At $25^{\circ}\mathrm{C}$, the solubility product of $\mathrm{Mg}(\mathrm{OH})_{2}$ is $1.0 \times 10^{-11}$. At which pH, will $\mathrm{Mg}^{2+}$ ions start precipitating in the form of $\mathrm{Mg}(\mathrm{OH})_{2}$ from a solution of $0.001\mathrm{M}$ $\mathrm{Mg}^{2+}$ ions?

JEE Main · 2010 · 11 Apr 2010 · Offline · Equilibrium

2 5^{\circ} C

, the solubility product of

Mg (OH)_{2}

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?

Answer: (B)

Precipitation just begins when the ionic product reaches the solubility product,

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

. With

[M g^{2 +}] = 1 0^{- 3}

[O H^{-}]^{2} = \frac{1.0 \times 1 0 ^{- 11}}{1 0 ^{- 3}} = 1 0^{- 8}

, so

[O H^{-}] = 1 0^{- 4}

M. That makes

pOH = 4

and therefore

pH = 14 - 4 = 10

, so

Mg (OH)_{2}

starts to precipitate at pH

10

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option