The molar solubility of $\mathrm{Cd}(\mathrm{OH})_{2}$ is $1.84 \times 10^{-5} \mathrm{M}$ in water. The expected solubility of $\mathrm{Cd}(\mathrm{OH})_{2}$ in a buffer solution of $\mathrm{pH}=12$ is:

(D) $2.49 \times 10^{-10} \mathrm{M}$

(A) $6.23 \times 10^{-11} \mathrm{M}$

(B) $1.84 \times 10^{-9} \mathrm{M}$

(C) $\frac{2.49}{1.84} \times 10^{-9} \mathrm{M}$

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Equilibrium

The molar solubility of

Cd (OH)_{2}

1.84 \times 1 0^{- 5} M

in water. The expected solubility of

Cd (OH)_{2}

in a buffer solution of

pH = 12

is:

Answer: (D)

From the solubility in pure water,

K_{s p} = 4 s^{3} = 4 (1.84 \times 1 0^{- 5})^{3} = 2.49 \times 1 0^{- 14}

. In a buffer fixed at

pH = 12

the hydroxide is held at

[OH X^{-}] = 1 0^{- 2}

M, so the cadmium that can dissolve is

s^{'} = \frac{K _{s p}}{[ OH X ^{-} ] ^{2}} = \frac{2.49 \times 1 0 ^{- 14}}{( 1 0 ^{- 2} ) ^{2}} = 2.49 \times 1 0^{- 10}

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option