Verified Solution • 2025–26 Session moderate

QJEEM25Ch9239: JEE Main 2025 Chemistry - Class 11 JEEM25Ch9239 Solution: Arrange the following in increasing order of solubility product :

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}

Multiple Choice Options

  • A PbS<HgS<Ca(OH)2<AgBr\mathrm{PbS}<\mathrm{HgS}<\mathrm{Ca}(\mathrm{OH})_2<\mathrm{AgBr}
  • B HgS<AgBr<PbS<Ca(OH)2\mathrm{HgS}<\mathrm{AgBr}<\mathrm{PbS}<\mathrm{Ca}(\mathrm{OH})_2
  • C HgS<PbS<AgBr<Ca(OH)2\mathrm{HgS}<\mathrm{PbS}<\mathrm{AgBr}<\mathrm{Ca}(\mathrm{OH})_2
  • D Ca(OH)2<AgBr<HgS<PbS\mathrm{Ca}(\mathrm{OH})_2<\mathrm{AgBr}<\mathrm{HgS}<\mathrm{PbS}

Step-by-Step Verified Solution

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

\mathrm{HgS}&lt;\mathrm{PbS}&lt;\mathrm{AgBr}&lt;\mathrm{Ca}(\mathrm{OH})_2

Ksp values

amp;HgS→4×10−53amp;PbS→8×10−28amp;AgBr→5×10−13amp;Ca(OH)2→5.5×10−6\begin{aligned}&amp; \mathrm{HgS} \rightarrow 4 \times 10^{-53} \\&amp; \mathrm{PbS} \rightarrow 8 \times 10^{-28} \\&amp; \mathrm{AgBr} \rightarrow 5 \times 10^{-13} \\&amp; \mathrm{Ca}(\mathrm{OH})_2 \rightarrow 5.5 \times 10^{-6}\end{aligned}

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