Question

If the solubility of $SrSO_4$ in water, 0.01 M $Na_2SO_4$ and 0.02 M $SrCl_2$ be $S_1$, $S_2$ and $S_3$ respectively then the correct order of solubility is

• A. $S_1 < S_2 < S_3$
• B. $S_3 < S_2 < S_1$
• C. $S_1 < S_3 < S_2$
• D. $S_3 < S_1 < S_2$

Answer 1

Answer: (B)

$S_3 < S_2 < S_1$

Answer 2

The solubility product, $K_{sp}$, is defined as:

$K_{sp} = [Sr^{2+}][SO_4^{2-}]$

1. In pure water ($S_1$):

   $K_{sp} = S_1^2$
   $S_1 = \sqrt{K_{sp}}$

2. In 0.01 M $Na_2SO_4$ ($S_2$):

   $K_{sp} = S_2 \times 0.01$
   $S_2 = \frac{K_{sp}}{0.01}$

3. In 0.02 M $SrCl_2$ ($S_3$):

   $K_{sp} = 0.02 \times S_3$
   $S_3 = \frac{K_{sp}}{0.02}$

Comparing the solubilities:

$S_1 = \sqrt{K_{sp}}$
$S_2 = \frac{K_{sp}}{0.01}$
$S_3 = \frac{K_{sp}}{0.02}$

Since $K_{sp}$ is a small value, the common ion effect reduces the solubility. Higher the concentration of the common ion, lower is the solubility.

Therefore, $S_3 < S_2 < S_1$.