Question

The solubility of $AgI$ in $NaI$ solution is less than that in pure water.
A. $AgI$ forms complex with $NaI$
B. Of common ion effects.
C. Solubility product of $AgI$ is less than that of $NaI$
D. The temperature of the solution decreases.

Answer 1

We know that $AgI$ and $NaI$ are electrolyte and when they dissolve in the solvent, after some time equilibrium is achieved. Now, if you are aware of the factors affecting the equilibrium as stated by the Le-Chatelier’s principle you can easily attempt the question.

Complete step-by-step answer:
We know that sodium iodide ($NaI$) is a strong electrolyte and when it is dissolved in water it undergoes complete dissociation. It gets completely dissociated in its corresponding ions.
$NaI\left( aq \right)\to N{{a}^{+}}\left( aq \right)+{{I}^{-}}\left( aq \right)$… (equation 1)

Now, we know that $AgI$ is a weak electrolyte and it undergoes partial dissociation. Also. After a few lapses of time there is an equilibrium established.
$AgI\left( aq \right)A{{g}^{+}}\left( aq \right)+{{I}^{-}}\left( aq \right)$…. (equation 2)

The above reaction occurs when both are dissolved individually in water. And the reaction will proceed always in the forward direction unless and until there is disturbance.

Now, if you dissolve $AgI$ in $NaI$ solution, then the solubility of $AgI$ will decrease. The reason for the same is explained below. $NaI$ solution already has sodium and iodide ions dissolved in it. Now if we dissolve $AgI$ in it, then it will also dissolve in the silver and iodide ions. Now, you can see that the concentration of iodide ion $\left( {{I}^{-}} \right)$is increased to a large extent.

From equation 2, we can clearly see that $\left( {{I}^{-}} \right)$ion is the product. So, we can say that product concentration for equation 2 has increased. This disturbs the equilibrium. Now, Le-Chatelier’s principle will come into play. As per the principle, if the product concentration is increased then the reaction moves in a backward direction to neutralize the change.

Therefore, equation 2 moves in a backward direction, and its solubility will decrease. Now, if we note carefully the full reaction, we can see that ${{I}^{-}}$was a common ion for both the reactions, and we also saw that in presence of common ion from a strong electrolyte, the dissociation of weak electrolyte is suppressed and in turn its solubility decreases. This is what exactly is a common-ion effect. The suppression in the dissociation of weak electrolyte by the addition of strong electrolyte having a common ion is called common ion effect.

Note: When the concentration of common ions is increased, the ionic product becomes greater than the solubility product. In order to make the ionic product equal to solubility product equilibrium will shift in the backward direction.