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

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(C) Given,$pH$ of $Ba(OH)_2 = 12$. Since $pH + pOH = 14$,we have $pOH = 14 - 12 = 2$. Therefore,$[OH^-] = 10^{-pOH} = 10^{-2} \, M$. The dissociation

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$5.00 	imes 10^{-7} \, M^3$

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(C) Given,$pH$ of $Ba(OH)_2 = 12$. Since $pH + pOH = 14$,we have $pOH = 14 - 12 = 2$. Therefore,$[OH^-] = 10^{-pOH} = 10^{-2} \, M$. The dissociation of $Ba(OH)_2$ is: $Ba(OH)_2 ightleftharpoons Ba^{2+} + 2OH^-$. From the stoichiometry,$[OH^-] = 2S$,where $S$ is the solubility of $Ba(OH)_2$. So,$S = \frac{[OH^-]}{2} = \frac{10^{-2}}{2} = 0.5 	imes 10^{-2} \, M$. The solubility product constant $K_{sp}$ is given by: $K_{sp} = [Ba^{2+}][OH^-]^2 = (S)(2S)^2 = 4S^3$. Substituting the value of $S$: $K_{sp} = 4 	imes (0.5 	imes 10^{-2})^3 = 4 	imes (0.125 	imes 10^{-6}) = 0.5 	imes 10^{-6} = 5.0 	imes 10^{-7} \, M^3$.

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

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