QJEEM10Ch5481: JEE Main 2010 Chemistry - Class 11 JEEM10Ch5481 Solution: At 25°C, the solubility product of Mg(OH) 2 is 1.0 × 10 –11 .…

Question

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At 25°C, the solubility product of Mg(OH)₂ is 1.0

\times

10^–11. At which pH, will Mg²⁺ ions start precipitating in the form of Mg(OH)₂ from a solution of 0.001 M Mg²⁺ ions?

Accepted Answer

M g {(O H)}_{2} ⇌ M g^{+ +} + 2 O H^{-}

K_{s p} = [M g^{+ +}] {[O H^{-}]}^{2}

1.0 \times 10^{- 11} = 10^{- 3} \times {[O H^{-}]}^{2}

[O H^{-}] = \sqrt{\frac{10^{- 11}}{10^{- 3}}} = 10^{- 4}

∴

p O H = 4

∴

p H + p O H = 14

∴

p H = 10

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