The correct increasing order of solubility of the following substances in is (), (), (), (). (Atomic — Ionic Equilibrium Chemistry Question
Question
The correct increasing order of solubility of the following substances in $\text{g}/100\text{ ml}$ is $\text{PbSO}_4$ ($K_{sp} = 2 \times 10^{-9}$), $\text{ZnS}$ ($K_{sp} = 1 \times 10^{-22}$), $\text{AgBr}$ ($K_{sp} = 4 \times 10^{-13}$), $\text{CuCO}_3$ ($K_{sp} = 1 \times 10^{-8}$). (Atomic masses: $\text{Pb}=208, \text{Zn}=65, \text{Ag}=108, \text{Br}=80, \text{Cu}=63$)
💡 Solution & Explanation
All four are 1:1 salts, so molar solubility $S = \sqrt{K_{sp}}$. For $\text{PbSO}_4$ (Molar mass 304), $S = \sqrt{2 \times 10^{-9}} = 4.47 \times 10^{-5}\text{ M}$, yielding $\approx 1.36 \times 10^{-2}\text{ g/L}$. For $\text{ZnS}$ (Molar mass 97), $S = \sqrt{10^{-22}} = 10^{-11}\text{ M}$, yielding $\approx 10^{-9}\text{ g/L}$. For $\text{AgBr}$ (Molar mass 188), $S = \sqrt{4 \times 10^{-13}} = 6.32 \times 10^{-7}\text{ M}$, yielding $\approx 1.2 \times 10^{-4}\text{ g/L}$. For $\text{CuCO}_3$ (Molar mass 123), $S = \sqrt{10^{-8}} = 10^{-4}\text{ M}$, yielding $\approx 1.23 \times 10^{-2}\text{ g/L}$. Comparing the values in g/L (and therefore g/100 ml), the order is $\text{ZnS} < \text{AgBr} < \text{CuCO}_3 < \text{PbSO}_4$. Therefore, correct answer is C.