alone, 0.05 M.

$$\alpha = \sqrt{\dfrac{K_a}{c}} = \sqrt{\dfrac{1.32 \times 10^{-5}}{0.05}} = \sqrt{2.64 \times 10^{-4}} \approx 1.63 \times 10^{-2}$$

$$[\mathrm{H^+}] = c\alpha = 0.05 \times 1.63 \times 10^{-2} = 8.12 \times 10^{-4}\,\mathrm{M}$$

$$\mathrm{pH} = -\log(8.12 \times 10^{-4}) = 4 - \log 8.12 = 4 - 0.910 = 3.09$$

With 0.01 M HCl (common-ion effect). $$[\mathrm{H^+}] \approx 0.01\,\mathrm{M}$$:

$$K_a = \dfrac{(0.01)(c\alpha')}{c} = 0.01\,\alpha' \;\Rightarrow\; \alpha' = \dfrac{K_a}{0.01} = \dfrac{1.32 \times 10^{-5}}{0.01} = 1.32 \times 10^{-3}$$

$$[\mathrm{H^+}] = 10^{-2.34} = 10^{-3} \times 10^{0.66} \approx 4.57 \times 10^{-3}\,\mathrm{M}$$

Degree of ionization:

$$\alpha = \dfrac{[\mathrm{H^+}]}{c} = \dfrac{4.57 \times 10^{-3}}{0.1} = 4.57 \times 10^{-2}$$ (≈ 4.57%)

Ionization constant (using the exact form):

$$K_a = \dfrac{[\mathrm{H^+}]^2}{c - [\mathrm{H^+}]} = \dfrac{(4.57 \times 10^{-3})^2}{0.1 - 4.57 \times 10^{-3}}$$

$$= \dfrac{2.088 \times 10^{-5}}{0.0954} \approx 2.19 \times 10^{-4}$$

$$\mathrm{NaNO_2}$$ is the salt of a strong base and a weak acid; the $$\mathrm{NO_2^-}$$ ion hydrolyses:

$$\mathrm{NO_2^- + H_2O \rightleftharpoons HNO_2 + OH^-}; \quad K_b = \dfrac{K_w}{K_a}$$

$$K_b = \dfrac{1.0 \times 10^{-14}}{4.5 \times 10^{-4}} \approx 2.22 \times 10^{-11}$$

Hydrolysis (small $$K_b$$): $$[\mathrm{OH^-}] = \sqrt{K_b c}$$

$$[\mathrm{OH^-}] = \sqrt{(2.22 \times 10^{-11})(0.04)} = \sqrt{8.88 \times 10^{-13}} \approx 9.42 \times 10^{-7}\,\mathrm{M}$$

$$\mathrm{pOH} = -\log(9.42 \times 10^{-7}) \approx 6.03$$

$$\mathrm{pH} = 14 - 6.03 = 7.97$$

Degree of hydrolysis:

$$h = \dfrac{[\mathrm{OH^-}]}{c} = \dfrac{9.42 \times 10^{-7}}{0.04} \approx 2.36 \times 10^{-5}$$

Pyridinium chloride is the salt of a weak base (pyridine) and a strong acid (HCl); the cation $$\mathrm{PyH^+}$$ hydrolyses:

$$\mathrm{PyH^+ + H_2O \rightleftharpoons Py + H_3O^+}; \quad K_a(\mathrm{PyH^+}) = \dfrac{K_w}{K_b(\mathrm{Py})}$$

$$[\mathrm{H^+}] = 10^{-3.44} = 10^{-4} \times 10^{0.56} \approx 3.63 \times 10^{-4}\,\mathrm{M}$$

Using $$[\mathrm{H^+}]^2 = K_a \cdot c$$ (small ionization):

$$K_a(\mathrm{PyH^+}) = \dfrac{[\mathrm{H^+}]^2}{c} = \dfrac{(3.63 \times 10^{-4})^2}{0.02} = \dfrac{1.318 \times 10^{-7}}{2 \times 10^{-2}}$$

$$K_a \approx 6.59 \times 10^{-6}$$

$$K_b(\mathrm{Py}) = \dfrac{K_w}{K_a} = \dfrac{1.0 \times 10^{-14}}{6.59 \times 10^{-6}} \approx 1.52 \times 10^{-9}$$

The nature of an aqueous salt solution depends on whether the parent acid/base is strong or weak.

