Question

In $1L$ saturated solution of \[AgCl\] $\left[ {{K_{sp}}(AgCl) = 1.6 \times {{10}^{ - 10}}} \right]$ , $0.1mol$ of $CuCl\left[ {{K_{sp}}(CuCl) = 1 \times {{10}^{ - 6}}} \right]$is added. The resultant concentration of $A{g^ + }$ in the solution is $1.6 \times {10^{ - x}}$ . The value of $x$ is:

Answer 1

Hint: ${K_{sp}}$ , also known as solubility product constant, is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. More soluble substances have higher ${K_{sp}}$ value.

Complete step by step answer:
We have given the solubility product constants for silver chloride and copper chloride which are as following:
\[{K_{sp}}(AgCl) = 1.6 \times {10^{ - 10}}\] and ${K_{sp}}(CuCl) = 1 \times {10^{ - 6}}$.
And the final or resultant concentration of $A{g^ + }$ is $1.6 \times {10^{ - x}}$. -------------(1)
So, we have to calculate the value of $x$.
Now, for copper chloride
Initially we have given $0.1M$of copper chloride which dissociates and becomes $0.1 - yM$at equilibrium.
                       $CuCl \to C{u^ + } + C{l^ - }$

initially	$0.1M$	$0$	$0$
At. eqm.	$0.1 - y$	$y$	$y$

So, we can write ${K_{sp}}$ for this,
 ${K_{sp}} = \left[ {C{u^ + }} \right]\left[ {C{l^ - }} \right]$
At equilibrium the concentrations of both copper ion and chloride ion becomes $y$.
$ \Rightarrow {K_{sp}} = y \times y$
$ \Rightarrow y = \sqrt {{K_{sp}}} = \sqrt {{{10}^{ - 6}}} = {10^{ - 3}}M$
Similarly the ${K_{sp}}$ for $AgCl$ will be
${K_{sp}} = \left[ {A{g^ + }} \right]\left[ {C{l^ - }} \right]$
$ \Rightarrow \left[ {A{g^ + }} \right] = \dfrac{{{K_{sp}}}}{{\left[ {C{l^ - }} \right]}} = \dfrac{{1.6 \times {{10}^{ - 10}}}}{{{{10}^{ - 3}}}} = 1.6 \times {10^{ - 7}}$ --------------(2)
Now comparing (1) with (2) we get,
$1.6 \times {10^{ - x}} = 1.6 \times {10^{ - 7}}$
$ \Rightarrow x = 7$
Hence the concentration of silver ion is $1.6 \times {10^{ - 7}}$ and the value of $x$ is $7$ .


Note:
${K_{sp}}$ is used to describe the saturated solution of ionic compounds. In order to calculate the solubility product constant we need to convert all the concentrations of each and every ion in terms of molarity or moles per liter.