Solubility And Complex Equilibria
The K_{sp} of Mg(OH)_2 is 1.2 \times 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?
Select the correct option:
Solution
3.46×10−5
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 \times 10^{-11}. Solving: [OH^-]^2 = 1.2 \times 10^{-11} / 10^{-2} = 1.2 \times 10^{-9}. Therefore [OH^-] = \sqrt{1.2 \times 10^{-9}} = \sqrt{12 \times 10^{-10}} = 3.46 \times 10^{-5} mol L^{-1}. Option 1.2 \times 10^{-9} is [OH^-]^2, not [OH^-]; the student forgot to take the square root. Option 1.1 \times 10^{-3} comes from computing \sqrt{K_{sp}} without incorporating [Mg^{2+}], effectively treating it as a 1:1 electrolyte. Option 1.2 \times 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 \times 10^{-5} M corresponds to pOH \approx 4.46, or pH \approx 9.54, which is reasonable for onset of Mg(OH)_2 precipitation.
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About This Question
- Subject
- chemistry
- Chapter
- equilibrium
- Topic
- solubility and complex equilibria
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
3.46×10−5
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 \times 10^{-11}. Solving: [OH^-]^2 = 1.2 \times 10^{-11} / 10^{-2} = 1.2 \times 10^{-9}. Therefore [OH^-] = \sqrt{1.2 \times 10^{-9}} = \sqrt{12 \times 10^{-10}} = 3.46 \times 10^{-5} mol L^{-1}. Option 1.2 \times 10^{-9} is [OH^-]^2, not [OH^-]; the student forgot to take the square root. Option 1.1 \times 10^{-3} comes from computing \sqrt{K_{sp}} without incorporating [Mg^{2+}], effectively treating it as a 1:1 electrolyte. Option 1.2 \times 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 \times 10^{-5} M corresponds to pOH \approx 4.46, or pH \approx 9.54, which is reasonable for onset of Mg(OH)_2 precipitation.
This hard difficulty chemistry question is from the chapter equilibrium, covering the topic of solubility and complex equilibria. It appeared in the 2025 exam.
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