Question

A sample of sodium carbonate contains sodium sulphate also. $1.5g$of the sample is dissolved in water and volume raised to $250mL$. $25mL$ of this solution requires $20mL$ of $\dfrac{N}{{10}}{H_2}S{O_4}$ solution for neutralization. Calculate percentage of sodium carbonate in the sample.

Answer 1

Normality is measure of concentration equal to gram equivalent per litre of solution, gram equivalent is the measure of the reactive capacity of a molecule.

Complete step by step answer: Only $N{a_2}C{O_3}$ will react with ${H_2}S{O_4}$.
Applying, ${\text{ }}{N_1}{V_1}{\text{ }} = {\text{ }}{N_2}{V_2} \\\ (N{a_2}C{O_{3)}}{\text{ }}\left( {{H_2}S{O_4}} \right) \\\$
$\Rightarrow$ ${N_1} \times 25 = 20 \times \dfrac{1}{{10}}$
$\Rightarrow$ ${N_1} = \dfrac{{20}}{{25 \times 10}} = 0.08N$
Equivalent Mass =$\dfrac{{Molar{\text{ }}Mass}}{{n{\text{ }}factor}} = \dfrac{{106}}{2} = 53$
Mass of $N{a_2}C{O_3}$ present in $250ml$ $0.08N$
Solution = $\dfrac{{N \times {\rm E} \times V}}{{1000}} = \dfrac{{0.08 \times 53 \times 250}}{{1000}} = 1.06gm$
Percentage of $N{a_2}C{O_3}$ in the mixture = $\dfrac{{1.06}}{{1.50}} \times 100 = 70.67\%$
The reaction is as follows.
$N{a_2}C{O_3} + {H_2}S{o_4} \to N{a_2}S{O_4} + C{O_2} + {H_2}O$ ….. (i)
Sodium carbonate reacts with sulphuric acid to produce sodium sulphate, carbon dioxide and water.
The carbon dioxide gas which is seen by the formation of bubbles.
Additional information –
For the one more reaction also, the products are the same as reaction (i).
$2NaHC{O_3} + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O + 2C{O_2}$
Also sodium sulphate (Glaublr’s salt) in the form of $N{a_2}S{O_4} \cdot 10{H_2}O$ also called salk cake.
It is used in the manufacture of glass and paper and in titration of Barium Nitrate.

Note: The indicator which is used in the titration of sodium carbonate with sulphuric acid is Methyl orange.