Question

Aqueous solution of $N{H_4}Cl$ is ______ in nature due to behaviour of _______ ion in solution.

A.Acidic, $N{H_4}^ + $
B.Alkaline, $N{H_4}^ + $
C.Acidic, $C{l^ - }$
D.Alkaline, $C{l^ - }$

Answer 1

Hint:$N{H_4}Cl$ is a salt formed by mixing of a weak base and strong acidic. To find behaviour of $N{H_4}Cl$, react it water and observe the product carefully. We know that when acid and base is reacted then salt and water is formed but here, we do the reverse process.

Complete step by step answer:
Aqueous solution of $N{H_4}Cl$ is formed by reacting it with excess water. The reaction of $N{H_4}Cl$ with water is as given below:
$N{H_4}Cl + {H_2}O \to N{H_4}OH + {H^ + } + C{l^ - }$
$N{H_4}Cl$ is a mixture of strong acid $(HCl)$ and weak base $\left( {N{H_4}OH} \right)$. Also, we can see in the above equation, when we add water in $N{H_4}Cl$, we get an acid and a base in solution.
Here acid is strong and base is weak means that solution is acidic in nature. Acid is formed by $C{l^ - }$ ions present in solution.
Or we also see that when we react $N{H_4}Cl$ with water we get ${H^ + }$ ions in solution which makes solution acidic. Here acid in solution is $HCl$ then acidic behaviour of solution is due to $C{l^ - }$ formed in solution.
Hence the correct answer is option C.



Note: $N{H_4}OH$ is a weak base then it does not dissociate into ions completely that is why it is present as a compound in the product of reaction of water with $N{H_4}Cl$. We know that $N{H_4}Cl$ is salt and formed by reacting strong acid $(HCl)$ and weak base $\left( {N{H_4}OH} \right)$ due to which some energy is released then when we react it with water we need to supply some energy to make reaction possible.