Chemistry · Ionic Equilibrium

JEE Main 2025 — 28 January, Evening Shift — Question 30

Arrange the following in increasing order of solubility product : Ca(OH)2,AgBr,PbS,HgS\mathrm{Ca}(\mathrm{OH})_{2}, \mathrm{AgBr}, \mathrm{PbS}, \mathrm{HgS}

  1. Option A:

    PbS<HgS<Ca(OH)2<AgBr\mathrm{PbS}<\mathrm{HgS}<\mathrm{Ca}(\mathrm{OH})_{2}<\mathrm{AgBr}

  2. Option B:

    HgS<PbS<AgBr<Ca(OH)2\mathrm{HgS}<\mathrm{PbS}<\mathrm{AgBr}<\mathrm{Ca}(\mathrm{OH})_{2}

    Correct
  3. Option C:

    Ca(OH)2<AgBr<HgS<PbS\mathrm{Ca}(\mathrm{OH})_{2}<\mathrm{AgBr}<\mathrm{HgS}<\mathrm{PbS}

  4. Option D:

    HgS<AgBr<PbS<Ca(OH)2\mathrm{HgS}<\mathrm{AgBr}<\mathrm{PbS}<\mathrm{Ca}(\mathrm{OH})_{2}

Answer: B

Step-by-step solution

Based on the Ksp values and salt analysis cation identification, we can say that order of Ksp value is:

HgS<PbS<AgBr<Ca(OH)2\mathrm{HgS}<\mathrm{PbS}<\mathrm{AgBr}<\mathrm{Ca}(\mathrm{OH})_{2}

Ksp values

HgS→4×10−53\mathrm{HgS} \rightarrow 4 \times 10^{-53}

PbS→8×10−28\mathrm{PbS} \rightarrow 8 \times 10^{-28}

AgBr→5×10−13\mathrm{AgBr} \rightarrow 5 \times 10^{-13}

Ca(OH)2→5.5×10−6\mathrm{Ca}(\mathrm{OH})_{2} \rightarrow 5.5 \times 10^{-6}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Chemistry
Chapter
Ionic Equilibrium
Topic
Sparingly Soluble Salts, Solubility Product & Precipitation Conditions
Arrange the following in increasing order of solubility product : Ca… | JEE Main 2025 PYQ with Solution · DhiX AI