Question

The correct order of decreasing solubility of $AgCl$ in
(A) Water
(B) $0.1\,M\,NaCl$
(C) $0.1\,BaC{l_2}$
(D) $0.1\,M\,N{H_3}$
(A) $D > A > B > C$
(B) $D > C > B > A$
(C) $B > A > D > C$
(D) $A > D > B > C$

Answer 1

In the given question, we are asked to tell the solubility order of Silver chloride in given various solvents. Solubility depends upon various factors like temperature, pressure, bonds, forces, and concentration of the same or different ions.

Complete step by step solution:
Solubility is the maximum amount of a substance (solute) that can be dissolved in any known quantity of solvent at a given temperature, as in this case silver chloride is an ionic compound, so it will get dissociated into $A{g^ + }$ and $C{l^ - }$ ions upon dissolving in any solvent.
So here we will apply the concept of the common ion effect, which says that adding a common ion decreases the solubility of a solute, so greater the presence of common ion, lesser will be the solubility or vice-versa. In the question we have $NaCl\,$ and $BaC{l_2}$ , both of them dissociates when dissolves and gives C{l^\\_} same ion as $AgCl$ so the solubility with these two solvents will be minimum and comparing in between them, $1$ molecule of $BaC{l_2}$ will give $2\,C{l^ - }$ while one molecule of $NaCl\,$ will give $1\,C{l^ - }$ hence, $BaC{l_2}$ will have the lowest solubility with silver chloride.
When silver chloride is dissolved in ammonia it forms a complex called diamine silver ion ${[Ag{(N{H_3})_2}]^ + }$ . So, it is highly soluble in ammonia.
Silver chloride also dissolves in water but not good as it dissolves in ammonia, as water has oxygen which is more electronegative than nitrogen, so water is more polar than ammonia.
Hence the correct decreasing order of solubility of silver chloride will be
Ammonia ( $N{H_3}$ ) > water ( ${H_2}O$ ) > $NaCl\,$ > $BaC{l_2}$ i.e. $D > A{{ }} > B > C$
So the correct option is (A).

Note:
If we have a solute and solution in equilibrium, adding a common ion (an ion that is common with the dissolving solid) decreases the solubility of the solute. This is because Le Chatelier’s principle states the reaction will shift toward the left (toward the reactants) to relieve the stress of the excess product.