0.1 M chloroacetic acid (HA). For $$\mathrm{ClCH_2COOH \rightleftharpoons H^+ + ClCH_2COO^-}$$ with $$K_a = 1.35 \times 10^{-3}$$ and $$c = 0.1\,\mathrm{M}$$. Since $$K_a$$ is not very small here, the approximation $$1-\alpha \approx 1$$ is avoided and the full quadratic is solved:

$$K_a = \dfrac{c\,\alpha^2}{1-\alpha} \;\Rightarrow\; 1.35 \times 10^{-3} = \dfrac{0.1\,\alpha^2}{1-\alpha}$$

$$0.1\,\alpha^2 = 1.35 \times 10^{-3}(1-\alpha) \;\Rightarrow\; \alpha^2 + 0.0135\,\alpha - 0.0135 = 0$$

The discriminant is $$(0.0135)^2 + 4(1)(0.0135) = 0.000182 + 0.054 = 0.054182$$, and $$\sqrt{0.054182} = 0.2328$$. Taking the positive root:

$$\alpha = \dfrac{-0.0135 + 0.2328}{2} = \dfrac{0.2193}{2} = 0.110$$

$$[\mathrm{H^+}] = c\,\alpha = 0.1 \times 0.110 = 1.10 \times 10^{-2}\,\mathrm{M}$$

$$\mathrm{pH} = -\log(1.10 \times 10^{-2}) = 2 - \log 1.10 = 2 - 0.041 = 1.96$$

0.1 M sodium chloroacetate (NaA). This salt of a strong base and a weak acid is basic, because the chloroacetate ion hydrolyses:

$$\mathrm{ClCH_2COO^- + H_2O \rightleftharpoons ClCH_2COOH + OH^-}$$

$$K_b = \dfrac{K_w}{K_a} = \dfrac{1.0 \times 10^{-14}}{1.35 \times 10^{-3}} = 7.41 \times 10^{-12}$$

Because $$K_b$$ is extremely small, $$[\mathrm{OH^-}] = \sqrt{K_b\,c} = \sqrt{(7.41 \times 10^{-12})(0.1)} = \sqrt{7.41 \times 10^{-13}} = 8.61 \times 10^{-7}\,\mathrm{M}$$

$$\mathrm{pOH} = -\log(8.61 \times 10^{-7}) = 6.06$$

$$\mathrm{pH} = 14 - 6.06 = 7.94$$

For neutral water the hydrogen-ion and hydroxide-ion concentrations are equal, $$[\mathrm{H^+}] = [\mathrm{OH^-}]$$. Substituting into the ionic product of water:

$$K_w = [\mathrm{H^+}][\mathrm{OH^-}] = [\mathrm{H^+}]^2 = 2.7 \times 10^{-14}$$

$$[\mathrm{H^+}] = \sqrt{2.7 \times 10^{-14}} = \sqrt{2.7} \times \sqrt{10^{-14}} = 1.643 \times 10^{-7}\,\mathrm{M}$$

$$\mathrm{pH} = -\log(1.643 \times 10^{-7}) = 7 - \log 1.643 = 7 - 0.216 = 6.78$$

Thus neutral water at 310 K has $$\mathrm{pH} \approx 6.78$$, which is below 7 because $$K_w$$ is larger than its value at 298 K. The water is nevertheless still neutral, since $$[\mathrm{H^+}] = [\mathrm{OH^-}]$$.

