Question

Given,$14.3g$ of $N{a_2}C{O_3}.x{H_2}O$ is dissolved in water and the volume is made up to $200ml$. $20ml$ Solution requires $40ml$ of $\dfrac{N}{4}HN{O_3}$ for complete neutralization. Calculate x.

Answer 1

We need to know that the hydrated sodium carbonate is a chemical compound having the formula, $N{a_2}C{O_3}.x{H_2}O$ which is used to prepare the hard water. The nitric acid is a strong acid having the chemical compound, $HN{O_3}$ and it is a highly corrosive mineral acid. By the decomposition of nitric acid, there is a formation of water and oxides of nitrogen. The nitric acid is used for the preparation of ammonium nitrate for fertilizers.

Complete answer:
According to the law of equivalent, the equality of normality is expressed as,
${N_1}{V_1} = {N_2}{V_2}$
Where, N is the normality and V is equal to volume. Volume of hydrated sodium carbonate is equal to $20ml$. Volume of nitric acid is equal to $0.25$and normality of nitric acid is equal to$40$. Substitute the values in the above equation.
$20\left( {{N_1}} \right) = 40 \times 0.25$
By rearranging the equation, ${N_1} = 0.5N$
Therefore, normality of sodium carbonate solution is equal to $0.5N$.
$0.5 = \dfrac{{14.3}}{{eq.wt}} \times \dfrac{{1000}}{{200}}$
Therefore, the equivalent is equal to,
$eq.wt = \dfrac{{14.3 \times 5}}{{0.5}} = 143$
But the equivalent weight of $N{a_2}C{O_3}.x{H_2}O$ is,
$\dfrac{{106 + x\left( {18} \right)}}{2} = 143$
Therefore, $x = \dfrac{{286 - 106}}{{18}} = 10$
Hence, the value of x in the hydrated sodium carbonate is equal to $10$.

Note:
We need to know that in the case of a neutralization reaction, the base and the acids are quantitatively reacted with each other. In the neutralization reaction with water, there is no excess of hydroxide ions or hydrogen ions present in that solution. . In the hydrated sodium carbonate, a particular amount of water molecules is attached with the solid structure of the sodium carbonate. And these hydrated water molecules can be removed by the heating process.