Question

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

Answer 1

$AgI$ and $NaI$ have something which is quite similar and just because of which it faces a problem and then the solubility is decreased or gets lesser than in pure water. Silver iodide is a compound which is inorganic in nature and Sodium iodide is a soluble compound in water and it is an ionic compound.

Complete step-by-step answer: The solubility of $AgI$ in $NaI$ solution is less than that in pure water because solubility is reduced because of common ion effect.
The common ion effect states the effect on ​ equilibrium which occurs when a common ion (an ion which is already stored/contained in the solution) is added to a solution. The common ion effect often decreases ​ the solubility of a solute.

ADDITIONAL INFORMATION:
Silver iodide is a compound which is inorganic in nature with the compound name $AgI$ . The compound is a bright yellow in colour and solid in nature, however samples mostly always contain impurities of metallic silver colour that gives a grey colour impact. The silver contamination comes as $AgI$ is highly photosensitive in nature.
Sodium iodide ( $NaI$ ) is soluble in water and it is an ionic compound which consists of an equal addition of sodium as a cation ( $N{a^ + }$ ) and iodide as an anion ( ${I^ - }$ ) in its crystal space. It is produced in the industries by creating the reaction of acidic iodides with sodium hydroxide ( $NaOH$ ) to form $NaI$ salts.

Hence, Option B is the right answer.

Note: Observe the hint carefully and check the similarity between these two compounds which can clearly tell the reason of lesser solubility and it is the common ion effect which is very known and common. The common ion effect often decreases ​ the solubility of a solute.