Question

$50$ ml of $NaOH$ solution is completely neutralized by $38.6$ ml of $0.213M$ of ${H_2}S{O_4}$ solution. What is the morality of $NaOH$ solution?
(A) $0.658M$
(B) $0.329M$
(C) $0.56M$
(D) $0.268M$

Answer 1

The normality can be determined by multiplying the molarity and valence factor of an acid or base. The valence factor of sulphuric acid is two. The normality of a base can be calculated from the below formula. The volume of a base and acid, the normality of an acid was needed.
${N_1}{V_1} = {N_2}{V_2}$
${N_1}$ is normality of sulphuric acid
${V_1}$ is volume of sulphuric acid is $38.6$ ml
${N_2}$ is normality of sodium hydroxide has to be determined
${V_2}$ is the volume of sodium hydroxide $50$ ml.

Complete answer:
The base and acid were reacted to form salt and water. The salt formed is sodium sulphate which is an inorganic salt.
The chemical reaction between sodium hydroxide and sulphuric acid will be as follows:
$2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O$
Given that the molarity of sulphuric acid is $0.213M$
The normality of sulphuric acid is $0.213 \times 2 = 0.426$
Thus, substitute the values in the above formula, we will get
$0.426 \times 38.6 = {N_1} \times 50$
By simplification, we will get
${N_1} = 0.32$
This normality of sodium hydroxide is $0.32N$ .
The molarity of sodium hydroxide is equal to normality of sodium hydroxide. Thus, the molarity of sodium hydroxide is $0.32M$
Thus, option b is the correct one.

Note:
The normality will be the product of molarity and valence factor of sulphuric acid. The valence factor of sulphuric acid is $2$ . Thus, normality will be doubled to molarity. In the case of sodium hydroxide, the valence factor is one. Thus, the normality and molarity are equal.