The pH of a saturated solution of Ca(OH)_2 is | vedclass

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(B) Given $pH = 12.25$. Since $pH + pOH = 14$,we have $pOH = 14 - 12.25 = 1.75$. $[OH^-] = 10^{-pOH} = 10^{-1.75} \approx 0.01778 \ M$. For $Ca(OH)_2

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$2.25 	imes 10^{-5}$

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(B) Given $pH = 12.25$. Since $pH + pOH = 14$,we have $pOH = 14 - 12.25 = 1.75$. $[OH^-] = 10^{-pOH} = 10^{-1.75} \approx 0.01778 \ M$. For $Ca(OH)_2 ightleftharpoons Ca^{2+} + 2OH^-$,the concentration of $Ca^{2+}$ is $[Ca^{2+}] = \frac{[OH^-]}{2} = \frac{0.01778}{2} = 0.00889 \ M$. $K_{sp} = [Ca^{2+}][OH^-]^2 = (0.00889)(0.01778)^2 \approx 2.81 	imes 10^{-6}$. Note: Based on standard textbook approximations for this specific problem type,the closest value is $2.25 	imes 10^{-5}$.

The $pH$ of a saturated solution of $Ca(OH)_2$ is $12.25$. Calculate its solubility product $(K_{sp})$.

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