Ionic EquilibriumHard

Question

The correct increasing order of solubility of the following substances in g/100 ml is PbSO4 (Ksp = 2 × 10−9), ZnS (Ksp = 1 × 10−22), AgBr (Ksp = 4 × 10−13), CuCO3 (Ksp = 1 × 10−8).

(Atomic masses: Pb = 208, Zn = 65, Ag = 108, Br = 80, Cu = 63)

Options

A.PbSO4 < ZnS < AgBr < CuCO3
B.PbSO4 < CuCO3 < AgBr < ZnS
C.ZnS < AgBr < CuCO3 < PbSO4
D.ZnS < AgBr < PbSO4 < CuCO3

Solution

$PbSO_{4} = \sqrt{2 \times 10^{- 9}} \times \frac{304}{10} \simeq 1.36 \times 10^{- 3}\text{ g/100ml}$

$ZaS = \sqrt{10^{- 22}} \times \frac{97}{10} = 9.7 \times 10^{- 11}\text{ g/100ml} $$${AgBr = \sqrt{4 \times 10^{- 13}} \times \frac{188}{10} = 1.19 \times 10^{- 5}\text{ g/100ml} }{\text{CuC}\text{O}_{3} = \sqrt{10^{- 8}} \times \frac{123}{10} = 1.23 \times 10^{- 8}\text{ g/100ml}}$$

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