The solubility of Al(OH)_3 in a buffer soluti | vedclass

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(B) The solubility equilibrium for $Al(OH)_3$ is given by: $Al(OH)_3(s) \rightleftharpoons Al^{3+}(aq) + 3OH^-(aq)$.Given $pH = 8.0$,we know $pOH = 14

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$2.7 	imes 10^{-9} \ M$

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(B) The solubility equilibrium for $Al(OH)_3$ is given by: $Al(OH)_3(s) ightleftharpoons Al^{3+}(aq) + 3OH^-(aq)$.Given $pH = 8.0$,we know $pOH = 14 - pH = 14 - 8 = 6.0$.Therefore,$[OH^-] = 10^{-pOH} = 10^{-6} \ M$.The solubility product expression is $K_{sp} = [Al^{3+}][OH^-]^3$.Let the solubility of $Al(OH)_3$ be $S$. Then $[Al^{3+}] = S$.Substituting the values: $2.7 	imes 10^{-27} = S 	imes (10^{-6})^3$.$2.7 	imes 10^{-27} = S 	imes 10^{-18}$.$S = \frac{2.7 	imes 10^{-27}}{10^{-18}} = 2.7 	imes 10^{-9} \ M$.

The solubility of $Al(OH)_3$ in a buffer solution at $pH = 8.0$ is [$K_{sp}$ of $Al(OH)_3 = 2.7 \times 10^{-27}$]

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