The $K_{sp}$ for bismuth sulphide $({Bi}_{2} S_{3})$ is $1.08 \times 10^{- 73}$ . The solubility of ${Bi}_{2} S_{3}$ in ${molL}^{- 1}$ at 298 K is

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$3.2 \times 10^{- 10}$

$4.2 \times 10^{- 8}$

$2.7 \times 10^{- 12}$

$1.0 \times 10^{- 15}$

answer is A.

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Detailed Solution

$\begin{aligned} {Bi}_{2} S_{3} & ⇌ 2 {Bi}^{3 +} + 3 S^{2 -} \\ k_{sp} & = (2 s)^{2} (3 s)^{3} \\ = 4 s \\ = 108 (s \times 27 (s)^{3} \\ (s)^{5} & = \frac{1.08 \times 10^{- 73}}{108} \\ \Rightarrow s = 10^{- 15} \end{aligned}$

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