Question

The concentrations of $KI$and $KCl$ in a certain solution containing both is 0.001 M each. If 20 mL of this solution is added to 20 mL of a saturated solution of $AgI$ in water. What will happen?
$({{K}_{sp}}AgCl={{10}^{-10}};{{K}_{sp}}AgI={{10}^{-16}})$
A. $AgI$will be precipitated.
B. $AgCl$ will not be precipitated.
C. There will be no precipitate.
D. Both $AgCl$ and $AgI$ will be precipitated.

Answer 1

Hint: A saturated solution is a solution that has been dissolved as much solute as it is capable of dissolving. In a saturated solution, no more solute can be dissolved at a given temperature. We are able to make a saturated solution by dissolving the solute until no more solute can be dissolved.

Complete Step by Step Answer:
Final concentration of ${{I}^{-}}$i.e. [${{I}^{-}}$] = $\dfrac{0.001\times 20}{40}M$ = $0.0005M$
Therefore, Final concentration $[C{{l}^{-}}]=0.0005M$
If the solution of $KCl$and $KI$is added to a saturated solution of $AgI$
$[A{{g}^{+}}]=\sqrt{1\times {{10}^{-16}}}=1\times {{10}^{-8}}$$AgI\rightleftharpoons A{{g}^{+}}+{{I}^{-}}$
Then ${{K}_{sp}}(AgI)=[A{{g}^{+}}][{{I}^{-}}]=1\times {{10}^{-16}}$
But $[A{{g}^{+}}]=[{{I}^{-}}]$
due to the lower ${{K}_{sp}}$ value of $AgI$.
${{K}_{sp}}$($AgI$) = ${{10}^{-16}}$
Hence $[A{{g}^{+}}]=\sqrt{1\times {{10}^{-16}}}=1\times {{10}^{-8}}$
Thus $AgI$will be precipitated, since here ${{I}^{-}}$> ${{10}^{-8}}$
Hence, Option (A) is correct.

Note: Solubility product is a type of equilibrium constant whose value depends on the temperature. It is denoted by ${{K}_{sp}}$. It depends upon the temperature. It usually increases with the increase in temperature because of the increased solubility. It is an equilibrium constant that gives the equilibrium between the solid solute and its ions that are dissolved in the solution.