JEE Advanced 2022 Paper 2, Chemistry Section 1 Q1: Solubility Equilibria Of Sparingly Soluble Salts

Since $K_{a1}$ of H${}_2$SO${}_4$ is very large, the first dissociation is complete: $[{\text{H}}^{+}]=1$ M and $[{\text{HSO}}_4^{-}]=1$ M from H${}_2$SO${}_4$.

For the second dissociation, $K_{a2}=1.2\times 10^{-2}$. With $[{\text{SO}}_4^{2-}]=1.8\times 10^{-2}$ M from Na${}_2$SO${}_4$, the reaction quotient $Q=\frac{(1.8\times 10^{-2})(1)}{1}=1.8\times 10^{-2}>K_{a2}$. The equilibrium therefore shifts backward, and SO${}_4^{2-}$ combines with H${}^{+}$ to form HSO${}_4^{-}$.

Let $x$ mol/L of SO${}_4^{2-}$ combine with H${}^{+}$. At equilibrium $[{\text{SO}}_4^{2-}]=1.8\times 10^{-2}-x$, $[{\text{H}}^{+}]\approx 1$, $[{\text{HSO}}_4^{-}]\approx 1$. Applying $K_{a2}$: $\frac{(1.8\times 10^{-2}-x)(1)}{1}=1.2\times 10^{-2}$, giving $x=0.6\times 10^{-2}$ and $[{\text{SO}}_4^{2-}]=1.2\times 10^{-2}$ M.

Let $s$ = molar solubility of PbSO${}_4$. Then $[{\text{Pb}}^{2+}]=s$ and $[{\text{SO}}_4^{2-}]\approx 1.2\times 10^{-2}$ (assuming $s\ll 1.2\times 10^{-2}$). Applying $K_{sp}$: $s\cdot 1.2\times 10^{-2}=1.6\times 10^{-8}$.

$s=\frac{1.6}{1.2}\times 10^{-6}=1.33\times 10^{-6}$ M. Comparing with $X\times 10^{-Y}$: $Y=6$.