Question

The equivalent weight of $NaH{{C}_{2}}{{O}_{4}}$ in reaction with HCl is----.
a.) 112
b.) 56
c.) 224
d.) 8

Answer 1

The equivalent weight of a chemical compound is the ratio of its gram atomic weight to its valence.
Equivalent weight of an acid = $\dfrac{\text{Molecular weight of acid}}{\text{valency of the acid}}$
Equivalent weight of base = $\dfrac{\text{Molecular weight of base}}{\text{valency of the base}}$

Complete step by step answer:
In the question it is given that sodium hydrogen oxalate ($NaH{{C}_{2}}{{O}_{4}}$) reacts with hydrochloric acid (HCl).

The chemical reaction of sodium hydrogen oxalate ($NaH{{C}_{2}}{{O}_{4}}$) with hydrochloric acid (HCl) is as follows.
$NaH{{C}_{2}}{{O}_{4}}+HCl\to NaCl+{{H}_{2}}{{C}_{2}}{{O}_{4}}$
Sodium hydrogen oxalate ($NaH{{C}_{2}}{{O}_{4}}$) reacts with hydrochloric acid and forms sodium chloride and oxalic acid are the products.

In the above chemical reaction one hydrogen atom is going to transfer from hydrochloric acid to sodium hydrogen oxalate.

Therefore acidity or basicity of sodium hydrogen oxalate is one.
So, the equivalent weight of $NaH{{C}_{2}}{{O}_{4}}$ reacts with hydrochloric acid is as follows.
Equivalent weight of sodium hydrogen oxalate $=\dfrac{\text{Molecular weight of }NaH{{C}_{2}}{{O}_{4}}}{\text{number of hydrogens transferred during the reaction}}$

Molecular weight of sodium hydrogen oxalate = 112
Number of hydrogen transferred during the reaction with hydrochloric acid is one.
Therefore Equivalent weight of sodium hydrogen oxalate $\begin{aligned} & =\dfrac{\text{Molecular weight of }NaH{{C}_{2}}{{O}_{4}}}{\text{number of hydrogens transferred during the reaction}} \\\ & =\dfrac{112}{1} \\\ & =112 \\\ \end{aligned}$
Therefore the equivalent weight of $NaH{{C}_{2}}{{O}_{4}}$ reacts with hydrochloric acid is 112.
So, the correct answer is “Option A”.

Note: Equivalent weight of oxalic acid is when it reacts with sodium hydroxide is as follows.
${{H}_{2}}{{C}_{2}}{{O}_{4}}+NaOH\to N{{a}_{2}}{{C}_{2}}{{O}_{4}}+{{H}_{2}}O$
In the above reaction two hydrogens are going to release from oxalic acid. So, the equivalent weight of oxalic acid when it reacts with sodium hydroxide is
Equivalent weight of oxalic acid in the above reaction = $\dfrac{\text{molecular weight of oxalic acid}}{\text{2}}$