Question

If 25 mL of a ${H_2}S{O_4}$ solution reacts completely with 1.06 gm of pure$N{a_2}C{O_3}$ , what is the normality of this acid solution?
A.1 N
B.0.5 N
C.1.8 N
D.0.8 N

Answer 1

In the given question, Moles or weight of acid is not given which will be used to find out normality. ${H_2}S{O_4}$ and $N{a_2}C{O_3}$ will react to give out salt and water and after looking at the reaction, look at the equimolar ratio of both which will tell the number of moles of ${H_2}S{O_4}$ and therefore, normality could be calculated further.

Complete step by step answer:
In the given question, the reaction is in between an acid and a base which finally forms salt and water and we will look upon the molar ratio in which they are reacting and which will ultimately give the number of moles of acid and from which we can find out normality.
In the question, volume of acid solution is given but weight or number of moles are not given which would be needed to calculate the normality.
Let us first see the reaction of sulfuric acid and sodium carbonate which is represented as: -
${H_2}S{O_4} + {\text{ }}N{a_2}C{O_3} \to N{a_2}S{O_4} + {\text{ }}{H_2}O{\text{ }} + {\text{ }}C{O_2}$
So, from the reaction, we see that $N{a_2}C{O_3}$ and ${H_2}S{O_4}$ are in the equimolar ratio i.e. 1:1 so they are reacting in equal number of moles. Now we will calculate the number of moles of $N{a_2}C{O_3}$ which will give number of moles of ${H_2}S{O_4}$
Given weight of $N{a_2}C{O_3}$ = 1.06 gm
Molecular mass of $N{a_2}C{O_3}$ = 160 gm
Number of moles = \dfrac{{given{\text{ }}mass}}{{molecular{\text{ }}mass}}$$$$ = \dfrac{{1.06}}{{160}}{\text{ }} = {\text{ }}0.01{\text{ }}moles
So,1 mole of $N{a_2}C{O_3}$ is reacting with 1 mole of ${H_2}S{O_4}$ and therefore, number of moles of ${H_2}S{O_4}$ = 0.01 moles
Now, Normality is defined as the number of gram equivalents of moles of solute present per litre of the solution.
Also, the normality = Molarity $\times$ Basicity where basicity is the number of ${H^ + }$ions given by the acid and sulfuric acid can give 2 Hydrogen ions. So, its basicity is 2
Normality = $\dfrac{{0.01}}{{0.025}} \times 2{\text{ }} = 0.80{\text{ }}N$

So, the correct answer is D i.e. 0.80 N

Note:
In normality there is the concept of acidity and basicity which is used to calculate gram equivalent weight of the compound which is molecular weight divided by acidity or basicity. If an acid is present then it will have basicity i.e. number of hydrogen ions it can give out and if a base is present, then it will have acidity i.e. number of OH- ions it can give out.