Question

The solubility product $\left(K_{sp}\right)$ of the following compounds are given at ${25}^{\circ }C$. The compound most soluble and least soluble coumpounds are:
Compounds $K_{sp}$
$AgCl1.1\times {10}^{-10}AgI1.1\times {10}^{-16}PbCr{O}_{4}4.0\times {10}^{-14}A{g}_{2}C{O}_{3}8.0\times {10}^{-12}$

• A. $AgClandPbCr{O}_4$
• B. $A{g}_{2}C{O}_{3}andPbCr{O}_4$
• C. $A{g}_{2}C{O}_{3}andAgI$
• D. $AgClandA{g}_{2}C{O}_3$

Answer 1

The correct option is C $A{g}_{2}C{O}_{3}andAgI$

The salts $AgCl,AgI,PbCr{O}_4$ are of same type AB.
So,
Solubility=${\sqrt{K}}_{sp}$
so,
${\text{Solubility}}_{AgCl}=1\times {10}^{-5}M$

${\text{Solubility}}_{AgI}=1\times {10}^{-8}M$

${\text{Solubility}}_{PbCr{O}_4}=2\times {10}^{-7}M$

For $A{g}_{2}C{O}_3$,
${\text{Solubility}}_{A{g}_{2}C{O}_3}=(\frac{K_{s}p}{4}{)}^{\frac{1}{3}}={10}^{-4}$


Thus, most soluble is $A{g}_{2}C{O}_3\text{and least soluble is}AgI$