JEE Challenger
More from Equilibrium
ChemMedAdv 2016 • Paper 1

Selective Precipitation of Sulfide Ions from Mixture

Which of the following reagent(s) can selectively precipitate S2−S^{2-} ions from an aqueous solution containing both S2−S^{2-} and SO42−SO_4^{2-}?

Options

A

CuCl2CuCl_2

Correct
B

BaCl2BaCl_2

C

Pb(OOCCH3)2Pb(OOCCH_3)_2

Correct
D

Na2[Fe(CN)5NO]Na_2[Fe(CN)_5NO]

Topics & Concepts

EquilibriumSolubility Product

Step-by-Step Solution

To determine which reagent(s) can selectively precipitate S2−\text{S}^{2-} ions from an aqueous solution containing a mixture of S2−\text{S}^{2-} and SO42−\text{SO}_4^{2-} ions, we analyze the reaction of each reagent with both anions:

1. Option (A): CuCl2\text{CuCl}_2 (Copper(II) chloride)

When CuCl2\text{CuCl}_2 is added to the mixture:

  • Reaction with S2−\text{S}^{2-}: Cu2+(aq)+S2−(aq)→CuS(s)↓(Black precipitate)\text{Cu}^{2+}(aq) + \text{S}^{2-}(aq) \rightarrow \text{CuS}(s) \downarrow \quad (\text{Black precipitate}) The solubility product of copper(II) sulfide is extremely small (Ksp(CuS)≈6×10−36K_{sp}(\text{CuS}) \approx 6 \times 10^{-36}), leading to immediate quantitative precipitation of S2−\text{S}^{2-} as CuS\text{CuS}.

  • Reaction with SO42−\text{SO}_4^{2-}: Cu2+(aq)+SO42−(aq)→CuSO4(aq)(Highly soluble)\text{Cu}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{CuSO}_4(aq) \quad (\text{Highly soluble}) Copper(II) sulfate is readily soluble in water and does not precipitate.

Thus, CuCl2\text{CuCl}_2 selectively precipitates S2−\text{S}^{2-} ions.

2. Option (B): BaCl2\text{BaCl}_2 (Barium chloride)

When BaCl2\text{BaCl}_2 is added to the mixture:

  • Reaction with S2−\text{S}^{2-}: Ba2+(aq)+S2−(aq)→BaS(aq)(Soluble in water)\text{Ba}^{2+}(aq) + \text{S}^{2-}(aq) \rightarrow \text{BaS}(aq) \quad (\text{Soluble in water})

  • Reaction with SO42−\text{SO}_4^{2-}: Ba2+(aq)+SO42−(aq)→BaSO4(s)↓(White precipitate)\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s) \downarrow \quad (\text{White precipitate})

Thus, BaCl2\text{BaCl}_2 selectively precipitates SO42−\text{SO}_4^{2-}, not S2−\text{S}^{2-}.

3. Option (C): Pb(OOCCH3)2\text{Pb(OOCCH}_3)_2 (Lead acetate)

When Pb(OOCCH3)2\text{Pb(OOCCH}_3)_2 is added:

  • Reaction with S2−\text{S}^{2-}: Pb2+(aq)+S2−(aq)→PbS(s)↓(Black precipitate,Ksp≈3×10−28)\text{Pb}^{2+}(aq) + \text{S}^{2-}(aq) \rightarrow \text{PbS}(s) \downarrow \quad (\text{Black precipitate}, K_{sp} \approx 3 \times 10^{-28})

  • Reaction with SO42−\text{SO}_4^{2-}: Pb2+(aq)+SO42−(aq)→PbSO4(s)↓(White precipitate,Ksp≈1.6×10−8)\text{Pb}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4(s) \downarrow \quad (\text{White precipitate}, K_{sp} \approx 1.6 \times 10^{-8})

Because Ksp(PbS)≪Ksp(PbSO4)K_{sp}(\text{PbS}) \ll K_{sp}(\text{PbSO}_4), by carefully controlling the concentration of Pb2+\text{Pb}^{2+} ions, S2−\text{S}^{2-} can be selectively precipitated as PbS\text{PbS} before PbSO4\text{PbSO}_4 reaches its ionic product threshold. However, since PbSO4\text{PbSO}_4 is also insoluble under general conditions, both (A) and (A, C) are valid answers.

4. Option (D): Na2[Fe(CN)5NO]\text{Na}_2[\text{Fe(CN)}_5\text{NO}] (Sodium nitroprusside)

When sodium nitroprusside is added to S2−\text{S}^{2-} in alkaline medium, it forms a purple/violet-colored soluble complex: S2−(aq)+[Fe(CN)5NO]2−(aq)→[Fe(CN)5NOS]4−(aq)(Purple solution)\text{S}^{2-}(aq) + [\text{Fe(CN)}_5\text{NO}]^{2-}(aq) \rightarrow [\text{Fe(CN)}_5\text{NOS}]^{4-}(aq) \quad (\text{Purple solution}) No precipitate is formed.

Conclusion:

The correct reagent(s) for selective precipitation of S2−\text{S}^{2-} are (A) and (C) [or (A) alone depending on standard analytical conditions].

Correct Options: A (or A, C)

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