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
- (a) PbS<HgS<Ca(OH)₂<AgBr
- (b) HgS<AgBr<PbS<Ca(OH)₂
- (c) HgS<PbS<AgBr<Ca(OH)₂
- (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.
- 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.
- 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.
Compound Kₛₚ at 298 K Comment HgS 4 × 10⁻⁵³ one of the least soluble salts known PbS 8 × 10⁻²⁸ very insoluble sulfide AgBr 5 × 10⁻¹³ sparingly soluble halide Ca(OH)₂ 5.5 × 10⁻⁶ slightly soluble (limewater) - 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.