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Q. Let the solubilities of $AgCl$ in $H _2 O$, and $0.01 \,M\, CaCl _2, 0.01 \,M \,NaCl$, and $0.05 \,M \,AgNO _3$ be $S_1, S_2, S_3, S_4$, respectively. What is the correct relationship between these quantities

Equilibrium

Solution:

i. $AgCl + H _2 O \longrightarrow Ag ^{\oplus}( aq )+ Cl ^{\ominus}( aq ) \ldots S_1$
ii. $AgCl$ in $0.01 \,M \,CaCl _2, \ldots S_2$
Concentration of $Cl ^{\Theta}=2 \times 0.01=0.02 \,M$
iii. $AgCl$ in $0.01\, M \,NaCl \ldots S_3$
Concentration of $Cl ^{\ominus}=0.01 \,M$
iv. $AgCl$ is $0.05 \,M , AgNO _3, \ldots S_4$
Concentration of $Ag ^{\oplus}=0.05$
Since both $Cl ^{\ominus}$ ion and $Ag ^{\oplus}$ ions acts as common ion. So larger the concentration of $Ag ^{\oplus}$ or $Cl ^{\ominus}$ ions, more is the suppression of ionization of $AgCl$ and hence less will be solubility of $AgCl$
$\therefore$ Solubility order
$S_1>S_3>S_2>S_4$