Question

Let the solubilities of $AgCl$ in $H_{2}O$; $0.01M$ $Ca{Cl}_2$; $0.01M$ $NaCl$ and $0.05M$ $Ag{NO}_3$ be $s_1,s_2,s_3$ and $s_4$ respectively. What is the correct relationship between these quantities?

• A. $s_1>s_2>s_3>s_4$
• B. $s_1>s_2=s_3>s_4$
• C. $s_1>s_3>s_2>s_4$
• D. $s_4>s_2>s_3>s_1$

Answer 1

Answer: (C)

$s_1>s_3>s_2>s_4$

Solubility of $AgCl$ inwater $=\sqrt{K_{sp}}=s_1$

In $0.01M$ $Ca{Cl}_2$, it is given by
$K_{sp}=s\times (0.01\times 2+s)$ ($\because s_2=\frac{K_{sp}}{0.02}$)

In $0.01M$ $NaCl$, it is given by
$K_{sp}=s\times (0.01+s)$ ($\because s_3=\frac{K_{sp}}{0.01}$)

In $0.05M$ $Ag{NO}_3$, it is given by
$K_{sp}=s\times (0.05+s)$ ($\because s_4=\frac{K_{sp}}{0.05}$)

The solubilities are derived by neglecting $s$ in comparison to $0.02,0.01$ and $0.05$.

So order is:
$s_1>s_3>s_2>s_4$

Option C is correct.