Chemical and Ionic Equilibrium 2 Question 50
50. 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 $\mathrm{CuCl}\left[K_{\text {sp }}(\mathrm{CuCl})=1.0 \times 10^{-6}\right]$ is added. The resultant concentration of $\mathrm{Ag}^{+}$in the solution is $1.6 \times 10^{-x}$. The value of ’ $x$ ’ is
(2011)
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Solution:
- 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 $\mathrm{CuCl}$.
$$ \left[\mathrm{Cl}^{-}\right]=\sqrt{K_{\mathrm{sp}}(\mathrm{CuCl})}=10^{-3} \mathrm{M} $$
Now, for $\mathrm{AgCl}, K_{\text {sp }}=1.6 \times 10^{-10}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]$
$$ \begin{aligned} & =\left[\mathrm{Ag}^{+}\right] \times 10^{-3} \ & \Rightarrow \quad\left[\mathrm{Ag}^{+}\right]=1.6 \times 10^{-7} \end{aligned} $$
