Question

The solubility product of a sparingly soluble salt $A_2B$ 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} $

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

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

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

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

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