The $K_{sp}$ of $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$ is $1.1 \times 10^{-12}$ at 298 K. The solubility (in $\mathrm{mol} / \mathrm{L}$) of $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$ in a $0.1 \mathrm{M} \mathrm{AgNO}_{3}$ solution is

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Equilibrium

The

K_{s p}

Ag_{2} CrO_{4}

1.1 \times 1 0^{- 12}

at 298 K. The solubility (in

mol / L

) of

Ag_{2} CrO_{4}

in a

0.1 M AgNO_{3}

solution is

Answer: (B)

For

Ag_{2} CrO_{4} ⇌ 2 Ag^{+} + CrO_{4}^{2 -}

K_{s p} = [Ag^{+}]^{2} [CrO_{4}^{2 -}]

.

In

0.1 M AgNO_{3}

[Ag^{+}] = 0.1 M

(common ion effect). Let solubility of

Ag_{2} CrO_{4}

s

, then

[CrO_{4}^{2 -}] = s

K_{s p} = (0.1)^{2} \times s = 1.1 \times 1 0^{- 12}

s = \frac{1.1 \times 1 0 ^{- 12}}{0.01} = 1.1 \times 1 0^{- 10} mol/L

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · One or more correct options