For each salt, write the dissolution equilibrium, express $$K_{sp}$$ in terms of the molar solubility $$s$$, and solve. The $$K_{sp}$$ values below are those listed in Table 6.9 (298 K).

(i) Silver chromate, $$\mathrm{Ag_2CrO_4}$$ — $$K_{sp} = 1.1 \times 10^{-12}$$. $$\mathrm{Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-}}$$, so $$[\mathrm{Ag^+}] = 2s$$ and $$[\mathrm{CrO_4^{2-}}] = s$$.

$$K_{sp} = (2s)^2(s) = 4s^3 \;\Rightarrow\; s^3 = \dfrac{1.1 \times 10^{-12}}{4} = 2.75 \times 10^{-13}$$

$$s = (2.75 \times 10^{-13})^{1/3} = 6.5 \times 10^{-5}\,\mathrm{M}$$

$$[\mathrm{Ag^+}] = 2s = 1.30 \times 10^{-4}\,\mathrm{M}, \quad [\mathrm{CrO_4^{2-}}] = s = 6.5 \times 10^{-5}\,\mathrm{M}$$

(ii) Barium chromate, $$\mathrm{BaCrO_4}$$ — $$K_{sp} = 1.2 \times 10^{-10}$$. $$\mathrm{BaCrO_4 \rightleftharpoons Ba^{2+} + CrO_4^{2-}}$$, so $$K_{sp} = (s)(s) = s^2$$.

$$s = \sqrt{1.2 \times 10^{-10}} = 1.1 \times 10^{-5}\,\mathrm{M}$$

$$[\mathrm{Ba^{2+}}] = [\mathrm{CrO_4^{2-}}] = 1.1 \times 10^{-5}\,\mathrm{M}$$

(iii) Ferric hydroxide, $$\mathrm{Fe(OH)_3}$$ — $$K_{sp} = 1.0 \times 10^{-38}$$. $$\mathrm{Fe(OH)_3 \rightleftharpoons Fe^{3+} + 3OH^-}$$, so $$[\mathrm{Fe^{3+}}] = s$$ and $$[\mathrm{OH^-}] = 3s$$.

$$K_{sp} = (s)(3s)^3 = 27s^4 \;\Rightarrow\; s^4 = \dfrac{1.0 \times 10^{-38}}{27} = 3.7 \times 10^{-40}$$

$$s = (3.7 \times 10^{-40})^{1/4} = 1.39 \times 10^{-10}\,\mathrm{M}$$

$$[\mathrm{Fe^{3+}}] = s = 1.39 \times 10^{-10}\,\mathrm{M}, \quad [\mathrm{OH^-}] = 3s = 4.17 \times 10^{-10}\,\mathrm{M}$$

(iv) Lead chloride, $$\mathrm{PbCl_2}$$ — $$K_{sp} = 1.6 \times 10^{-5}$$. $$\mathrm{PbCl_2 \rightleftharpoons Pb^{2+} + 2Cl^-}$$, so $$[\mathrm{Pb^{2+}}] = s$$ and $$[\mathrm{Cl^-}] = 2s$$.

$$K_{sp} = (s)(2s)^2 = 4s^3 \;\Rightarrow\; s^3 = \dfrac{1.6 \times 10^{-5}}{4} = 4.0 \times 10^{-6}$$

$$s = (4.0 \times 10^{-6})^{1/3} = 1.59 \times 10^{-2}\,\mathrm{M}$$

$$[\mathrm{Pb^{2+}}] = s = 1.59 \times 10^{-2}\,\mathrm{M}, \quad [\mathrm{Cl^-}] = 2s = 3.17 \times 10^{-2}\,\mathrm{M}$$

(v) Mercurous iodide, $$\mathrm{Hg_2I_2}$$ — $$K_{sp} = 4.5 \times 10^{-29}$$. $$\mathrm{Hg_2I_2 \rightleftharpoons Hg_2^{2+} + 2I^-}$$, so $$[\mathrm{Hg_2^{2+}}] = s$$ and $$[\mathrm{I^-}] = 2s$$.

