Question

What is the ‘n’ factor for acid and how is it calculated for oxalic acid dihydrate?

Answer 1

Hint: The n-factor is defined as the number of $H^{+}$ ions replaced by $1$ mole of acid in a reaction and the number of $O{H}^{-}$ replaced by $1$ mole of base in a reaction.The molecular formula of oxalic acid dihydrate is $C_2{H}_2{O}_4.2H_{2}O$ .

Complete answer:
For an acid ‘n’ factor is the number of $H^{+}$ ions replaced by $1$ mole of an acid
The factors related to oxalic acid are:
Oxalic acid is a dibasic acid.
Its basicity is 2 because it has two protons that can be replaced.
For an oxalic acid let us consider the following reaction as example:
 $COOH-COOH\to 2C{O}_2+2e^{-}+2H^{+}$
From the above reaction we can say that the “n” factor for oxalic acid dihydrate is $2$ as $2$ hydrogen ions are replaced in the reaction.
Equivalent weight of acid = ${\displaystyle \frac{\text{Molecular weight of acid}}{\text{Number of Replaceable hydrogen ions}}}$
In Oxalic acid, the number of replaceable hydrogen ions is $2$ .
Molecular weight of oxalic acid is $126$ .
Equivalent weight of acid $={\displaystyle \frac{126}{2}}$
Equivalent weight of oxalic acid $=63$
Oxalic acid is acidic in nature. As oxalic acid has $2$ moles of acidic $H^{+}$ ions, which means one mole of oxalic acid can neutralize two moles of $O{H}^{-}$ .
Hence, the n-factor of oxalic acid is $2$ .

Note:
Oxalic acid is a dicarboxylic acid. It is also known as Ethanedioic acid. It is called a dibasic acid because it has the ability to donate two $H^{+}$ ions. One molecule of oxalic acid releases two hydrogen ions. Hence, n-factor is two.