Question

In 1 L saturated solution of $AgCl$( ${{K}_{sp}}AgCl\text{ = 1}\text{.6 x 1}{{\text{0}}^{-10}}$), 0.1 mol of $CuCl$(${{K}_{sp}}CuCl\text{ = 1}\text{.0 x 1}{{\text{0}}^{-6}}$) is added. The resultant concentration of \[A{{g}^{+}}\] in the solution is $1.6\text{ x 1}{{\text{0}}^{-x}}$ . Calculate the value of x.

Answer 1

Hint: Try to recall the concept of partial dissolution shown by weak electrolytes given in electrochemistry chapter. The solubility of weak electrolytes is given by the value of their solubility product (${{K}_{sp}}$). Strong electrolytes completely dissociate in a solution.

Complete step by step solution:
Solubility is defined as the property of a substance (solute) to get dissolved in a solvent to form a solution.
The solubility product constant is the equilibrium constant for the dissolution of a solid solute in a solvent to form a solution. It is denoted by the symbol ${{K}_{sp}}$. We can form an idea about the dissolution of an ionic compound by the value of its solubility product.
In the above question, we see that both the compounds have chloride ions in common. This leads to a common ion effect which will be explained later in the note section.
We will first find the concentration of chloride ions from the solubility product of $CuCl$and then find the concentration of $A{{g}^{+}}$.
                    $\text{CuC}{{\text{l}}_{(s)}}\text{ }\to \text{ C}{{\text{u}}^{+}}\text{ + C}{{\text{l}}^{-}}$
Initial concentration(M): 0.1 0 0
Eq. concentration (M): 0.1-x x x
$\begin{align}
  & {{K}_{sp,CuCl}}\text{ = }\!\![\!\!\text{ C}{{\text{u}}^{+}}]\text{ }\!\![\!\!\text{ C}{{\text{l}}^{-}}] \\
 & {{K}_{sp}}\text{ = }X\text{ x }X\text{ } \\
\end{align}$
The chloride ion concentration will be, $[C{{l}^{-}}]$ = X = $\sqrt{{{K}_{sp,CuCl}}}$= $\sqrt{1.0\text{ x 1}{{\text{0}}^{-6}}}\text{ M }$= $1.0\text{ x 1}{{\text{0}}^{-3}}\text{ M}$
${{K}_{sp,CuCl}}\text{ = }\!\![\!\!\text{ A}{{\text{g}}^{+}}]\text{ }\!\![\!\!\text{ C}{{\text{l}}^{-}}]$
The silver ion concentration is, $\text{ }\!\![\!\!\text{ A}{{\text{g}}^{+}}]$ = $\dfrac{{{K}_{sp,AgCl}}}{[C{{l}^{-}}]}$= $\dfrac{1.6\text{ x 1}{{\text{0}}^{-10}}}{1.0\text{ x 1}{{\text{0}}^{-3}}}$= $1.6\text{ x 1}{{\text{0}}^{-7}}\text{ M}$

Therefore, the value of x is 7.

Note: The common ion effect describes the effect of adding a common ion on the ​equilibrium of the new solution. The common ion effect generally decreases ​solubility of a solute. The equilibrium shifts to the left to relieve off the excess product formed.
When a strong electrolyte is added to a solution of weak electrolyte, the solubility of weak electrolyte decreases leading to the formation of precipitate at the bottom of the testing apparatus.