Question 4 of 6: Divalent-metal hydroxide speciation and solubility diagram

Problem 4 — Divalent-metal hydroxide speciation and solubility diagram (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) $K_1$, $K_2$, $K_3$ for the three stated reactions; (b) the solubility diagram of M(OH)₂(s), with units and values.

Approach. At a crossover point the two adjacent species are present in EQUAL amounts (fractions and absolute concentrations coincide, since both are normalized by the same total), so each equilibrium constant reduces to $1/[\text{OH}^-]$ (or a ratio of two such constants) evaluated at that crossover pH via $[\text{OH}^-]=10^{\text{pH}-14}$.

1. K1: M2+ + OH- = M(OH)+. At the crossover, $[\text{M(OH)}^+]=[\text{M}^{2+}]$, so $K_1=\dfrac{[\text{M(OH)}^+]}{[\text{M}^{2+}][\text{OH}^-]}=\dfrac{1}{[\text{OH}^-]_{7.8}}$. With $[\text{OH}^-]_{7.8}=10^{7.8-14}=10^{-6.2}$: $$K_1=10^{6.2}=\boxed{1.585\times10^{6}}$$
2. K2: M(OH)+ + OH- = M(OH)2(aq). Same logic at pH 9.5, $[\text{OH}^-]_{9.5}=10^{-4.5}$: $$K_2=10^{4.5}=\boxed{3.162\times10^{4}}$$
3. K3: M(OH)+ + 2OH- = M(OH)3-. This spans TWO steps from M(OH)⁺, so first find the stepwise constant $K_2'$ for M(OH)₂(aq) + OH⁻ = M(OH)₃⁻ the SAME way as $K_1$/$K_2$: at the (1,2)/(1,3) crossover (pH 11.0, equal fractions), $K_2'=1/[\text{OH}^-]_{11.0}$. With $[\text{OH}^-]_{11.0}=10^{11.0-14}=10^{-3}$: $$K_2'=\frac{1}{10^{-3}}=10^{3}=1000$$ $K_3$ as DEFINED in the question (2 OH⁻ added directly to M(OH)⁺) is the PRODUCT of the two consecutive steps: $$K_3=K_2\cdot K_2'=10^{4.5}\times10^{3}=10^{7.5}=\boxed{3.162\times10^{7}}$$

Part (b) — Given. $K_{sp}=1.2\times10^{-5}$ for M(OH)₂(s), plus $K_1$, $K_2$, $K_3$ from part (a).

Find. The solubility diagram (log[species] vs. pH) of M(OH)₂(s), with units and values, over pH 0-14.

Approach. Anchor $[\text{M}^{2+}]=K_{sp}/[\text{OH}^-]^2$ from $K_{sp}$ alone, then propagate through $K_1$, $K_2$, $K_2'=K_3/K_2$ to get each higher species' absolute concentration; sum for the total, and identify the classical three-branch (slope -2, flat, slope +1) amphoteric-hydroxide envelope.

1. Propagate the four species. $$[\text{M}^{2+}]=\frac{K_{sp}}{[\text{OH}^-]^2},\quad [\text{M(OH)}^+]=K_1[\text{M}^{2+}][\text{OH}^-],\quad [\text{M(OH)}_2(\text{aq})]=K_1K_2K_{sp}\ (\text{pH-independent!}),\quad [\text{M(OH)}_3^-]=K_1K_3K_{sp}[\text{OH}^-]$$ The middle species collapses to a CONSTANT ($K_1K_2K_{sp}$, no $[\text{OH}^-]$ left after substitution) — exactly the pH-independent plateau shape seen on the printed mol% diagram, and the same structural result as the Problem 2(b)-cited Ni(OH)₂ example.
2. Evaluate the plateau. $$[\text{M(OH)}_2(\text{aq})]=K_1K_2K_{sp}=(1.585\times10^{6})(3.162\times10^{4})(1.2\times10^{-5})=\boxed{6.01\times10^{5}\ \text{mol/L}}$$
3. Check the physically-meaningful window. $[\text{M}^{2+}]$ alone only drops to a realistic 1 mol/L where $[\text{OH}^-]=\sqrt{K_{sp}}=3.46\times10^{-3}$, i.e. at $\text{pH}=14-(-\log_{10}3.46\times10^{-3})=\boxed{11.54}$. Below that pH, $[\text{M}^{2+}]$ — and, propagated through $K_1K_2$, the "plateau" species too — both exceed 1 mol/L, rising to absurd (>10⁹ mol/L) values at low pH. The mathematical minimum of the total (Step 1's sum, unphysical scale aside) sits near $\text{pH}\approx10.25$, at $\approx8.16\times10^{5}$ mol/L — still nowhere near a realizable concentration.
4. Interpret. The given $K_{sp}$ describes an unusually SOLUBLE hydroxide (comparable in magnitude to Ca(OH)₂, $K_{sp}\approx5.5\times10^{-6}$) combined with very strong hydrolysis constants (log $K_1=6.2$ matches literature Pb²⁺ hydrolysis closely). For any REALISTIC total-metal loading (say ≤1 mol/L), M(OH)₂(s) therefore never actually limits solubility anywhere in the practical pH 0-11.5 window — the "solubility curve" for such a loading is simply FLAT at the total dissolved metal concentration across that whole range, only turning down (becoming genuinely $K_{sp}$-limited) above pH ≈ 11.5. This is stated explicitly as the honest engineering conclusion rather than forcing a numerically "clean" but physically meaningless minimum-solubility figure.