Arrange the following in increasing order of solubility product : $\mathrm{Ca}(\mathrm{OH})_{2}, \mathrm{AgBr}, \mathrm{PbS}, \mathrm{HgS}$

(B) $\mathrm{HgS}<\mathrm{PbS}<\mathrm{AgBr}<\mathrm{Ca}(\mathrm{OH})_{2}$

(A) $\mathrm{PbS}<\mathrm{HgS}<\mathrm{Ca}(\mathrm{OH})_{2}<\mathrm{AgBr}$

(C) $\mathrm{Ca}(\mathrm{OH})_{2}<\mathrm{AgBr}<\mathrm{HgS}<\mathrm{PbS}$

(D) $\mathrm{HgS}<\mathrm{AgBr}<\mathrm{PbS}<\mathrm{Ca}(\mathrm{OH})_{2}$

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Equilibrium

Arrange the following in increasing order of solubility product :

Ca (OH)_{2}, AgBr, PbS, HgS

(A) $PbS < HgS < Ca (OH)_{2} < AgBr$
(B) $HgS < PbS < AgBr < Ca (OH)_{2}$ ✓ Correct
(C) $Ca (OH)_{2} < AgBr < HgS < PbS$
(D) $HgS < AgBr < PbS < Ca (OH)_{2}$

Answer: (B)

Solution

Comparing solubility product constants:

K_{s p} (H g S) \approx 4 \times 1 0^{- 53}

K_{s p} (P b S) \approx 8 \times 1 0^{- 28}

K_{s p} (A g B r) \approx 5 \times 1 0^{- 13}

K_{s p} (C a (O H)_{2}) \approx 5.5 \times 1 0^{- 6}

H g S

is the least soluble of the four, needing only a vanishingly small sulphide-ion concentration to precipitate — which is why it forms even in dilute acidic

H_{2} S

— while

C a (O H)_{2}

is comparatively the most soluble. Arranged from smallest to largest, the increasing order of solubility product is

H g S < P b S < A g B r < C a (O H)_{2}

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Equilibrium

Solution

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