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

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(B) Given,$pH = 12$.Since $pH + pOH = 14$,we have $pOH = 14 - 12 = 2$.The concentration of hydroxide ions is $[OH^-] = 10^{-pOH} = 10^{-2} \ M$.The di

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$5.0 	imes 10^{-7}$

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(B) Given,$pH = 12$.Since $pH + pOH = 14$,we have $pOH = 14 - 12 = 2$.The concentration of hydroxide ions is $[OH^-] = 10^{-pOH} = 10^{-2} \ M$.The dissociation of $Ba(OH)_2$ is represented as: $Ba(OH)_2 ightleftharpoons Ba^{2+} + 2OH^-$.If the solubility of $Ba(OH)_2$ is $s$,then $[Ba^{2+}] = s$ and $[OH^-] = 2s$.From the given concentration,$2s = 10^{-2} \ M$,which implies $s = 0.5 	imes 10^{-2} = 5 	imes 10^{-3} \ M$.The solubility product expression is $K_{sp} = [Ba^{2+}][OH^-]^2$.Substituting the values,$K_{sp} = (s)(2s)^2 = 4s^3$.$K_{sp} = 4 	imes (5 	imes 10^{-3})^3 = 4 	imes 125 	imes 10^{-9} = 500 	imes 10^{-9} = 5.0 	imes 10^{-7}$.

The $pH$ of a saturated solution of $Ba(OH)_2$ is $12$. What is the value of the solubility product $(K_{sp})$ of $Ba(OH)_2$?

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