At $298 \ K$ temperature,the $K_{sp}$ of $Mg(OH)_2$ is $1.8 \times 10^{-11}$. If $0.1 \ M$ $NaOH$ solution is added to it,what is the concentration of $Mg^{2+}$ ions? Calculate its solubility in water.
(N/A) $1$. Solubility in water $(S)$: For $Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^-$,$K_{sp} = (S)(2S)^2 = 4S^3$. $S = (K_{sp}/4)^{1/3} = (1.8 \times 10^{-11} / 4)^{1/3} = (4.5 \times 10^{-12})^{1/3} \approx 1.65 \times 10^{-4} \ M$.
$2$. Solubility in $0.1 \ M$ $NaOH$: $[OH^-] = 0.1 \ M$. $K_{sp} = [Mg^{2+}][OH^-]^2$. $1.8 \times 10^{-11} = [Mg^{2+}](0.1)^2$. $[Mg^{2+}] = 1.8 \times 10^{-11} / 0.01 = 1.8 \times 10^{-9} \ M$.
$2$. Solubility in $0.1 \ M$ $NaOH$: $[OH^-] = 0.1 \ M$. $K_{sp} = [Mg^{2+}][OH^-]^2$. $1.8 \times 10^{-11} = [Mg^{2+}](0.1)^2$. $[Mg^{2+}] = 1.8 \times 10^{-11} / 0.01 = 1.8 \times 10^{-9} \ M$.
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