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In which of the following, the solubility of \(\mathrm{AgCl}\) will be minimum?
Options:
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2849 Upvotes
Verified Answer
The correct answer is:
\(0.1 \mathrm{M} \mathrm{KCl}\)
Solubility of \(\mathrm{AgCl}\) in \(0.1 \mathrm{M} \mathrm{KCl}\) will be less because of common ion effect of \(\mathrm{Cl}^{-}\)
\(\begin{aligned}
& \mathrm{AgCl} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Cl}^{-\prime} \quad \text{(Common ion)} \\
& \mathrm{KCl} \rightleftharpoons \mathrm{K}^{+}+{ }^{\prime} \mathrm{Cl}^{-}{ }^{\prime} \\
\end{aligned}\)
Hence, option (2) is correct.
\(\begin{aligned}
& \mathrm{AgCl} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Cl}^{-\prime} \quad \text{(Common ion)} \\
& \mathrm{KCl} \rightleftharpoons \mathrm{K}^{+}+{ }^{\prime} \mathrm{Cl}^{-}{ }^{\prime} \\
\end{aligned}\)
Hence, option (2) is correct.
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