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pH of a saturated solution of $Ba(OH)_2$ is 12. The value of solubility product ($K_{sp}$) of $Ba(OH)_2$ is
A
$5.0\times10^{-6}$
B
$3.3\times10^{-7}$
C
$5.0\times10^{-7}$
D
$4.0\times10^{-6}$
Detailed Solution
pH = 12, so pOH = 2 and $[OH^-] = 10^{-2}$ M
$Ba(OH)_2 \rightleftharpoons Ba^{2+} + 2OH^-$; if solubility is x, $[Ba^{2+}] = x$ and $[OH^-] = 2x$
$2x = 10^{-2} \Rightarrow x = 5\times10^{-3}$ M
$K_{sp} = [Ba^{2+}][OH^-]^2 = x(2x)^2 = 5\times10^{-3}\times(10^{-2})^2 = 5\times10^{-7}$
$Ba(OH)_2 \rightleftharpoons Ba^{2+} + 2OH^-$; if solubility is x, $[Ba^{2+}] = x$ and $[OH^-] = 2x$
$2x = 10^{-2} \Rightarrow x = 5\times10^{-3}$ M
$K_{sp} = [Ba^{2+}][OH^-]^2 = x(2x)^2 = 5\times10^{-3}\times(10^{-2})^2 = 5\times10^{-7}$