Question

The volume of 0.1M ${H_2}S{O_4}$ required to neutralize 30 ml of 2.0M $NaOH$ is:
a.) 100 ml
b.) 300 ml
c.) 400 ml
d.) 200 ml

Answer 1

In order to solve the given problem and to find out the volume of sulphuric acid of given molarity required to neutralize the 30 ml of sodium hydroxide of the given molarity, first we will write the balanced chemical equation for the given neutralization reaction. Further we will use the formula relating the molarity and the volume which we will deduce out of the relation between the normality and the volume in a chemical reaction.

Complete step by step answer:
Let us first see the chemical reaction between the sodium hydroxide and sulphuric acid.
$2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O$
Given in the problem that:
Molarity of ${H_2}S{O_4}$ is = 0.1 M
Molarity of $NaOH$ is = 2 M
Volume of $NaOH$ is = 30 ml
Let the volume of ${H_2}S{O_4}$ required to neutralize the sodium hydroxide is V ml.
As we know the relation between the Normality of two solutions and the volume of the solutions in the case of neutralization reaction is given as:
${N_1}{V_1} = {N_2}{V_2} - - - - - - (1)$

In the above equation ${N_1}$ represents the normality of the first solution and the term ${N_2}$ represents the normality of the second solution. Whereas the terms ${V_1}$ and ${V_2}$ represents the volume of the first and the second solution respectively.

But in the given problem we don’t know the value of normality for the given solutions. So we will use the relation between the normality and the molarity to find the value of molarity.
We know that the molarity and the normality are related in the terms of valence factor which can be found out by the balanced reaction.
The relation between the molarity and the normality is:
$N = nM - - - - - - (2)$

In the above reaction “N” is the normality, “n” is the balance factor and the term “M” is the molarity.
Now, let us combine equation (1) and equation (2)
$\because {N_1}{V_1} = {N_2}{V_2}\& N = nM \\\ \Rightarrow {n_1}{M_1}{V_1} = {n_2}{M_2}{V_2} - - - - - - - (3) \\\$
From the balanced chemical reaction we know that
For ${H_2}S{O_4}$ , n = 2
For $NaOH$ , n = 1
Also we have other values of the equation (3), so let us substitute the values in order to find the volume of the ${H_2}S{O_4}$ solution.
$\because {n_1}{M_1}{V_1} = {n_2}{M_2}{V_2} \\\ \Rightarrow 2 \times 0.1 \times V = 1 \times 2 \times 30 \\\$
Let us now solve the above equation to get the value of Volume.

$$\because 2 \times 0.1 \times V = 1 \times 2 \times 30 \\\ \Rightarrow V = \dfrac{{1 \times 2 \times 30}}{{2 \times 0.1}} \\\ \Rightarrow V = \dfrac{{60}}{{0.2}} \\\ \Rightarrow V = 300ml \\\$$

Hence, the volume of 0.1M ${H_2}S{O_4}$ required to neutralize 30 ml of 2.0M $NaOH$ is 300 ml.
So, the correct answer is “Option B”.

Note: In order to solve such types of problem students must start with the balanced chemical equation as the problem becomes even harder to solve without the reaction. Students must remember the relation between the molarity and the volume for different solutions in a reaction. Students must remember that this relation has been derived from the ideal gas equation and can be derived by students as well.