Question

The solubility product of a sparingly soluble salt $A{X}_2$ is $3.2\times {10}^{11}$. Its solubility $\left(moles/litre\right)$ is:

• A. $3.1\times {10}^{-4}$
• B. $2\times {10}^3$
• C. $4\times {10}^{-4}$
• D. $5.6\times {10}^{-6}$

Answer 1

Answer: (B)

$2\times {10}^3$

Let 's' be the solubility of a salt such as $A{X}_2$ which undergoes dissociation as follows :-

$A{X}_2\to A^{+2}+2X^{-}$

(Solubility) 1-s s 2s

Thus, the solubility product $K_{sp}$ is $s\times (2s{)}^2$

$\therefore K_{sp}=4s^3$

$\therefore 3.2\times {10}^{10}=4s^3$

$\therefore s^3=\frac{3.2\times {10}^{10}}{4}$

$\therefore s^3=0.8\times {10}^{10}$

Thus taking cube root on both the sides we get,

$s=2\times {10}^3$