JEE Advanced 2022 — Paper 2 — CHEM-I solution

JEE Advanced2022Paper 2CHEM-I

Concentration of H $_{2}$ SO $_{4}$ and Na $_{2}$ SO $_{4}$ in a solution is 1 M and $1.8 \times 1 0^{- 2}$ M, respectively. Molar solubility of PbSO $_{4}$ in the same solution is $X \times 1 0^{- Y}$ M (expressed in scientific notation). The value of Y is _________.

[Given: Solubility product of PbSO $_{4}$ ( $K_{s p}$ ) = $1.6 \times 1 0^{- 8}$ . For H $_{2}$ SO $_{4}$ , $K_{a 1}$ is very large and $K_{a 2} = 1.2 \times 1 0^{- 2}$ ]

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Solution

Since $K_{a 1}$ of H $_{2}$ SO $_{4}$ is very large, the first dissociation is complete: $[H^{+}] = 1$ M and $[HSO_{4}^{-}] = 1$ M from H $_{2}$ SO $_{4}$ .

For the second dissociation, $K_{a 2} = 1.2 \times 1 0^{- 2}$ . With $[SO_{4}^{2 -}] = 1.8 \times 1 0^{- 2}$ M from Na $_{2}$ SO $_{4}$ , the reaction quotient $Q = \frac{( 1.8 \times 1 0 ^{- 2} ) ( 1 )}{1} = 1.8 \times 1 0^{- 2} > K_{a 2}$ . 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 $[SO_{4}^{2 -}] = 1.8 \times 1 0^{- 2} - x$ , $[H^{+}] \approx 1$ , $[HSO_{4}^{-}] \approx 1$ . Applying $K_{a 2}$ : $\frac{( 1.8 \times 1 0 ^{- 2} - x ) ( 1 )}{1} = 1.2 \times 1 0^{- 2}$ , giving $x = 0.6 \times 1 0^{- 2}$ and $[SO_{4}^{2 -}] = 1.2 \times 1 0^{- 2}$ M.

Let $s$ = molar solubility of PbSO $_{4}$ . Then $[Pb^{2 +}] = s$ and $[SO_{4}^{2 -}] \approx 1.2 \times 1 0^{- 2}$ (assuming $s ≪ 1.2 \times 1 0^{- 2}$ ). Applying $K_{s p}$ : $s \cdot 1.2 \times 1 0^{- 2} = 1.6 \times 1 0^{- 8}$ .

$s = \frac{1.6}{1.2} \times 1 0^{- 6} = 1.33 \times 1 0^{- 6}$ M. Comparing with $X \times 1 0^{- Y}$ : $Y = 6$ .

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