$$K_{sp} = (s)(2s)^2 = 4s^3 \;\Rightarrow\; s^3 = \dfrac{4.5 \times 10^{-29}}{4} = 1.125 \times 10^{-29}$$

$$s = (1.125 \times 10^{-29})^{1/3} = 2.24 \times 10^{-10}\,\mathrm{M}$$

$$[\mathrm{Hg_2^{2+}}] = s = 2.24 \times 10^{-10}\,\mathrm{M}, \quad [\mathrm{I^-}] = 2s = 4.48 \times 10^{-10}\,\mathrm{M}$$

$$\mathrm{Ag_2CrO_4}$$: $$K_{sp} = (2s_1)^2(s_1) = 4 s_1^{3}$$.

$$s_1 = \left(\dfrac{K_{sp}}{4}\right)^{1/3} = \left(\dfrac{1.1 \times 10^{-12}}{4}\right)^{1/3} = (2.75 \times 10^{-13})^{1/3}$$

$$\log s_1 = \tfrac{1}{3}\log(2.75 \times 10^{-13}) = \tfrac{1}{3}(-12.56) = -4.187$$

$$s_1 \approx 6.5 \times 10^{-5}\,\mathrm{M}$$

$$\mathrm{AgBr}$$: $$K_{sp} = s_2^2$$.

$$s_2 = \sqrt{5.0 \times 10^{-13}} \approx 7.07 \times 10^{-7}\,\mathrm{M}$$

Ratio:

$$\dfrac{s_1}{s_2} = \dfrac{6.5 \times 10^{-5}}{7.07 \times 10^{-7}} \approx 91.9$$

So $$\mathrm{Ag_2CrO_4}$$ is roughly 92 times more soluble than AgBr.

On mixing equal volumes, each ion's concentration is halved:

$$[\mathrm{Cu^{2+}}] = \dfrac{0.002}{2} = 1.0 \times 10^{-3}\,\mathrm{M}$$
$$[\mathrm{IO_3^-}] = \dfrac{0.002}{2} = 1.0 \times 10^{-3}\,\mathrm{M}$$

Cupric iodate dissolves as $$\mathrm{Cu(IO_3)_2 \rightleftharpoons Cu^{2+} + 2IO_3^-}$$, so the ionic product is

$$Q = [\mathrm{Cu^{2+}}]\,[\mathrm{IO_3^-}]^2 = (1.0 \times 10^{-3})(1.0 \times 10^{-3})^2 = 1.0 \times 10^{-9}$$

Compare with $$K_{sp} = 7.4 \times 10^{-8}$$. Since $$Q (10^{-9}) < K_{sp} (7.4 \times 10^{-8})$$, the solution is unsaturated and no precipitation occurs.

In pure water: $$\mathrm{AgC_6H_5COO \rightleftharpoons Ag^+ + C_6H_5COO^-}$$.

Let $$s_0$$ be the molar solubility. $$K_{sp} = s_0^2$$.

$$s_0 = \sqrt{2.5 \times 10^{-13}} = 5.0 \times 10^{-7}\,\mathrm{M}$$

In buffer of pH 3.19: $$[\mathrm{H^+}] = 10^{-3.19} = 6.46 \times 10^{-4}\,\mathrm{M}$$.

The benzoate ion is partly protonated. At equilibrium

$$\dfrac{[\mathrm{C_6H_5COO^-}]}{[\mathrm{C_6H_5COOH}]} = \dfrac{K_a}{[\mathrm{H^+}]} = \dfrac{6.46 \times 10^{-5}}{6.46 \times 10^{-4}} = 0.1$$

