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If pH of a saturated solution of $Ba(OH)_2$ is 12, the value of its $K_{sp}$ is
A
$5.00\times10^{-7}\ M^3$
B
$4.00\times10^{-6}\ M^3$
C
$4.00\times10^{-7}\ M^3$
D
$5.00\times10^{-6}\ M^3$
Detailed Solution
pH = 12, so pOH = 14 − 12 = 2 and $[OH^-] = 10^{-2}$ M
$Ba(OH)_2 \rightleftharpoons Ba^{2+} + 2OH^-$; if the solubility is S, then $[Ba^{2+}] = S$ and $[OH^-] = 2S$
$2S = 10^{-2}$, so $S = \frac{10^{-2}}{2} = 5\times10^{-3}$ M
$K_{sp} = [Ba^{2+}][OH^-]^2$
$K_{sp} = (5\times10^{-3})(10^{-2})^2$
$K_{sp} = 5.00\times10^{-7}\ M^3$
$Ba(OH)_2 \rightleftharpoons Ba^{2+} + 2OH^-$; if the solubility is S, then $[Ba^{2+}] = S$ and $[OH^-] = 2S$
$2S = 10^{-2}$, so $S = \frac{10^{-2}}{2} = 5\times10^{-3}$ M
$K_{sp} = [Ba^{2+}][OH^-]^2$
$K_{sp} = (5\times10^{-3})(10^{-2})^2$
$K_{sp} = 5.00\times10^{-7}\ M^3$