Question

You have sodium hydroxide solution. If you want to perform the experiment of neutralisation, which among the following will you use?

• A. ${\text{NH}}_4\text{OH}$ (Ammonium hydroxide)
• B. $\text{LiOH}$ (Lithium hydroxide)
• C. $\text{HCl}$ (Hydrochloric acid)
• D. ${\text{H}}_2{\text{SO}}_4$ (Sulphuric acid )

Answer 1

Answer: (C), (D)

The correct options are
C $\text{HCl}$ (Hydrochloric acid)
D ${\text{H}}_2{\text{SO}}_4$ (Sulphuric acid )

An acid reacts with a base to form salt and water. This is known as neutralisation reaction. Sodium hydroxide is a base. It reacts with an acid to form salt and water respectively.

Here, hydrochloric acid $\left(\text{HCl}\right)$ and sulphuric acid $\left({\text{H}}_2{\text{SO}}_4\right)$ are acids. So, sodium hydroxide $\left(\text{NaOH}\right)$ will react with them to form respective salts and water. The reactions involved are:

$\text{HCl (aq)}+\text{NaOH (aq)}\to \text{NaCl (aq)}+{\text{H}}_2\text{O (l)}$

${\text{H}}_2{\text{SO}}_4\text{(aq)}+2\text{NaOH (aq)}\to {\text{Na}}_2{\text{SO}}_4\text{(aq)}+2{\text{H}}_2\text{O (l)}$

Two bases do not react. So, sodium hydroxide $\left(\text{NaOH}\right)$ will not react with ammonium hydroxide $\left({\text{NH}}_4\text{OH}\right)$ and lithium hydroxide $\left(\text{LiOH}\right)$.