At what minimum pH will go into solution as and at what maximum pH, it will dissolved as ? Given: <b — Ionic Equilibrium Chemistry Question
Question
At what minimum pH will $10^{-3} \text{ M} - \text{Al(OH)}_3$ go into solution $(V = 1 \text{ L})$ as $\text{Al(OH)}_4^-$ and at what maximum pH, it will dissolved as $\text{Al}^{3+}$? Given: $\log 2 = 0.3$<br><br>$\text{Al(OH)}_4^- \rightleftharpoons \text{Al}^{3+} + 4\text{OH}^-; K_{eq} = 1.6 \times 10^{-34}$<br>$\text{Al(OH)}_3 \rightleftharpoons \text{Al}^{3+} + 3\text{OH}^-; K_{eq} = 8.0 \times 10^{-33}$
💡 Solution & Explanation
For dissolution as $\text{Al}^{3+}$: $[\text{Al}^{3+}][\text{OH}^-]^3 = 8.0 \times 10^{-33} \implies (10^{-3})[\text{OH}^-]^3 = 8.0 \times 10^{-33} \implies [\text{OH}^-] = 2.0 \times 10^{-10} \text{ M}$. Thus, maximum $\text{pH} = 14 - 9.7 = 4.3$. For dissolution as $\text{Al(OH)}_4^-$: Combining the equations gives $\text{Al(OH)}_3 + \text{OH}^- \rightleftharpoons \text{Al(OH)}_4^-$ with $K = \frac{8.0 \times 10^{-33}}{1.6 \times 10^{-34}} = 50$. We need $[\text{Al(OH)}_4^-] = 10^{-3} \text{ M}$, so $10^{-3} / [\text{OH}^-] = 50 \implies [\text{OH}^-] = 2.0 \times 10^{-5} \text{ M}$. Thus, minimum $\text{pH} = 14 - 4.7 = 9.3$. The requested values are 9.3 and 4.3, which match the values in Option C. Therefore, correct answer is C.