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← JEE Advanced 2022 · Paper 2
ChemistryInteger+3 / -1Question 37 of 54

JEE Advanced 2022 · Paper 2, Chemistry question 1

Concentration of HX2SOX4\ce{H2SO4} and NaX2SOX4\ce{Na2SO4} in a solution is 1 M and 1.8×10−21.8 \times 10^{-2} M, respectively. Molar solubility of PbSOX4\ce{PbSO4} in the same solution is X×10−YX \times 10^{-Y} M (expressed in scientific notation). The value of YY is ________.

[Given: Solubility product of PbSOX4\ce{PbSO4} (Ksp)=1.6×10−8(K_{sp}) = 1.6 \times 10^{-8}. For HX2SOX4\ce{H2SO4}, Ka1K_{a1} is very large and Ka2=1.2×10−2K_{a2} = 1.2 \times 10^{-2}]

Answer and worked solution

Official answer: 6

Sulphate concentration. The first dissociation of HX2SOX4\ce{H2SO4} is complete, giving [HX+]≈1[\ce{H+}]\approx1 M and [HSOX4X−]≈1[\ce{HSO4-}]\approx1 M. Let the final sulphate concentration be ss. Starting from 0.018 M from NaX2SOX4\ce{Na2SO4}, the equilibrium HSOX4X−⇌HX++SOX4X2−\ce{HSO4- <=> H+ + SO4^{2-}} gives
Ka2=(0.982+s) s1.018−s=1.2×10−2  ⟹  s≈1.2×10−2 M.K_{a2}=\frac{(0.982+s)\,s}{1.018-s}=1.2\times10^{-2}\implies s\approx1.2\times10^{-2}\ \text{M}.
So the large [HX+][\ce{H+}] pushes some of the added sulphate back to HSOX4X−\ce{HSO4-}.

Solubility of PbSOX4\ce{PbSO4}.
S=Ksp[SOX4X2−]=1.6×10−81.2×10−2≈1.3×10−6 M.S=\frac{K_{sp}}{[\ce{SO4^{2-}}]}=\frac{1.6\times10^{-8}}{1.2\times10^{-2}}\approx1.3\times10^{-6}\ \text{M}.
So Y=6Y=6.

Topics:Solubility productCommon ion effectpH and ionisation
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