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In $1 \mathrm{~L}$ saturated solution of $\mathrm{AgCl}\left[K_{\mathrm{sp}} \mathrm{AgCl}=1.6 \times 10^{-10}\right], 0.1$ mole of $\mathrm{CuCl}\left[K_{\mathrm{sp}}(\mathrm{CuCl})=1.0 \times 10^{-6}\right]$ is added. The resultant concentration of $\mathrm{Ag}^{+}$in the solution is $1.6 \times 10^{-x}$. The value of $x$ is
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It is a case of simultaneous solubility of salts with a common ion. Here solubility product of $\mathrm{CuCl}$ is much greater than that of $\mathrm{AgCl}$, it can be assumed that $\mathrm{Cl}^{-}$in solution comes mainly from $\mathrm{CuCl}$.
$$
\Rightarrow \quad\left[\mathrm{Cl}^{-}\right]=\sqrt{K_{s p}(\mathrm{CuCl})}=10^{-3} \mathrm{M}
$$
Now for $\mathrm{AgCl}: K_{s p}=1.6 \times 10^{-10}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=\left[\mathrm{Ag}^{+}\right] \times 10^{-3} \Rightarrow\left[\mathrm{Ag}^{+}\right]=1.6 \times 10^{-7}$
$$
\Rightarrow \quad\left[\mathrm{Cl}^{-}\right]=\sqrt{K_{s p}(\mathrm{CuCl})}=10^{-3} \mathrm{M}
$$
Now for $\mathrm{AgCl}: K_{s p}=1.6 \times 10^{-10}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=\left[\mathrm{Ag}^{+}\right] \times 10^{-3} \Rightarrow\left[\mathrm{Ag}^{+}\right]=1.6 \times 10^{-7}$
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