JEE MAIN - Chemistry (2025 - 28th January Evening Shift - No. 5)
Arrange the following in increasing order of solubility product :
$\mathrm{Ca}(\mathrm{OH})_2, \mathrm{AgBr}, \mathrm{PbS}, \mathrm{HgS}$
$\mathrm{PbS}<\mathrm{HgS}<\mathrm{Ca}(\mathrm{OH})_2<\mathrm{AgBr}$
$\mathrm{HgS}<\mathrm{AgBr}<\mathrm{PbS}<\mathrm{Ca}(\mathrm{OH})_2$
$\mathrm{HgS}<\mathrm{PbS}<\mathrm{AgBr}<\mathrm{Ca}(\mathrm{OH})_2$
$\mathrm{Ca}(\mathrm{OH})_2<\mathrm{AgBr}<\mathrm{HgS}<\mathrm{PbS}$
Explanation
Based on the Ksp values and salt analysis cation identification, we can say that order of Ksp value is:
$$\mathrm{HgS}<\mathrm{PbS}<\mathrm{AgBr}<\mathrm{Ca}(\mathrm{OH})_2$$
Ksp values
$$\begin{aligned} & \mathrm{HgS} \rightarrow 4 \times 10^{-53} \\ & \mathrm{PbS} \rightarrow 8 \times 10^{-28} \\ & \mathrm{AgBr} \rightarrow 5 \times 10^{-13} \\ & \mathrm{Ca}(\mathrm{OH})_2 \rightarrow 5.5 \times 10^{-6} \end{aligned}$$
Comments (0)
