At 25^ , the solubility product of Mg(OH)_2 is 1.0 × 10^-11 . At which pH, will Mg^2+ ions start precipitating in the form of Mg(OH)_2 from a solution - Chemistry MCQ AIEEE 2026 | ExamTest.live

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At 2 5^{\circ} C, the solubility product of Mg(OH)_{2} is 1.0 \times 1 0^{- 11} . At which pH, will Mg^{2 +} ions start precipitating in the form of Mg(OH)_{2} from a solution of 0.001 M Mg^{2 +} ions?

Solution & Step-by-step Explanation

K_{s p} = [Mg^{2 +}] [OH^{-}]^{2} $ 1.0 \times 1 0^{- 11} = (0.001) \times [OH^{-}]^{2} $ [OH^{-}]^{2} = 1 0^{- 8} [OH^{-}] = 1 0^{- 4} M pOH = - lo g (1 0^{- 4}) = 4 pH = 14 - 4 = 10

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At 2 5^{\circ} C, the solubility product of Mg(OH)_{2} is 1.0 \times 1 0^{- 11} . At which pH, will Mg^{2 +} ions start precipitating in the form of Mg(OH)_{2} from a solution of 0.001 M Mg^{2 +} ions?

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