Question

If 1 mole of \[{{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}\]is reacted with 1 mole of \[\text{X}{{\left( \text{OH} \right)}_{\text{2}}}\] as:
\[{{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}\text{+ X}{{\left( \text{OH} \right)}_{\text{2}}}\to \text{XHP}{{\text{O}}_{\text{4}}}\text{ + 2}{{\text{H}}_{\text{2}}}\text{O}\], then: This question has multiple correct options
A) The equivalent weight of the base is $\dfrac{\text{Mol}\text{.wt}}{\text{2}}$
B) The equivalent weight of \[{{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}\] is $\dfrac{98}{3}$
C) The resulting solution is required 1 mole \[\text{NaOH}\]for complete neutralization
D) Minimum 1 mole of \[\text{X}{{\left( \text{OH} \right)}_{\text{2}}}\]more required for complete neutralization of $\text{XHP}{{\text{O}}_{\text{4}}}$

Answer 1

Hint: A neutralization reaction is a reaction when one mole of acid reacts with one mole of the base to form one mole of salt and water. The phosphoric acid reacts with a base to generate its corresponding phosphate salt and water.

Complete answer:
We know that one mole \[{{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}\] is reacting with\[\text{X}{{\left( \text{OH} \right)}_{\text{2}}}\]. The reaction is as follows;
${{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}\text{+X(OH}{{\text{)}}_{\text{2}}}\to \text{XHP}{{\text{O}}_{\text{4}}}\text{+2}{{\text{H}}_{\text{2}}}\text{O}$
The phosphoric acid \[{{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}\]reacts with the hydroxide \[\text{X}{{\left( \text{OH} \right)}_{\text{2}}}\] to give a salt of phosphate$\text{XHP}{{\text{O}}_{\text{4}}}$.
It is a simple acid-base reaction to give the salt. Let's find out which of the above statements are true.
A) We know that the equivalent weight of the solution is described by the molecular weight of solute divided by the valence number of solute. It is used to determine the amount of mass of substance in reaction. For acid-base reactions, it is usually considered as the amount of substance reacts with the one atom of hydrogen.
$\text{Equivalent weight=}\dfrac{\text{Molecular weight}}{\text{Valence number}}$
Let us write the reaction for the base.
$\text{X(OH}{{\text{)}}_{\text{2}}}\to {{\text{X}}^{+}}+2\text{O}{{\text{H}}^{\text{-}}}$
So in reaction, every mole of the base \[\text{X}{{\left( \text{OH} \right)}_{\text{2}}}\]gives 2 moles of$\text{O}{{\text{H}}^{\text{-}}}$
Therefore, the valence number for the base is 2
Thus, the $\text{Equivalent weight of base=}\dfrac{\text{mol}\text{.wt}}{2}$

Option (A) is the correct option.

B) In reaction,
${{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}\to \text{HPO}_{4}^{2-}+2{{\text{H}}^{\text{+}}}$
Phosphoric acid loses its two protons.
Therefore the valence factor is 2.
The molecular weight ${{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}$is equal to $\text{98g/mol}$.
Therefore, the equivalent weight of acid is as:
$\text{Equivalent weight of acid=}\dfrac{\text{98}}{2}$
But in option, the valence factor is considered as 3.

Therefore, option (B) is an incorrect option.

C) The salt of phosphate i.e. $\text{XHP}{{\text{O}}_{\text{4}}}$is acidic. It further undergoes the neutralization reaction as follows,
$\text{XHP}{{\text{O}}_{\text{4}}}+\text{NaOH}\to \text{NaXP}{{\text{O}}_{\text{4}}}+{{\text{H}}_{\text{2}}}\text{O}$
Here, one mole of acid requires the one mole of the base for complete neutralization. This can be justified as ${{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}$is triprotic. It can lose it's 3${{\text{H}}^{\text{+}}}$to undergo complete neutralization. Since $\text{XHP}{{\text{O}}_{\text{4}}}$is the monobasic acid is derived from the phosphoric acid it requires the one mole of the base for neutralization.

Thus, option (C) is the correct option.

D) Let’s find out how many moles \[\text{X}{{\left( \text{OH} \right)}_{\text{2}}}\]requires the complete neutralization of$\text{XHP}{{\text{O}}_{\text{4}}}$.
The reaction between \[\text{X}{{\left( \text{OH} \right)}_{\text{2}}}\]and $\text{XHP}{{\text{O}}_{\text{4}}}$is as follows,
$\text{2XHP}{{\text{O}}_{\text{4}}}\text{+X(OH}{{\text{)}}_{\text{2}}}\to {{\text{X}}_{\text{3}}}{{\text{(P}{{\text{O}}_{\text{4}}}\text{)}}_{\text{2}}}\text{+2}{{\text{H}}_{\text{2}}}\text{O}$
Here, 2 moles $\text{XHP}{{\text{O}}_{\text{4}}}$require 1mole of\[\text{X}{{\left( \text{OH} \right)}_{\text{2}}}\].

Therefore, D) is an incorrect option.

Additional information:
Phosphoric acid ${{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}$is also known as orthophosphoric acid. It is triprotic acid. The ${{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}$reacts with the base in three steps.
$\begin{align}
  & \text{1)}{{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}\text{+NaOH}\to \text{Na}{{\text{H}}_{\text{2}}}\text{P}{{\text{O}}_{\text{4}}}\text{+}{{\text{H}}_{\text{2}}}\text{O} \\
 & \text{2)Na}{{\text{H}}_{\text{2}}}\text{P}{{\text{O}}_{\text{4}}}\text{+NaOH}\to \text{N}{{\text{a}}_{\text{2}}}\text{HP}{{\text{O}}_{\text{4}}}\text{+}{{\text{H}}_{\text{2}}}\text{O} \\
 & \text{3)N}{{\text{a}}_{\text{2}}}\text{HP}{{\text{O}}_{\text{4}}}\text{+NaOH}\to \text{N}{{\text{a}}_{\text{2}}}\text{P}{{\text{O}}_{\text{4}}}\text{+}{{\text{H}}_{\text{2}}}\text{O} \\
\end{align}$

Note:
In such types of questions, Sometimes the term equivalent weight can be confusing. The number of gram equivalents depends on the characteristic properties of the substance. This is also called a valence factor.
For acids, the valence factor for acid=basicity=removal no.of proton (${{\text{H}}^{\text{+}}}$)
For bases, the valence factor for base=acidity=removal no.of hydroxide (\[O{{H}^{-}}\])