JEE Main 2023 — Ionic Equilibrium Question with 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:

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

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

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

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