Question

If the solubility product of PbS is 8 × 10–28, then the solubility of PbS in pure water at 298 K is x × 10–16 mol L–1. The value of x is ________. (Nearest integer)
[given : $\sqrt2=1.41$]

Answer 1

$\text{PbS(s)} \rightleftharpoons \text{Pb}^{2+}(\text{aq}) + \text{S}^{2-}(\text{aq})$
$Ksp = S2$
$8 \times 10^{-28} = \text{S}^{2}$
$S = 2\sqrt{2} \times 10^{-14} \, \text{mol/L}$
$2.82 \times 10^{-14} \, \text{mol/L} = 282 \times 10^{-16} \, \text{mol/L}$
Hence, $x = 282$$$