Question

The salt which will give an acidic solution on dissolving in water is:
A.KCl
B.$N{H_4}Cl$
C.$N{a_2}C{O_3}$
D.$C{H_3}COONa$

Answer 1

Hint: Ammonium Chloride $(N{H_4}Cl)$ on dissolving in water gives an acidic solution because in water it gives hydrochloric acid which makes the solution acidic.

Complete step by step answer:
KCl will dissociate in water. It does not react with water. It will dissolve resulting in ionisation.
Ammonium Chloride $(N{H_4}Cl)$ is formed by the combination of a weak base ammonium hydroxide $N{H_4}OH$ and a very strong acid i.e. Hydrochloric acid (HCl). Chloride ion is a very weak base that it does not react with water whereas the cations of the salt is a weak acid that reacts with water.
 \[N{H_4}^ + (aq) + {H_2}O(l) \to N{H_3}(aq) + {H_3}{O^ + }(aq)\]
 \[{H_3}{O^ + }\] is a stronger acid than \[N{H_4}^ + \] and also ammonia is a stronger base than water? The equilibrium will therefore lie far to the left side in this case, thereby favouring the weaker acid - base pair. The \[{H_3}{O^ + }\] concentration produced by the reactions is great, however it decreases the pH of the solution thus making the solution acidic.
Sodium carbonate $(N{a_2}C{O_3})$ when dissolved in water produces 2 Na ions and one $C{O_3}$ ions, and forms carbonic acid which is a weak acid.
Sodium acetate $C{H_3}COONa$ when dissolved in water ionises to form $C{H_3}CO{O^ - }$ anion which accepts H+ ion from water and undergoes hydrolysis.
Thus, $(N{H_4}Cl)$ will give an acidic solution on dissolving in water.

Therefore, the correct answer is option (B)

Note: When it is dissolved in water, Ammonium chloride $(N{H_4}Cl)$ hydrolysed to form more ${H^ + }$ ions than $O{H^ - }$ ions. Thus, the solution becomes acidic with pH less than 7.