If the pH of a saturated solution of Ba(OH)_2 | vedclass

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(B) Given $pH = 12$,so $pOH = 14 - 12 = 2$. Therefore,$[OH^-] = 10^{-pOH} = 10^{-2} \ M$. The dissociation of $Ba(OH)_2$ is: $Ba(OH)_2 \rightleftharpo

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$5.0 	imes 10^{-7}$

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(B) Given $pH = 12$,so $pOH = 14 - 12 = 2$. Therefore,$[OH^-] = 10^{-pOH} = 10^{-2} \ M$. The dissociation of $Ba(OH)_2$ is: $Ba(OH)_2 ightleftharpoons Ba^{2+} + 2OH^-$. Since $[OH^-] = 10^{-2} \ M$,the concentration of $Ba^{2+}$ is $[Ba^{2+}] = \frac{1}{2} [OH^-] = 0.5 	imes 10^{-2} = 5.0 	imes 10^{-3} \ M$. The solubility product $K_{sp}$ is given by: $K_{sp} = [Ba^{2+}][OH^-]^2$. Substituting the values: $K_{sp} = (5.0 	imes 10^{-3}) 	imes (10^{-2})^2 = 5.0 	imes 10^{-3} 	imes 10^{-4} = 5.0 	imes 10^{-7}$.

If the $pH$ of a saturated solution of $Ba(OH)_2$ is $12$,what is the value of its solubility product constant $K_{sp}$?

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