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JEE Chemistry 2026 Question with Solution

Consider two Group IV metal ions X2+X^{2+} and Y2+Y^{2+}. A solution containing 0.01 M X2+0.01 \, \text{M} \, X^{2+} and 0.01 M Y2+0.01 \, \text{M} \, Y^{2+} is saturated with H2SH_2S. The pH at which the metal sulphide YS will form as a precipitate is _____ (Nearest integer)

Given: Ksp(XS)=1×10−22K_{sp}(\mathrm{XS}) = 1 \times 10^{-22} at 25∘C25^\circ \text{C} Ksp(YS)=4×10−16K_{sp}(\mathrm{YS}) = 4 \times 10^{-16} at 25∘C25^\circ \text{C} [H2S]=0.1 M[\mathrm{H_2S}] = 0.1 \, \text{M} Ka1×Ka2(H2S)=1.0×10−21K_{a1} \times K_{a2}(\mathrm{H_2S}) = 1.0 \times 10^{-21} log⁡2=0.30, log⁡3=0.48, log⁡5=0.70\log 2 = 0.30,\ \log 3 = 0.48,\ \log 5 = 0.70

Answer

Correct answer:4

Step-by-step solution

Standard Method

Given:

  • [Y2+]=0.01 M[Y^{2+}] = 0.01 \, \text{M}
  • Ksp(YS)=4×10−16K_{sp}(\mathrm{YS}) = 4 \times 10^{-16}
  • [H2S]=0.1 M[\mathrm{H_2S}] = 0.1 \, \text{M}
  • Ka1Ka2=1.0×10−21K_{a1}K_{a2} = 1.0 \times 10^{-21}

Find: The pH at which YS starts precipitating.

From the solution, the sulphide ion concentration is written as

[S2−]=Ka1Ka2[H2S][H+]2[\mathrm{S^{2-}}] = \frac{K_{a1}K_{a2}[\mathrm{H_2S}]}{[\mathrm{H^+}]^2}

For the precipitation of YS,

Ksp(YS)=[Y2+][S2−]K_{sp}(\mathrm{YS}) = [Y^{2+}][S^{2-}]

So,

4×10−16=0.01×[S2−]4 \times 10^{-16} = 0.01 \times [S^{2-}]

Hence,

[S2−]=4×10−14[S^{2-}] = 4 \times 10^{-14}

Substituting into the sulphide ion expression,

4×10−14=(1×10−21)(0.1)[H+]24 \times 10^{-14} = \frac{(1 \times 10^{-21})(0.1)}{[\mathrm{H^+}]^2} [H+]2=2.5×10−9[\mathrm{H^+}]^2 = 2.5 \times 10^{-9} [H+]=5×10−5[\mathrm{H^+}] = 5 \times 10^{-5}

Now,

pH=−log⁡(5×10−5)=5−0.70=4.3\text{pH} = -\log(5 \times 10^{-5}) = 5 - 0.70 = 4.3

Therefore, the nearest integer pH is 44.

The solution shows a final printed value of 2, but this contradicts its own calculation and the listed correct answer 4. Using the working, the answer is 44.

Common mistakes

  • Using the precipitation condition for XS instead of YS. This is wrong because the question asks for the pH at which YS begins to precipitate. Use Ksp(YS)=[Y2+][S2−]K_{sp}(\mathrm{YS}) = [Y^{2+}][S^{2-}].

  • Forgetting that sulphide concentration depends on acidity as [S2−]∝1[H+]2[S^{2-}] \propto \frac{1}{[H^+]^2}. This gives an incorrect pH dependence. Start from [S2−]=Ka1Ka2[H2S][H+]2[\mathrm{S^{2-}}] = \frac{K_{a1}K_{a2}[\mathrm{H_2S}]}{[\mathrm{H^+}]^2}.

  • Making an error in logarithms while evaluating −log⁡(5×10−5)-\log(5 \times 10^{-5}). This is wrong because −log⁡(5×10−5)=5−log⁡5=5−0.70=4.3-\log(5 \times 10^{-5}) = 5 - \log 5 = 5 - 0.70 = 4.3, not 2. Use the given log values carefully.

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