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The $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{CuS}, \mathrm{Ag}_{2} \mathrm{~S}$ and $\mathrm{HgS}$ are $10^{-31}, 10^{-44}$ and $10^{-54}$ respectively. The solubility of these sulphides are in the order :
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Verified Answer
The correct answer is:
$\quad \mathrm{Ag}_{2} \mathrm{~S}>\mathrm{CuS}>\mathrm{HgS}$
For CuS, solubility $=\left(10^{-31}\right)^{1 / 2}$;
$$
\begin{array}{l}
\text { For } \mathrm{Ag}_{2} \mathrm{~S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{4}\right)^{\frac{1}{3}}=\left(\frac{10^{-44}}{4}\right)^{\frac{1}{3}} \text { and } \\
\text { for } \mathrm{HgS}=\left(10^{-54}\right)^{\frac{1}{2}}
\end{array}
$$
$$
\begin{array}{l}
\text { For } \mathrm{Ag}_{2} \mathrm{~S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{4}\right)^{\frac{1}{3}}=\left(\frac{10^{-44}}{4}\right)^{\frac{1}{3}} \text { and } \\
\text { for } \mathrm{HgS}=\left(10^{-54}\right)^{\frac{1}{2}}
\end{array}
$$
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