Calculate the molar solubility of calcium hyd | vedclass

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(C) The dissociation of $Ca(OH)_2$ is given by: $Ca(OH)_2(s) ightleftharpoons Ca^{2+}(aq) + 2OH^-(aq)$.$K_{sp} = [Ca^{2+}][OH^-]^2 = 5.5 	imes 10^{

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$5.5 	imes 10^{-4}$

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(C) The dissociation of $Ca(OH)_2$ is given by: $Ca(OH)_2(s) ightleftharpoons Ca^{2+}(aq) + 2OH^-(aq)$.$K_{sp} = [Ca^{2+}][OH^-]^2 = 5.5 	imes 10^{-6}$.In $0.10 \ M \ NaOH$ solution, the concentration of $OH^-$ ions is $0.10 \ M$ due to complete dissociation of $NaOH$.Let the molar solubility of $Ca(OH)_2$ be $S$.Then $[Ca^{2+}] = S$ and $[OH^-] = (0.10 + 2S) \approx 0.10 \ M$ (since $S$ is very small).Substituting these values into the $K_{sp}$ expression:$5.5 	imes 10^{-6} = S 	imes (0.10)^2$.$5.5 	imes 10^{-6} = S 	imes 0.01$.$S = \frac{5.5 	imes 10^{-6}}{0.01} = 5.5 	imes 10^{-4} \ M$.

Calculate the molar solubility of calcium hydroxide $Ca(OH)_2$ in $0.10 \ M \ NaOH$ solution. The solubility product constant $(K_{sp})$ of calcium hydroxide is $5.5 \times 10^{-6}$.

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