Question

The correct order of increasing solubility of $AgCl$ is in
A. Water
B. $0.1M$ $NaCl$
C. $0.1M$ $BaC{l_2}$
D. $0.1M$ $N{H_3}$
1. $A < B < C < D$
2. $B < C < A < D$
3. $C < B < D < A$
4. $C < B < A < D$

Answer 1

We know that solubility is the ability of a substance, the solute to dissolve in a solvent. Here to determine the increasing solubility of a solution we will use the concept of common ion effect which states that on the addition of a solution into a soluble compound, the solubility will decrease if common ions are present.

Complete answer:
$AgCl$ will dissociate into $A{g^ + }$ and $C{l^ - }$ when dissolved in a solution.
Here as we can see that $NaCl$ and $BaC{l_2}$, both contain the chloride ion which is common to $AgCl$ so the solubility of $AgCl$ will decrease. Common ion effects will take place in this case and the reaction will shift in the backward direction. Greater the concentration of chloride ions, greater will be the decrease in the solubility. Here $BaC{l_2}$ will dissociate into two chloride ions hence the decrease will be more when $AgCl$ will be dissolved in it. When $AgCl$ is dissolved in $N{H_3}$ ,it will form a complex with ammonia i.e. ${[Ag{(N{H_3})_2}]^ + }$ so it will be highly soluble in ammonia. $AgCl$ will also be soluble in water but not more than ammonia as we know that oxygen is more electronegative than nitrogen so ammonia will be less polar than water.
Hence the order of increasing solubility will be:
$BaC{l_2} < NaCl < water < N{H_3}$

**Therefore, the correct answer is Option 4. i.e. $C < B < A < D$

Note:**
The common ion effect is the consequence of Le chatelier’s principle for equilibrium reactions of the ionic dissociation and association. This principle states that if on changing the conditions the dynamic equilibrium is disturbed so to counteract the change the position of the equilibrium will also shift in order to reestablish the equilibrium.