Question

The equivalent weight of water in a neutralization reaction between a dibasic acid and triacidic is

Answer 1

Step 1: Explanation of concept :

1. An acid that can donate two protons is called dibasic.
2. Acids that can donate three hydroxide ions are called triacidic.
3. The formula of equivalent weight is $\mathrm{Equivalent}\mathrm{weight}=\frac{\mathrm{Molecular}\mathrm{weight}}{\mathrm{Valance}\mathrm{factor}}$

Step 2: Neutralization reaction between dibasic acid and triacidic :

We will take sulfuric acid $\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$ as dibasic acid and aluminium hydroxide $\left(\mathrm{Al}{\left(\mathrm{OH}\right)}_3\right)$.

The neutralization reaction of dibasic sulfuric acid and triacidic aluminium hydroxide is :

$$\begin{gathered} 3{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+2\mathrm{Al}{\left(\mathrm{OH}\right)}_3\left(\mathrm{aq}\right)\to {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3\left(\mathrm{aq}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ (\mathrm{Sulfuric}(\mathrm{Aluminium}(\mathrm{Aluminium}(\mathrm{Water}\left) \\ \mathrm{acid}\right)\mathrm{hydroxide})\mathrm{sulphate}) \end{gathered}$$

Water dissociates in ${\mathrm{H}}^{+}$ and ${\mathrm{OH}}^{-}$ ions, so the valence factor for the water is $1$.
The molecular weight of water is $18$.
So, the equivalent weight of water is :

$\mathrm{Equivalent}\mathrm{weight}=\frac{18}{1}=18$

The equivalent weight of water in a neutralization reaction between a dibasic acid and triacidic is $18$.