So out of every 11 parts of dissolved benzoate-containing species, 1 part is free $$\mathrm{C_6H_5COO^-}$$ and 10 parts are $$\mathrm{C_6H_5COOH}$$. Hence if the total solubility is $$s$$ (which equals $$[\mathrm{Ag^+}]$$ by mass balance on silver):

$$[\mathrm{C_6H_5COO^-}] = \dfrac{s}{11}$$

Apply $$K_{sp}$$:

$$K_{sp} = [\mathrm{Ag^+}][\mathrm{C_6H_5COO^-}] = s \cdot \dfrac{s}{11} = \dfrac{s^2}{11}$$

$$s^2 = 11 \times K_{sp} = 11 \times 2.5 \times 10^{-13} = 2.75 \times 10^{-12}$$

$$s \approx 1.66 \times 10^{-6}\,\mathrm{M}$$

Ratio of solubilities:

$$\dfrac{s}{s_0} = \dfrac{1.66 \times 10^{-6}}{5.0 \times 10^{-7}} \approx 3.32$$

When equal volumes are mixed, each ion's concentration is halved. Let the original equimolar concentrations be $$c$$:

$$[\mathrm{Fe^{2+}}]_{\text{mix}} = [\mathrm{S^{2-}}]_{\text{mix}} = c/2$$

To prevent precipitation, the ionic product must not exceed $$K_{sp}$$:

$$[\mathrm{Fe^{2+}}][\mathrm{S^{2-}}] \le K_{sp}$$

$$\left(\dfrac{c}{2}\right)\left(\dfrac{c}{2}\right) = \dfrac{c^2}{4} \le 6.3 \times 10^{-18}$$

$$c^2 \le 4 \times 6.3 \times 10^{-18} = 2.52 \times 10^{-17}$$

$$c \le 5.02 \times 10^{-9}\,\mathrm{M}$$

So the maximum permissible initial concentration is about $$5.0 \times 10^{-9}\,\mathrm{mol\,L^{-1}}$$.

Molar solubility of $$\mathrm{CaSO_4}$$:

$$K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{SO_4^{2-}}] = s^2 \Rightarrow s = \sqrt{9.1 \times 10^{-6}} \approx 3.02 \times 10^{-3}\,\mathrm{mol\,L^{-1}}$$

Molar mass of $$\mathrm{CaSO_4} = 40 + 32 + 4(16) = 136\,\mathrm{g\,mol^{-1}}$$.

Moles in 1 g:

$$n = \dfrac{1}{136} \approx 7.35 \times 10^{-3}\,\mathrm{mol}$$

To just dissolve this amount in a saturated solution:

$$V_{\min} = \dfrac{n}{s} = \dfrac{7.35 \times 10^{-3}}{3.02 \times 10^{-3}} \approx 2.43\,\mathrm{L}$$

When the two solutions are mixed the total volume is $$10 + 5 = 15\,\mathrm{mL}$$. The concentrations in the mixture are found by dilution:

$$[\mathrm{S^{2-}}] = \dfrac{10}{15} \times (1.0 \times 10^{-19}) = 6.67 \times 10^{-20}\,\mathrm{M}$$

$$[\mathrm{M^{2+}}] = \dfrac{5}{15} \times 0.04 = 1.33 \times 10^{-2}\,\mathrm{M}$$

Each metal sulphide dissolves as $$\mathrm{MS \rightleftharpoons M^{2+} + S^{2-}}$$, so the ionic product has the same value for every cation:

$$Q_{sp} = [\mathrm{M^{2+}}][\mathrm{S^{2-}}] = (1.33 \times 10^{-2})(6.67 \times 10^{-20}) = 8.9 \times 10^{-22}$$

Precipitation takes place only when $$Q_{sp} \gt K_{sp}$$. Comparing $$Q_{sp}$$ with the solubility products from Table 6.9:

$$Q_{sp}$$ exceeds $$K_{sp}$$ only for ZnS and CdS, so precipitation occurs in the $$\mathrm{ZnCl_2}$$ solution (as ZnS) and the $$\mathrm{CdCl_2}$$ solution (as CdS). For FeS and MnS, $$Q_{sp} \lt K_{sp}$$, so no precipitate forms.