Question

The equivalent weight of water in a neutralization reaction between a dibasic acid and triacidic base is:
A. $9$
B. $18$
C. $6$
D. $3$

Answer 1

Hint: To answer we should know what equivalent weight is. First, we will choose a dibasic acid and triacidic base then we will write the neutralization reaction. Then by diving the molecular formula with the valence factor of water we will determine the equivalent weight.
\[{\text{equivalent weight}}\,{\text{ = }}\,\dfrac{{{\text{Molecular weight}}}}{{{\text{valance}}\,{\text{factor}}}}\]

Complete solution:
The formula to determine the equivalent weight is as follows:
\[{\text{equivalent weight}}\,{\text{ = }}\,\dfrac{{{\text{Molecular weight}}}}{{{\text{valance}}\,{\text{factor}}}}\]
A dibasic acid means the acid can donate two protons so, we will select the very famous sulphuric acid, ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$.
A triacidic base means the base can donate three hydroxide ions so, we select the aluminium hydroxide, ${\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}$.
Now, the neutralization reaction of dibasic sulphuric acid, ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$and triacidic aluminium hydroxide, ${\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}$is as follows:
${\text{3}}\,{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\, + \,2\,{\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}\, \to {\text{A}}{{\text{l}}_2}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}}\, + \,6{{\text{H}}_{\text{2}}}{\text{O}}$
So, the neutralization reaction of dibasic sulphuric acid and triacidic aluminium hydroxide gives the aluminium sulphate and water.
Water dissociates as ${{\text{H}}^{\text{ + }}}$ and ${\text{O}}{{\text{H}}^ - }$ions. The oxidation state of the proton or hydroxide ion is one, so the valence factor for the water is one.
The valence factor for the water is $1$ and the molecular weight of water is$18$.
So, on substituting $1$ for valance factor and $18$ for molecular weight in equivalent weight formula,
\[{\text{equivalent weight}}\,{\text{ = }}\,\dfrac{{{\text{18}}}}{{\text{1}}}\]
\[{\text{equivalent weight}}\,{\text{ = }}\,18\]
So, the equivalent weight of water in a neutralization reaction is\[18\].
Therefore, the correct answer is (B).


Note: Molecular weight is the product of equivalent weight and balance factor. Molecular weight is determined by adding the atomic mass of each constituting atom. The valence factor is also known as n-factor. The valence factor is the number of electrons gained or lost by a species or the oxidation number or charge of the species In the case of acid, the valence factor is determined as the number of protons donated by the species.