Chemical And Ionic Equilibrium Result Question 50
50. In $1$ 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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Answer:
Correct Answer: 50. ( $ 1.6 \times 10^{-7} $ )
Solution:
- It is a case of simultaneous solubility of salts with a common ion. Here, solubility product of $CuCl$ is much greater than that of $AgCl$ , it can be assumed that $Cl -$ in solution comes mainly from $CuCl$ .
$ \left[\mathrm{Cl}^{-}\right]=\sqrt{K_{\text {sp }}(\mathrm{CuCl})}=10^{-3} \mathrm{M} $
Now, for $\mathrm{AgCl}, K_{\mathrm{sp}}=1.6 \times 10^{-10}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]$
$ \begin{aligned} =\left[\mathrm{Ag}^{+}\right] \times 10^{-3} \ \Rightarrow \left[\mathrm{Ag}^{+}\right] & =1.6 \times 10^{-7} \end{aligned} $
BETA
