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The $\mathrm{K_{s p}}$ for bismuth sulphide $\left(\mathrm{Bi}_{2} \mathrm{S}_{3}\right)$ is $1.08 \times 10^{-73}$ . The solubility of $\mathrm{Bi}_{2} \mathrm{S}_{3}$ in $\mathrm{mol L^{-1}}$ at 298 K is
Option: 1
$1.0 \times 10^{-15}$

Option: 2
$2.7 \times 10^{-12}$

Option: 3
$3.2 \times 10^{-10}$

Option: 4
$4.2 \times 10^{-8}$

Answers (1)

best_answer

Let, the solubility of $\mathrm{Bi_{2}S_{3}}$ be $\mathrm{s\ mol\ L^{-1}}$

So,

$\\\mathrm{Bi}_{2} \mathrm{~S}_{3} \rightleftharpoons 2 \mathrm{Bi}^{3+}+\mathrm{3S^{2-}}$

$\text{2s}$ $\text{3s}$

$\begin{aligned} \mathrm{k_{s p}=(2 s)^{2}(3 s)^{3} }& \Rightarrow 108 \mathrm{s^{5}}=1.08 \times 10^{-73} \\ \end{aligned}$

$\begin{aligned} & \Rightarrow \mathrm{s^{5}}=\frac{1.08 \times 10^{-73}}{108} \\ & \Rightarrow \mathrm{s=10^{-15}=1.0 \times 10^{-15}} \end{aligned}$

Hence, the correct answer is Option (1)

Posted by

Ritika Kankaria

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