Question

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 1

Answer: (C)

HgS < PbS < AgBr < $Ca(OH)_2$

Answer 2

Known approximate Kₛₚ values:

• $Ca(OH)_2$: ≈10⁻⁶
• AgBr: ≈10⁻¹³
• PbS: ≈10⁻²⁸
• HgS: ≈10⁻⁵⁴

In increasing order (lowest Kₛₚ means lower solubility):

$\text{HgS} < \text{PbS} < \text{AgBr} < \text{Ca(OH)}_2$

Thus, the correct answer is Option C.