Question

Molarity of 0.1 N of oxalic acid?

Answer 1

Step 1 n-factor of Oxalic acid:

The chemical formula of Oxalic acid is${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4$

Normality the of the Oxalic acid solution is $0.1\mathrm{N}$

The acids are formed when the hydrogen (${\mathrm{H}}^{+}$) ions are dissolved in a solvent.

In a reaction, the number of ${\mathrm{H}}^{+}$ ions that get replaced by one mole of acid is known as the $\mathrm{n}$-factor of that acid.

The chemical reaction of oxalic acid can be written as

$\mathrm{HOOC}-\mathrm{COOH}\to 2{\mathrm{CO}}_{{}^2}+2\mathrm{H}+2{\mathrm{e}}^{-}$

Since two ${\mathrm{H}}^{+}$ is involved in this reaction, the $\mathrm{n}$-factor is found as 2.

Step 2 Molarity of Oxalic acid:

Molarity can be defined as the number of moles of solute per liters of a solution.

$\mathrm{Molarity}=\frac{\mathrm{Normality}}{\mathrm{n}-\mathrm{factor}}$

Therefore, substituting the values we get,

$$\begin{gathered} \mathrm{Molarity}=\frac{\mathrm{Normality}}{\mathrm{n}-\mathrm{factor}} \\ \mathrm{Molarity}=\frac{0.1}{2} \\ \mathrm{Molarity}=0.05\mathrm{M} \end{gathered}$$

That is, the molarity of $0.1\mathrm{N}$ Oxalic acid is $0.05\mathrm{M}$.