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

Hint: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.