Question

If the solubility product of $PbS$ is $8 \times 10^{-28}$, then the solubility of $PbS$ in pure water at $298\, K$ is $x \times 10^{-16} \, mol\, L ^{-1}$. The value of $x$ is ___ . (Nearest Integer)
[Given $\sqrt{2}=1.41$ ]

Answer 1

$ K _{ sp }= S ^2 $
$S =\sqrt{ K _{ sp }}=\sqrt{8 \times 10^{-28}}=2 \sqrt{2} \times 10^{-14}$
$ =2.82 \times 10^{-14} $
$ =282 \times 10^{-16} $
Ans. =$282$