The K_{sp} of Mg(OH)_2 is 1.2 times 10^{-11}. | vedclass

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(A) The dissociation of $Mg(OH)_2$ is given by: $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$.Let the solubility be $s \ mol/L$.Then,$[Mg^{

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$1.44 	imes 10^{-4} \ M$

Vedclass · Answer

(A) The dissociation of $Mg(OH)_2$ is given by: $Mg(OH)_2(s) ightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$.Let the solubility be $s \ mol/L$.Then,$[Mg^{2+}] = s$ and $[OH^-] = 2s$.The solubility product expression is $K_{sp} = [Mg^{2+}][OH^-]^2$.Substituting the values: $1.2 	imes 10^{-11} = (s)(2s)^2 = 4s^3$.$s^3 = \frac{1.2 	imes 10^{-11}}{4} = 0.3 	imes 10^{-11} = 3.0 	imes 10^{-12}$.$s = (3.0 	imes 10^{-12})^{1/3} \approx 1.442 	imes 10^{-4} \ M$.

The $K_{sp}$ of $Mg(OH)_2$ is $1.2 \times 10^{-11}$. Calculate its solubility in pure water.

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