Question

The solubility of $A{s}_2{S}_3$ is observed to be $\frac{1}{48}\times {10}^{-9}M$ in a solution of ${10}^{–2}M$ $N{a}_{2}S$ solution at $25°\text{C}$, assuming no hydrolysis of cationic or anionic part. The $K_{sp}$ value of$A{s}_2{S}_3$ at $25°\text{C}$ is:

• A. $\frac{1}{576}\times {10}^{-24}$
• B. $\frac{1}{36}\times {10}^{-10}$
• C. $\frac{1}{44}\times {10}^{-22}$
• D. $\frac{1}{24}\times {10}^{-9}$

Answer 1

The correct option is A $\frac{1}{576}\times {10}^{-24}$

$A{s}_2{S}_3\rightleftharpoons 2A{s}^{3+}+3S^{2-}2s3s+{10}^{-2}\approx {10}^{-2}$
$\therefore K_{sp}=(2s{)}^2\times ({10}^{-2}{)}^3$
$=4\times {10}^{-6}\times {(\frac{1}{48}\times {10}^{-9})}^2=\frac{1}{576}\times {10}^{-24}$