What is the value of K_{sp} for a saturated s | vedclass

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Question

(D) For a saturated solution of $Ba(OH)_2$,the dissociation is $Ba(OH)_2 \rightleftharpoons Ba^{2+} + 2OH^-$. Given $pH = 12$,then $pOH = 14 - 12 = 2$

Vedclass · Accepted Answer

$5 	imes 10^{-7}$

Vedclass · Answer

(D) For a saturated solution of $Ba(OH)_2$,the dissociation is $Ba(OH)_2 ightleftharpoons Ba^{2+} + 2OH^-$. Given $pH = 12$,then $pOH = 14 - 12 = 2$. Therefore,$[OH^-] = 10^{-pOH} = 10^{-2} \ M$. From the stoichiometry,$[Ba^{2+}] = \frac{1}{2} [OH^-] = \frac{1}{2} 	imes 10^{-2} = 0.5 	imes 10^{-2} \ M$. The solubility product constant $K_{sp}$ is given by $K_{sp} = [Ba^{2+}][OH^-]^2$. Substituting the values: $K_{sp} = (0.5 	imes 10^{-2}) 	imes (10^{-2})^2 = 0.5 	imes 10^{-2} 	imes 10^{-4} = 0.5 	imes 10^{-6} = 5 	imes 10^{-7}$. Thus,the correct option is $D$.

What is the value of $K_{sp}$ for a saturated solution of $Ba(OH)_2$ having $pH = 12$?

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