Question

The solubility of $AgI$ in $Nal$ solution is less than that in pure water because:

• A. $AgI$ forms complex with $NaI$
• B. le-chatelier's principle
• C. solubility product of $AgI$ is less than that of $Nal$
• D. the temperature of the solution decreases

Answer 1

Answer: (B)

le-chatelier's principle

The reaction for the dissociation of silver iodide is,
$AgI\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+I^{-}\left(aq\right)$
The expression for the dissociation of silver iodide is, $NaI\to N{a}^{+}+I^{-}$.
Thus iodide ion acts as common ion. Due to presence of the common ion, the equilibrium for the dissociation of AgI is shifted backwards.
This is because, the increase in $\left[I^{-}\right]$ brings in an increase in the solubility product for AgI which is $\left[A{g}^{+}\right]\left[I^{-}\right]$.
This is in accordance with Le-chatlier's principle.