The K_{sp} of M(OH)_2 is 3.2 times 10^{-11}. | vedclass

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(C) For a salt of the type $M(OH)_2$,the solubility product is given by $K_{sp} = [M^{2+}][OH^-]^2 = (S)(2S)^2 = 4S^3$.Given $K_{sp} = 3.2 	imes 10^{

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$10.60$

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(C) For a salt of the type $M(OH)_2$,the solubility product is given by $K_{sp} = [M^{2+}][OH^-]^2 = (S)(2S)^2 = 4S^3$.Given $K_{sp} = 3.2 	imes 10^{-11}$,we have $4S^3 = 3.2 	imes 10^{-11}$,which implies $S^3 = 0.8 	imes 10^{-11} = 8 	imes 10^{-12}$.Thus,$S = 2 	imes 10^{-4} \ M$.The concentration of hydroxide ions is $[OH^-] = 2S = 2 	imes (2 	imes 10^{-4}) = 4 	imes 10^{-4} \ M$.$pOH = -\log[OH^-] = -\log(4 	imes 10^{-4}) = 4 - \log 4 = 4 - 0.60 = 3.40$.Since $pH + pOH = 14$,we have $pH = 14 - 3.40 = 10.60$.

The $K_{sp}$ of $M(OH)_2$ is $3.2 \times 10^{-11}$. The $pH$ of its saturated solution in water is:

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