What is the solubility of solid zinc hydroxide at a pH of 13? Given that<br><br><br><br><br><br> — Ionic Equilibrium Chemistry Question
Question
What is the solubility of solid zinc hydroxide at a pH of 13? Given that<br><br>$\text{Zn(OH)}_2(g) \rightleftharpoons \text{Zn(OH)}_2(aq); K_1 = 10^{-6} \text{ M}$<br>$\text{Zn(OH)}_2(aq) \rightleftharpoons \text{Zn(OH)}^+ + \text{OH}^-; K_2 = 10^{-7} \text{ M}$<br>$\text{Zn(OH)}^+ \rightleftharpoons \text{Zn}^{2+} + \text{OH}^-; K_3 = 10^{-4} \text{ M}$<br>$\text{Zn(OH)}_2(aq) + \text{OH}^- \rightleftharpoons \text{Zn(OH)}_3^-; K_4 = 10^3 \text{ M}^{-1}$<br>$\text{Zn(OH)}_3^- + \text{OH}^- \rightleftharpoons \text{Zn(OH)}_4^{2-}; K_5 = 10 \text{ M}^{-1}$
💡 Solution & Explanation
At pH 13, $[\text{OH}^-] = 0.1 \text{ M}$. The total solubility $S = [\text{Zn}^{2+}] + [\text{Zn(OH)}^+] + [\text{Zn(OH)}_2(aq)] + [\text{Zn(OH)}_3^-] + [\text{Zn(OH)}_4^{2-}]$. From the constants: $[\text{Zn(OH)}_2(aq)] = 10^{-6} \text{ M}$. $[\text{Zn(OH)}_3^-] = K_4[\text{Zn(OH)}_2(aq)][\text{OH}^-] = 10^3 \times 10^{-6} \times 0.1 = 10^{-4} \text{ M}$. $[\text{Zn(OH)}_4^{2-}] = K_5[\text{Zn(OH)}_3^-][\text{OH}^-] = 10 \times 10^{-4} \times 0.1 = 10^{-4} \text{ M}$. The cationic species are negligibly small at this high pH ($[\text{Zn}^{2+}] \approx 10^{-15} \text{ M}$). Thus, the solubility is dominated by the anionic complexes: $S \approx [\text{Zn(OH)}_3^-] + [\text{Zn(OH)}_4^{2-}] = 10^{-4} + 10^{-4} = 2 \times 10^{-4} \text{ M}$. Therefore, correct answer is D.