Question

How to calculate the equivalent weight of ${\mathrm{Na}}_2{\mathrm{CO}}_3$?

Answer 1

Step 1: formula of equivalent weight:-

The equivalent weight of any element can be calculated using the formula;

$\mathrm{Equivalent}\mathrm{weight}=\frac{\mathrm{Molecular}\mathrm{weight}}{\mathrm{X}}$

Where X= valence factor

In the case of acid and bases, the valence factor is represented as the number of ${\mathrm{H}}^{+}\mathrm{and}{\mathrm{OH}}^{-}$gets dissociates in the solution.

It is also denoted as the number of ions formed as cations and anions in the solution.

Step 2: Calculation of the molecular mass for ${\mathrm{Na}}_2{\mathrm{CO}}_3$:-

The molecular mass/weight of 2 $\mathrm{Na}$ atoms = $2\times 23=46\mathrm{g}{\mathrm{mol}}^{-1}$.

The molecular mass/weight of $\mathrm{C}$= $12\mathrm{g}{\mathrm{mol}}^{-1}$.

For 3 Oxygen atoms present, the molecular weight= $16\times 3=48\mathrm{g}{\mathrm{mol}}^{-1}$.

Hence the molecular weight is given by;

$$\begin{gathered} =46+12+48 \\ =106\mathrm{g}{\mathrm{mol}}^{-1} \end{gathered}$$

Step 3; Calculation of the equivalent weight for ${\mathrm{Na}}_2{\mathrm{CO}}_3$:-

Hence, the equivalent weight of ${\mathrm{Na}}_2{\mathrm{CO}}_3$= $\frac{106}{2}=53\mathrm{g}{\mathrm{eq}}^{-1}$ ( $\because$charge on ions = 2).

Hence, the calculated equivalent weight of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ is found to be $53\mathrm{g}{\mathrm{eq}}^{-1}$.