What is the molar solubility of Al(OH)_3 in 0 | vedclass

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(D) The dissociation of $Al(OH)_3$ is given by: $Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^-$$K_{sp} = [Al^{3+}][OH^-]^3$In $0.2 \ M \ NaOH$,the conce

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$3 	imes 10^{-22}$

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(D) The dissociation of $Al(OH)_3$ is given by: $Al(OH)_3 ightleftharpoons Al^{3+} + 3OH^-$$K_{sp} = [Al^{3+}][OH^-]^3$In $0.2 \ M \ NaOH$,the concentration of $OH^-$ is $0.2 \ M$ (assuming complete dissociation of $NaOH$ and neglecting the contribution from $Al(OH)_3$ due to the common ion effect).Let $s$ be the molar solubility of $Al(OH)_3$.Then $[Al^{3+}] = s$ and $[OH^-] = 0.2 + 3s \approx 0.2$.Substituting these values into the $K_{sp}$ expression:$2.4 	imes 10^{-24} = s(0.2)^3$$2.4 	imes 10^{-24} = s(0.008)$$s = \frac{2.4 	imes 10^{-24}}{0.008} = 3 	imes 10^{-22} \ M$.

What is the molar solubility of $Al(OH)_3$ in $0.2 \ M \ NaOH$ solution? Given that,the solubility product of $Al(OH)_3 = 2.4 \times 10^{-24}$.

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