Question

Increasing order of solubility of AgCl in (i) H$_2$O, (ii) 1M NaCl (aq.), (iii) 1M CaCl$_2$ (aq.), and (iv) 1M NaNO$_3$ (aq.) solution

• A. CaCl$_2$ $<$ NaNO$_3$ $<$ NaCl $<$ H$_2$O
• B. CaCl$_2$ $>$ H$_2$O $>$ NaCl $>$ NaNO$_3$
• C. CaCl$_2$ $>$ NaCl $>$ H$_2$O $>$ NaNO$_3$
• D. CaCl$_2$ $<$ NaCl $<$ H$_2$O $<$ NaNO$_3$

Hint:

Solubility rules: \begin{itemize} \item Common ion $\downarrow$ solubility \item Higher concentration common ion $\Rightarrow$ stronger effect \item Inert salts may increase solubility via ionic strength \end{itemize}

Answer 1

The Correct Option is D

Concept: Solubility of sparingly soluble salts depends on:

• Common ion effect (reduces solubility)
• Ionic strength (inert electrolyte may increase solubility)

Step 1: Identify common ions AgCl \(\rightleftharpoons\) Ag$^+$ + Cl$^-$ Solutions containing Cl$^-$ reduce solubility.

• 1M CaCl$_2$ \(\Rightarrow\) highest Cl$^-$ (strongest suppression)
• 1M NaCl \(\Rightarrow\) moderate suppression
• Pure water \(\Rightarrow\) normal solubility
• 1M NaNO$_3$ \(\Rightarrow\) no common ion

Step 2: Effect of inert electrolyte NaNO$_3$ increases ionic strength \(\Rightarrow\) slightly increases solubility. Conclusion: \[ \text{CaCl}_2 < \text{NaCl} < \text{H}_2\text{O} < \text{NaNO}_3 \]