Solubility And Complex Equilibria

Question

The K_{sp} of Mg(OH)_2 is 1.2 	imes 10^{-11} at 25°C. What is the minimum concentration of OH^- ions required (in mol L^{-1}) to just begin precipitating Mg^{2+} from a 0.01 M MgCl_2 solution?

RankGuru · Accepted Answer

Precipitation of Mg(OH)_2 begins when the ionic product Q = [Mg^{2+}][OH^-]^2 just exceeds K_{sp}. To find the minimum [OH^-] required to initiate precipitation, we set the ionic product equal to K_{sp}: [Mg^{2+}][OH^-]^2 = K_{sp}. Given [Mg^{2+}] = 0.01 M = 10^{-2} M (from complete dissociation of MgCl_2): (10^{-2})[OH^-]^2 = 1.2 	imes 10^{-11}. Solving: [OH^-]^2 = 1.2 	imes 10^{-11} / 10^{-2} = 1.2 	imes 10^{-9}. Therefore [OH^-] = \sqrt{1.2 	imes 10^{-9}} = \sqrt{12 	imes 10^{-10}} = 3.46 	imes 10^{-5} mol L^{-1}. Option 1.2 	imes 10^{-9} is [OH^-]^2, not [OH^-]; the student forgot to take the square root. Option 1.1 	imes 10^{-3} comes from computing \sqrt{K_{sp}} without incorporating [Mg^{2+}], effectively treating it as a 1:1 electrolyte. Option 1.2 	imes 10^{-11} is just K_{sp} itself and has wrong units for concentration. This type of selective precipitation problem is a hallmark of JEE Advanced ionic equilibrium questions requiring careful stoichiometric treatment of K_{sp}. Plausibility check: [OH^-] = 3.46 	imes 10^{-5} M corresponds to pOH \approx 4.46, or pH \approx 9.54, which is reasonable for onset of Mg(OH)_2 precipitation.

Solubility And Complex Equilibria

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