Question

The solubility product of a sparingly soluble salt $A_{2}B$ is $3.2\times 10^{-11}$. Its solubility in $mol{L}^{-1}$ is

• A. $4\times 10^{-4}$
• B. $2\times 10^{-4}$
• C. $6\times 10^{-4}$
• D. $3\times 10^{-4}$

Answer 1

Answer: (B)

For the salt of type $A_{2}B$, solubility product

$\left(K_{sp}\right)=x^x\cdot y^y\cdot S^{x+y}$

where, $x$ and $y$ are the number of moles of $A$ and $B$ respectively.

$S=$ solubility.

Thus, $K_{sp}=2^2\times 1^1\cdot S^{2+1}$

$\Rightarrow K_{sp}=4S^3$

Given, $K_{cp}$ for $A_{2}B=3.2\times 10^{-11}$

$\therefore 3.2\times 10^{-11}=4S^3$

or, $S^3=\frac{3.2}{4}\times 10^{-11}=8\times 10^{-12}$

Thus, $S=2\times 10^{-4}$