QJEEM23Chc494: JEE Main 2023 Chemistry - Class 11 JEEM23Chc494 Solution: The solubility product of is at . The solubility of in solution …

Barium sulfate, BaSO₄, is a sparingly soluble salt. Its dissolution can be represented by the following equilibrium reaction:

$\mathrm{BaSO}_4 (\mathrm{s}) \rightleftharpoons \mathrm{Ba}^{2+}(\mathrm{aq}) + \mathrm{SO}_4^{2-}(\mathrm{aq})$

The solubility product constant, Ksp, is given by:

$\mathrm{K}_{sp} = [\mathrm{Ba}^{2+}][\mathrm{SO}_4^{2-}]$

In this case, the salt is being dissolved in a solution that already contains sulfate ions, $\mathrm{SO}_4^{2-}$ , from the $\mathrm{K}_2\mathrm{SO}_4$ .

When $\mathrm{K}_2\mathrm{SO}_4$ dissolves completely in water, it forms $\mathrm{K}^+$ and $\mathrm{SO}_4^{2-}$ ions. Because its concentration is 0.1 M, the $\mathrm{SO}_4^{2-}$ concentration due to $\mathrm{K}_2\mathrm{SO}_4$ is 0.1 M.

Therefore, the concentration of $\mathrm{SO}_4^{2-}$ ions is now not just due to the BaSO₄ dissolving, but also the added $\mathrm{K}_2\mathrm{SO}_4$ . Hence, the total concentration of $\mathrm{SO}_4^{2-}$ ions is $\mathrm{S} + 0.1$ where S is the solubility of $\mathrm{BaSO}_4$ .

We then substitute into the Ksp expression:

$1 \times 10^{-10} = [\mathrm{S}][\mathrm{S} + 0.1]$

However, because BaSO₄ is sparingly soluble, S is very small compared to 0.1. Therefore, we can make the approximation that $\mathrm{S} + 0.1$ is approximately 0.1. This simplifies the equation to:

$1 \times 10^{-10} = 0.1[\mathrm{S}]$

Solving for S (the molar solubility of BaSO₄) gives:

$S = \frac{1 \times 10^{-10}}{0.1} = 1 \times 10^{-9} \, \mathrm{mol/L}$

To convert this molarity to a mass per volume concentration, we multiply by the molar mass of BaSO₄, which is 233 g/mol:

$\mathrm{Solubility} = S \times \text{Molar mass of BaSO4} = 1 \times 10^{-9} \, \mathrm{mol/L} \times 233 \, \mathrm{g/mol} = 233 \times 10^{-9} \, \mathrm{g/L}$

So, the solubility of BaSO₄ in a 0.1 M K₂SO₄ solution is 233 x $10^{-9}$ g/L, or 233 ng/L.