Question

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

• 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)

$\left[\text{I}^{-}\right]_{\text{solution}_{\text{I}}} = 1 0^{- 3} \text{M}$
$\left[\text{Cl}^{-}\right]_{\text{solution}_{\text{I}}} = 1 0^{- 3} \text{M}$
$\left(\left[\left(\text{Ag}\right)^{+}\right]\right)_{\text{solution } \left(\text{II}\right)} =$ $\sqrt{\text{K}_{\text{sp}}} = 1 0^{- 8} \text{M}$
$\left(\left[\left(\text{I}\right)^{-}\right]\right)_{\text{solution } \left(\text{II}\right)} =$ $\sqrt{\text{K}_{\text{sp}}} = 1 0^{- 8} \text{M}$ . on mixing the equal volume of two solutions.
$\left[\text{I}^{-}\right] = \frac{1 0^{- 3} + 1 0^{- 8}}{2} = \frac{1 0^{- 3} \left[1 + 1 0^{- 5}\right]}{2}$ $\approx \, 5\times 10^{- 4} \, \text{M}$
$\left[\text{Cl}^{-}\right] = \frac{1 0^{- 3}}{2} \text{M}$
$\left[\text{Ag}^{+}\right] =$ $ \, \frac{10^{- 16}}{5 \times 10^{- 4}} \, \, =2\times 10^{- 13}$
$\left[\text{K}_{\text{IP}}\right]_{\text{AgI}} = \frac{1 0^{- 8}}{2} \times \frac{1 0^{- 3} \left[1 + 1 0^{- 5}\right]}{2} \approx \frac{1 0^{- 1 1}}{4} > \text{K}_{\text{sp}} \text{AgI}$
$\left[\text{K}_{\text{IP}}\right]_{\text{AgCl}} = \frac{1 0^{- 8}}{2} \times \frac{1 0^{- 3}}{2} \approx \frac{1 0^{- 1 1}}{4} < \text{K}_{\text{sp}} \text{AgCl}$
Hence $\text{AgI}$ will be precipitated, but $\text{AgCl}$ will not be precipitated.