Question 3 of 6: Idealized Ni(OH)2 solubility diagram

Problem 3 — Idealized Ni(OH)2 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) $K_{sp}$ of Ni(OH)2(s); (b) which curve is Ni2+, Ni(OH)2(aq), Ni(OH)3-; (c) the shape of the real (total) solubility curve.

[Figure not reproduced: Fig. 2 — the three Ni-species curves labelled per part (b), plus the total (real) solubility curve of part (c) as the bold dashed envelope. Compare against the printed (unlabelled) graph on the exam page. See the official exam paper.]

Approach. Identify each line's controlling equilibrium from its slope (a species formed by losing/gaining $n$ OH^- ions from Ni2+ has slope $\pm n$ in $\log[\text{species}]$ vs. pH), then back out $K_{sp}$ from one point on the Ni2+ line using $K_{sp}=[\text{Ni}^{2+}][\text{OH}^-]^2$; the real (total) solubility is the sum of all dissolved species, which on a log-log plot is graphically just the UPPER envelope of the three individual lines.

1. (a) Identify the Ni2+ line and its equilibrium. Ni2+ + 2OH^- $\rightleftharpoons$ Ni(OH)2(s) governs the descending line ($K_{sp}=[\text{Ni}^{2+}][\text{OH}^-]^2$), since $\log[\text{Ni}^{2+}]=\log K_{sp}-2\log[\text{OH}^-]=\log K_{sp}-2(\text{pH}-14)$ has slope $-2$ in pH – matching the printed line's slope exactly. Reading the printed corner point (pH = 6.0, [Ni2+] = 0.1 molal): $$\log K_{sp}=\log(0.1)+2(6.0-14)=-1-16=\boxed{-17.0},\quad K_{sp}=1\times10^{-17}.$$
2. (b) Assign the other two curves. The FLAT line (constant $\approx7\times10^{-9}$ molal, independent of pH) is Ni(OH)2(aq): it is the neutral hydrolysis species in equilibrium with the solid, Ni(OH)2(s) $\rightleftharpoons$ Ni(OH)2(aq), whose equilibrium constant does not involve H⁺/OH^- at all, so its solubility is pH-independent (a horizontal line on this plot). The ASCENDING line (slope $+1$) is Ni(OH)3^-, from Ni(OH)2(s) + OH^- $\rightleftharpoons$ Ni(OH)3^-: $\log[\text{Ni(OH)}_3^-]=\log K_3+\log[\text{OH}^-]=\log K_3+(\text{pH}-14)$, slope $+1$, matching the printed line. As a check, the Ni2+ and Ni(OH)2(aq) lines should cross at $\log[\text{Ni}^{2+}]=\log(7\times10^{-9})$: solving $-17.0-2(\text{pH}-14)=\log(7\times10^{-9})$ gives $\text{pH}=\boxed{9.58}$, matching the printed crossing point (≈ pH 9.3-9.6) to within graph-reading precision.
3. (c) The real (total) solubility curve. Total dissolved Ni is the SUM of all three species, $[\text{Ni}]_T=[\text{Ni}^{2+}]+[\text{Ni(OH)}_2(\text{aq})]+[\text{Ni(OH)}_3^-]$. On a log-log plot, a sum of terms that differ by orders of magnitude is graphically indistinguishable from whichever term is LARGEST at that pH — so the real solubility curve simply traces the UPPER ENVELOPE of the three lines: it follows the steep Ni2+ line at low pH (pH < 9.58), flattens along the Ni(OH)2(aq) line through the minimum-solubility "trough" (9.58 < pH < ≈10.8, where the ascending Ni(OH)3^- line takes over), then follows the Ni(OH)3^- line at high pH. The MINIMUM total solubility of Ni(OH)2 is therefore $\boxed{\approx7\times10^{-9}\ \text{molal}}$, occurring anywhere in the flat mid-pH band — the classic amphoteric-hydroxide "solubility trough" shape (see Fig. 2, bold dashed envelope).