JEE Main · 2025 · Shift-IIeasyIEQ-083

Arrange the following in increasing order of solubility product: {Ca(OH)}_2,\;{AgBr},\;{PbS},\;{HgS}

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

Arrange the following in increasing order of solubility product:

Ca(OH)2,  AgBr,  PbS,  HgS\text{Ca(OH)}_2,\;\text{AgBr},\;\text{PbS},\;\text{HgS}

Options
  1. a

    PbS<HgS<Ca(OH)2<AgBr\text{PbS} < \text{HgS} < \text{Ca(OH)}_2 < \text{AgBr}

  2. b

    HgS<PbS<AgBr<Ca(OH)2\text{HgS} < \text{PbS} < \text{AgBr} < \text{Ca(OH)}_2

    ✓
  3. c

    Ca(OH)2<AgBr<HgS<PbS\text{Ca(OH)}_2 < \text{AgBr} < \text{HgS} < \text{PbS}

  4. d

    HgS<AgBr<PbS<Ca(OH)2\text{HgS} < \text{AgBr} < \text{PbS} < \text{Ca(OH)}_2

Correct Answerb

HgS<PbS<AgBr<Ca(OH)2\text{HgS} < \text{PbS} < \text{AgBr} < \text{Ca(OH)}_2

Detailed Solution

KspK_{sp} values (approximate):

| Compound | KspK_{sp} | |---|---| | HgS\text{HgS} | 4×10−534 \times 10^{-53} | | PbS\text{PbS} | 8×10−288 \times 10^{-28} | | AgBr\text{AgBr} | 5×10−135 \times 10^{-13} | | Ca(OH)2\text{Ca(OH)}_2 | 5.5×10−65.5 \times 10^{-6} |

Increasing order: HgS<PbS<AgBr<Ca(OH)2\text{HgS} < \text{PbS} < \text{AgBr} < \text{Ca(OH)}_2

Answer: (b)

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