The solubility product of $\mathrm{BaSO}_{4}$ is $1 \times 10^{-10}$ at 298 K . The solubility of $\mathrm{BaSO}_{4}$ in $0.1 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})$ solution is \_\_\_\_ $\times 10^{-9} \mathrm{~g} \mathrm{~L}^{-1}$ (nearest integer). Given : Molar mass of $\mathrm{BaSO}_{4}$ is $233 \mathrm{~g} \mathrm{~mol}^{-1}$

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Equilibrium

The solubility product of

BaSO_{4}

1 \times 1 0^{- 10}

at 298 K . The solubility of

BaSO_{4}

0.1 M K_{2} SO_{4} (aq)

solution is ____

\times 1 0^{- 9} g L^{- 1}

(nearest integer).
Given : Molar mass of

BaSO_{4}

233 g mol^{- 1}

Answer: 233

In the presence of

0.1 M

potassium sulphate the sulphate ion is already present, so by the common ion effect the solubility of barium sulphate is set by

K_{s p} = [Ba^{2 +}] [SO_{4}^{2 -}]

. Taking the solubility as

S

, the barium concentration is

S

while the sulphate concentration stays close to

0.1 M

, so

S \times 0.1 = 1 \times 1 0^{- 10}

and

S = 1 \times 1 0^{- 9} mol L^{- 1}

. Multiplying by the molar mass

233 g mo l^{- 1}

gives

233 \times 1 0^{- 9} g L^{- 1}

, so the required integer is 233.

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Integer answer