Question

$\begin{matrix} {{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}} & \text{+} & \text{NaOH} & \to & \text{NaHC}{{\text{O}}_{\text{3}}} & \text{+} & {{\text{H}}_{\text{2}}}\text{O} \\\ \end{matrix}$
The equivalent weight of ${{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$in the above reaction is (Molar mass of ${{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$is$\text{62 g/mol}$).
A) $\text{60}$
B) $\text{32}$
C) $62$
D) $63$

Answer 1

The equivalent weight is obtained by dividing the molecular weight of species by the valence factor. For the neutralization reaction, the valence factor is the number of protons donated by the acid or the basicity of acid. Therefore the equivalent weight for acids the species can be given as:
$\text{Equivalent weight of atom or molecule =}\dfrac{\text{Molecular weight of atom or molecule}}{\text{Basicity of acid}}$

Complete step by step answer:
The equivalent weight of the compound is defined as the molecular weight of the compound per number of equivalent mole of it. Equivalent moles are also called the valence factor which depends upon the chemical properties of the compound.
Therefore, the equivalent weight of the compound or species can be given as:
$\text{Equivalent weight=}\dfrac{\text{Molecular weight}}{\text{No of equivalent moles}}$
For acid it is given as:
$\text{Equivalent weight of acid =}\dfrac{\text{Molecular weight of acid}}{\text{Basicity of acid}}$
In a neutralization reaction, the acid loses the proton ${{\text{H}}^{\text{+}}}$ .thus the number of protons donated by the acid is equal to the valence factor of acid.
We are given the following reaction:
$\begin{matrix} {{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}} & \text{+} & \text{NaOH} & \to & \text{NaHC}{{\text{O}}_{\text{3}}} & \text{+} & {{\text{H}}_{\text{2}}}\text{O} \\\ \end{matrix}$
The carbonic acid ${{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$ undergoes the neutralization reaction with the$\text{NaOH}$. The carbonic acid loses one of its protons as shown below:

{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}} & \rightleftharpoons & \text{HCO}_{\text{3}}^{\text{-}} & \text{+} \\\ \end{matrix}\begin{matrix} {{\text{H}}^{\text{+}}} & {} \\\ \end{matrix}$$ Since we know that the valence factor for acid is equal to the number of protons lose by acid. Thus the valence factor for carbonic acid ${{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$ is one. We are provided with the molecular weight $\text{62 g/mol}$ . Then the equivalent weight of the ${{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$can be given as,$$\begin{aligned} \text{Equivalent weight of }{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ = }\dfrac{\text{Molecular weight of }{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}}{\text{Basicity of }{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}} & \\\ \text{i}\text{.e}\text{. Equivalent weight of }{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ =}\dfrac{62\text{ g/mol}}{\text{1}} & \\\ & \\\ \Rightarrow \text{ Equivalent weight of }{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{= }62\text{ g/mol} & \\\ \end{aligned}$$ **Hence, (C) is the correct option.** **Note:** In such types of questions, sometimes it can be confusing between moles and the number of equivalent mole. No moles are given as the ratio of the weight of substance by molecular weight however no of equivalent moles is dependent on the characteristic properties of the substance. This is also called a valence factor. 1\. For acids, valence factor for acid=basicity=removal no.of proton (${{\text{H}}^{\text{+}}}$) 2\. For bases, valence factor for base=acidity=removal no.of hydroxide ($$O{{H}^{-}}$$)