JEE Main 2025 · 28 January · evening · Single correct · +4 / −1

Arrange the following in increasing order of solubility product :
Ca(OH)₂, AgBr, PbS, HgS

  1. (a) PbS<HgS<Ca(OH)₂<AgBr
  2. (b) HgS<AgBr<PbS<Ca(OH)₂
  3. (c) HgS<PbS<AgBr<Ca(OH)₂
  4. (d) Ca(OH)₂<AgBr<HgS<PbS
Answer and solution

Answer (c) HgS<PbS<AgBr<Ca(OH)₂

Solution

Chemistry ma’am · 3 steps

Metal sulfides sit at the very bottom of the Kₛₚ ladder, and Ca(OH)₂ sits near the top.

  1. What Ksp tells you. A smaller solubility product means fewer ions go into solution, so the salt is less soluble. Here we only need to rank how tiny each Kₛₚ is.
  2. Use known values. These four are standard textbook values. The sulfides of heavy metals are extremely insoluble, which is why H₂S can precipitate them even in acidic solution. Ca(OH)₂ dissolves enough to make limewater.
    CompoundKₛₚ at 298 KComment
    HgS4 × 10⁻⁵³one of the least soluble salts known
    PbS8 × 10⁻²⁸very insoluble sulfide
    AgBr5 × 10⁻¹³sparingly soluble halide
    Ca(OH)₂5.5 × 10⁻⁶slightly soluble (limewater)
  3. So. Increasing Kₛₚ runs HgS < PbS < AgBr < Ca(OH)₂, which is option (c).

The idea to keep. Remember the rough ladder: heavy-metal sulfides (around 10⁻⁵⁰ to 10⁻²⁸) are below silver halides (around 10⁻¹⁰ to 10⁻¹⁷), which are below hydroxides like Ca(OH)₂ (around 10⁻⁶).

Answer from the NTA final answer key. Solution by PrepWiser.

Study Ionic Equilibrium

Asked about 100 times in JEE Main since 2020. This question is from 2025.