Question

$K_{sp}$ of salts $AB,A{B}_2$ and $A_{3}B$ are $4.0\times {10}^{-8},3.2\times {10}^{-14}$ and $2.7\times {10}^{-15}$ respectively at temperature T. The solubility order of these salts in water at temperature T (in mole litre${}^{-1}$) is:

• A. $AB>A{B}_2>A_{3}B$
• B. $A_{3}B>A{B}_2>AB$
• C. $A{B}_2>A_{3}B>AB$
• D. $AB>A_{3}B>A{B}_2$

Answer 1

The correct option is D $AB>A_{3}B>A{B}_2$

Let us assume that $s_1,s_2,s_3$ represents the solubility of $AB,A{B}_2$ and $A_{3}B$ respectively.
The expression for $s_1$ is $s_1=\sqrt{K_{sp}}$
Substitute values in the above expression,
$s_1=\sqrt{4\times {10}^{-8}}=2\times {10}^{-4}M$
The expression for $s_2$ is $s_2=\sqrt[3]{3\frac{K_{sp}}{4}}$
Substitute values in the above expression,
$s_2=\sqrt[3]{3\frac{3.2\times 10-14}{4}}=2\times {10}^{-5}M$
The expression for $s_3$ is $s_3=\sqrt[4]{4\frac{K_{sp}}{27}}$
Substitute values in the above expression.
$s_3=\sqrt[4]{4\frac{2.7\times {10}^{-15}}{27}}=1\times {10}^{-4}M$

The correct order is $AB>A_{3}B>A{B}_2$