Question

K$_{sp}$ of AgCl is . Its solubility in 0.1 M KNO$_3$ will be:
A.10$^{-5}$ moles/litre
B.>10$^{-5}$ moles/litre
C.<10$^{-5}$ moles/litre
D.None of these

Answer 1

Hint: We know that on dissolving, most salts dissociate into ions. The solubility product constant depends on the dissociation. So, dissociate the AgCl into the ions, and solubility could be known.

Complete step by step answer:
-Firstly, let us talk about the solubility product constant (K$_{sp}$). It tells us that the solid AgCl when in equilibrium with its saturated solution, the product of concentrations of ions of both silver and chloride is equal to the solubility product constant.
-Now, on the basis of definition we will find the solubility of AgCl in 0.1 M KNO$_3$.
-As we know, KNO$_3$ is salt of strong acid, and strong base, whereas AgCl is salt of strong acid, and strong base. Thus, we can say that it shows no common ion effect.
-According to the definition, AgCl is dissociating into silver and chloride ions because KNO$_3$ doesn’t affect the equilibrium process. We can write it as
            AgCl $\rightleftharpoons$ Ag$^{+}$ + Cl$^{-}$
Now, consider the solubility of both the dissociated ions.
Thus, K$_{sp}$ = s$\times$s,
       1 $\times$ $^{-10}$ = s$^{2}$
So, s= 10$^{-5}$ moles/litre.

Therefore, the solubility of AgCl in 0.1 M KNO$_3$ is 10$^{-5}$ moles/litre. The correct option is A.

Note: Don’t get confused about why there was no effect of KNO$_3$, it doesn’t show any effect because there are no common ions in the formation, or in the dissociation, termed as the common ion effect. Thus, it represents no common ion effect. The solubility constant for silver, and chloride ion is considered to be s, as mentioned in the equilibrium concentration of ions is equal.