JEE Advanced 2022 Paper 2 · Q01 · Buffers & Solubility Product

Question

Concentration of H$_2$SO$_4$ and Na$_2$SO$_4$ in a solution is 1 M and $1.8 	imes 10^{-2}$ M, respectively. Molar solubility of PbSO$_4$ in the same solution is $X 	imes 10^{-Y}$ M (expressed in scientific notation). The value of $Y$ is _____.

[Given: Solubility product of PbSO$_4$ ($K_{sp}$) = $1.6 	imes 10^{-8}$. For H$_2$SO$_4$, $K_{a1}$ is very large and $K_{a2} = 1.2 	imes 10^{-2}$.]

SolveKar AI · Accepted Answer

Compute the actual [SO$_4^{2-}$] in solution from the H$_2$SO$_4$ + Na$_2$SO$_4$ buffer, then use $K_{sp}$ to get the PbSO$_4$ solubility.

Step 1 — sulfate equilibrium. H$_2$SO$_4$ is a strong first-step diprotic; second step has $K_{a2} = 1.2 	imes 10^{-2}$: HSO$_4^-$ $ightleftharpoons$ H$^+$ + SO$_4^{2-}$. With [H$_2$SO$_4$] = 1 M (fully ionised once: [HSO$_4^-$] $\approx$ 1 M, [H$^+$] $\approx$ 1 M from first step) and added Na$_2$SO$_4$ giving 0.018 M SO$_4^{2-}$, set $[SO_4^{2-}] = 0.018 + a$, $[HSO_4^-] = 1 - a$, $[H^+] = 1 + a$. From $K_{a2}$:
$\frac{(1+a)(0.018+a)}{1-a} = 1.2 	imes 10^{-2}$.
Solving with the approximation $a \ll 1$: $0.018 + a \approx 0.012 \Rightarrow a \approx -0.006$. Hence [SO$_4^{2-}$] $\approx 0.012$ M.

Step 2 — PbSO$_4$ solubility $s$:
$K_{sp} = [Pb^{2+}][SO_4^{2-}] = s \cdot (0.012 + s) \approx 0.012\,s$ (s very small).
$s = \frac{1.6 	imes 10^{-8}}{0.012} = 1.33 	imes 10^{-6}$ M.

Written as $X 	imes 10^{-Y}$ M with $X$ between 1 and 10, $s = 1.33 	imes 10^{-6}$, so $Y = 6$.

JEE Advanced 2022 Paper 2 · Q01 · Buffers & Solubility Product

Solution