The solubility product of Mg(OH)_2 is 1.0 tim | vedclass

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(D) For the precipitation of $Mg(OH)_2$,the ionic product must exceed the solubility product $(K_{sp})$.The equilibrium is: $Mg(OH)_2(s) \rightlefthar

Vedclass · Accepted Answer

$9.0$

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(D) For the precipitation of $Mg(OH)_2$,the ionic product must exceed the solubility product $(K_{sp})$.The equilibrium is: $Mg(OH)_2(s) ightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$$K_{sp} = [Mg^{2+}][OH^-]^2 = 1.0 	imes 10^{-12}$Given $[Mg^{2+}] = 0.01 \ M = 10^{-2} \ M$.Substituting the values: $10^{-2} 	imes [OH^-]^2 = 10^{-12}$$[OH^-]^2 = 10^{-10}$$[OH^-] = 10^{-5} \ M$$pOH = -\log[OH^-] = -\log(10^{-5}) = 5$Since $pH + pOH = 14$ at $25^{\circ}C$,$pH = 14 - 5 = 9$.

The solubility product of $Mg(OH)_2$ is $1.0 \times 10^{-12}$. Concentrated aqueous $NaOH$ solution is added to a $0.01 \ M$ aqueous solution of $MgCl_2$. The $pH$ at which precipitation occurs is

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