Question

The concentration of $KI$ and $KCl$ in a certain solution containing both is $0.001M$ each. If $20mL$ of this solution is added to $20mL$ of a saturated solution of $AgI$ in water. What will happen ?
$\left(K_{sp}ofAgCl={\left(10\right)}^{-10};K_{sp}ofAgI={\left(10\right)}^{-16}\right)$

• A. $\text{AgI}$ will be precipitated
• B. $\text{AgCl}$ will be precipitated
• C. Both $\text{AgCl}$ and $\text{AgI}$ will be precipitated
• D. There will be no precipitate

Answer 1

Answer: (A)

${\left[{\text{I}}^{-}\right]}_{{\text{solution}}_{\text{I}}}=10^{-3}\text{M}$
${\left[{\text{Cl}}^{-}\right]}_{{\text{solution}}_{\text{I}}}=10^{-3}\text{M}$
${\left(\left[{\left(\text{Ag}\right)}^{+}\right]\right)}_{\text{solution }\left(\text{II}\right)}=$ $\sqrt{{\text{K}}_{\text{sp}}}=10^{-8}\text{M}$
${\left(\left[{\left(\text{I}\right)}^{-}\right]\right)}_{\text{solution }\left(\text{II}\right)}=$ $\sqrt{{\text{K}}_{\text{sp}}}=10^{-8}\text{M}$ . on mixing the equal volume of two solutions.
$\left[{\text{I}}^{-}\right]=\frac{10^{-3}+10^{-8}}{2}=\frac{10^{-3}\left[1+10^{-5}\right]}{2}$ $\approx 5\times 10^{-4}\text{M}$
$\left[{\text{Cl}}^{-}\right]=\frac{10^{-3}}{2}\text{M}$
$\left[{\text{Ag}}^{+}\right]=$ $\frac{10^{-16}}{5\times 10^{-4}}=2\times 10^{-13}$
${\left[{\text{K}}_{\text{IP}}\right]}_{\text{AgI}}=\frac{10^{-8}}{2}\times \frac{10^{-3}\left[1+10^{-5}\right]}{2}\approx \frac{10^{-11}}{4}>{\text{K}}_{\text{sp}}\text{AgI}$
${\left[{\text{K}}_{\text{IP}}\right]}_{\text{AgCl}}=\frac{10^{-8}}{2}\times \frac{10^{-3}}{2}\approx \frac{10^{-11}}{4}<{\text{K}}_{\text{sp}}\text{AgCl}$
Hence $\text{AgI}$ will be precipitated, but $\text{AgCl}$ will not be precipitated.