Question

What will be the solubility of ${\text{AgCl}}$ in a 0.1 M ${\text{KCl}}$ solution? (${{\text{K}}_{{\text{sp}}}}{\text{ AgCl = 1}}{\text{.20}} \times {\text{1}}{{\text{0}}^{{\text{ - 10}}}}$)
A. $\text{0.1 M}$
B. ${\text{1}}{\text{.2}} \times {\text{1}}{{\text{0}}^{{\text{ - 4}}}}$M
C. ${\text{1}}{\text{.2}} \times {\text{1}}{{\text{0}}^{{\text{ - 9}}}}$ M
D. ${\text{1}}{\text{.2}} \times {\text{1}}{{\text{0}}^{{\text{ - 10}}}}$M

Answer 1

Write the dissociation reaction of ${\text{AgCl(s)}}$. Using the concentration of ${\text{KCl}}$ solution calculate the initial concentration of ${\text{C}}{{\text{l}}^{\text{ - }}}$ ions. Using the solubility product of ${\text{AgCl(s)}}$ and initial concentration of ${\text{C}}{{\text{l}}^{\text{ - }}}$ ions calculate the solubility of ${\text{AgCl(s)}}$.

Formula Used: ${{\text{K}}_{{\text{sp}}}}{\text{ = }}[{\text{A}}{{\text{g}}^ + }]{\text{[C}}{{\text{l}}^{\text{ - }}}]$

Complete answer:
For the given ${\text{AgCl(s)}}$ salt we have to write the balanced dissociation reaction.
${\text{AgCl(s)}} \rightleftharpoons {\text{ A}}{{\text{g}}^ + }{\text{(aq)}} + {\text{ C}}{{\text{l}}^ - }{\text{(aq)}}$
We have to determine the solubility of ${\text{AgCl(s)}}$ in 0.1M ${\text{KCl}}$ solution. As we know ${\text{KCl}}$ is a strong electrolyte so will completely dissociate into ${{\text{K}}^{\text{ + }}}$ and ${\text{C}}{{\text{l}}^{\text{ - }}}$ ions.
So, the initial concentration of ${\text{C}}{{\text{l}}^{\text{ - }}}$ ions = 0.1M
The solubility product of ${\text{AgCl(s)}}$ given to us is ${\text{1}}{\text{.20}} \times {\text{1}}{{\text{0}}^{{\text{ - 10}}}}$. The smaller value of the solubility product indicates that ${\text{AgCl(s)}}$ is slightly soluble.
Assume solubility of ${\text{AgCl(s)}}$ as ‘s’ M
So,‘s’ M of ${\text{AgCl(s)}}$ after dissolution will give ‘s’ M${\text{A}}{{\text{g}}^ + }$ and ‘s’ M ${\text{C}}{{\text{l}}^ - }$.
We have 0.1M ${\text{C}}{{\text{l}}^{\text{ - }}}$ions from ${\text{KCl}}$ also so at equilibrium concentration of ${\text{C}}{{\text{l}}^ - }$ is (0.1+s) M
So, at equilibrium we have
$[{\text{A}}{{\text{g}}^ + }]{\text{ = ‘s’M}}$
${\text{[C}}{{\text{l}}^{\text{ - }}}] = (0.1 + s){\text{M}}$
Now, we will set up the solubility product equation for ${\text{AgCl(s)}}$ as follows:
${\text{Ksp = }}[{\text{A}}{{\text{g}}^ + }]{\text{[C}}{{\text{l}}^{\text{ - }}}]$
Here, ${\text{Ksp}}$ = solubility product
Now we have to substitute ${\text{1}}{\text{.20}} \times {\text{1}}{{\text{0}}^{{\text{ - 10}}}}$ for solubility product, ‘s’ M for the concentration of ${\text{A}}{{\text{g}}^ + }$ ion and ($0.1 + {\text{s}}){\text{M}}$ for the concentration of ${\text{C}}{{\text{l}}^ - }$ .
${\text{1}}{\text{.20}} \times {\text{1}}{{\text{0}}^{{\text{ - 10}}}} = ({\text{s M}})(0.1 + {\text{s}}){\text{M}}$
Since the solubility of ${\text{AgCl(s)}}$ is very less, we can neglect’ from 0.1+s
So, the equation will become as follows
${\text{1}}{\text{.20}} \times {\text{1}}{{\text{0}}^{{\text{ - 10}}}} = ({\text{‘s’M}})0.1{\text{M}}$
So, ${\text{s = 1}}{\text{.20}} \times {\text{1}}{{\text{0}}^{{\text{ - 9}}}}{\text{M}}$
Thus, the solubility of ${\text{AgCl(s)}}$ is ${\text{1}}{\text{.2}} \times {\text{1}}{{\text{0}}^{{\text{ - 9}}}}{\text{M}}$.

**Hence, the correct option is (C) ${\text{1}}{\text{.2}} \times {\text{1}}{{\text{0}}^{{\text{ - 9}}}}{\text{M}}$

Note:**
Due to the presence of common ion solubility of salt decreases. Here due to the presence of a common ion ${\text{C}}{{\text{l}}^ - }$ the solubility of ${\text{AgCl(s)}}$ decreases. To simplify the calculation we have neglected ’s’ from 0.1+s else we would have ended up with a quadratic equation.