Question

Assertion: The addition of silver ions to a mixture of aqueous sodium chloride and sodium bromide solution will first precipitate $AgBr$ rather than $AgCl$.
Reason: ${K_{SP}}$ of $AgBr$$ < {K_{SP}}$ of $AgCl$
A.Both assertion and reason are correct and the reason is the correct explanation of assertion.
B.Both assertion and reason are correct and the reason is not the correct explanation of assertion.
C.The assertion is correct but the reason is incorrect.
D.Both assertion and reason are incorrect.

Answer 1

Hint: Solubility product of an electrolyte at a particular temperature is defined as the product of the molar concentration of its ion in a saturated solution, each concentration raised to the power equal to several ions produced on the dissociation of an electrolyte.

Complete step by step answer:
As we know the solubility product of any less soluble salt can be determined with the knowledge of solubility of that salt at a particular temperature. Solubility products help in predicting ionic or precipitation of reaction. If the ionic product exceeds the solubility product of sparingly soluble salt the excess ion will combine resulting in the formation of a precipitate or we can say the species having a minimum value of solubility product will precipitate first. In the case of $AgBr$ and $AgCl$, the solubility product $AgBr$ is lesser than the solubility product of $AgCl$ hence on adding the silver ion in a mixture of sodium bromide and sodium chloride solution, $AgBr$ will precipitate first rather than $AgCl$.
Hence assertion and reason both are correct and the reason is the explanation of assertion.

Hence, option A is correct.


Note:
There are some other applications of solubility product given as :
 Solubility product helps to find the solubility of sparingly soluble salts at a particular temperature.
We can also purify the common salt with the help of solubility product as:
In the purification of common salt, $HCl$ gas is circulated through the saturated solution of common salt. Here $HCl$ and $NaCl$ dissociated into ions as:
$NaCl \to N{a^ + } + C{l^ - }$
$HCl \to {H^ + } + C{l^ - }$
As the concentration of chloride ions increases in the solution due to dissociation of $HCl$ and due to common ion effect, dissection of sodium chloride decreases. As a result, the ionic product of sodium and chloride ions exceeds the solubility product of $NaCl$ and hence pure $NaCl$ will precipitate out.