What is the solubility product of Al(OH) in water. Given:<br>Al(OH)(aq) Al(aq) + 4OH(aq); <br>Al(OH) — Ionic Equilibrium Chemistry Question
Question
What is the solubility product of Al(OH)$_3$ in water. Given:<br>Al(OH)$_4^-$(aq) $\rightleftharpoons$ Al$^{3+}$(aq) + 4OH$^-$(aq); $K = 1.3 \times 10^{-34}$<br>Al(OH)$_3$(s) + OH$^-$(aq) $\rightleftharpoons$ Al(OH)$_4^-$(aq); $K = 38.5$
Answer: B
💡 Solution & Explanation
The required dissolution equation is $\text{Al(OH)}_3\text{(s)} \rightleftharpoons \text{Al}^{3+}\text{(aq)} + 3\text{OH}^-\text{(aq)}$. This is obtained by adding the two given equilibrium reactions. When reactions are added, their equilibrium constants multiply. $K_{sp} = (1.3 \times 10^{-34}) \times (38.5) = 50.05 \times 10^{-34} = 5.005 \times 10^{-33} \approx 5 \times 10^{-33}$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes