In 1L saturated solution of $AgCl\ \lbrack K_{sp}(AgCl)1.6\ 10^{10}\rbrack$, 0.1 mole of $CuCl\ \lbrack K_{sp}(CuCl)1.0\ 10^{6}\rbrack$ is added. The resultant concentration of Ag in the solution is $1.6\ 10^{x}$. The value of $'x'$ is

Question

Dr. Abubacker · Accepted Answer

(c)
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 $CuCl$.
$\Rightarrow$ $\left\lbrack Cl^{-} ightbrack = \sqrt{K_{sp}(CuCl)} = 10^{- 3}\ M$
Now for $AgCl\ :K_{sp} = 1.6 	imes 10^{- 10}$
$\  = \left\lbrack Ag^{+} ightbrack\lbrack Cl^{-}brack$
$= \left\lbrack Ag^{+} ightbrack 	imes 10^{- 3}$
$\Rightarrow$ $\left\lbrack Ag^{+} ightbrack = 1.6 	imes 10^{- 7}$

Q. In 1L saturated solution of $AgCl\ \lbrack K_{sp}(AgCl)1.6\ 10^{10}\rbrack$, 0.1 mole of $CuCl\ \lbrack K_{sp}(CuCl)1.0\ 10^{6}\rbrack$ is added. The resultant concentration of Ag in the solution is $1.6\ 10^{x}$. The value of $'x'$ is

Explanation

Reviewed by Dr. Abubacker

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