Equivalent weight of ${H_3}P{O_4}$ in each of the reaction will be respectively:
${H_3}P{O_4} + O{H^ - } \to {H_2}PO_4^{2 - } + 2{H_2}O$
${H_3}P{O_4} + 2O{H^ - } \to HPO_4^{2 - } + 2{H_2}O$
${H_3}P{O_4} + 3O{H^ - } \to PO_4^{3 - } + 3{H_2}O$
A. 98, 49, 32.67
B. 49, 98, 32.67
C. 98, 32.67, 49
D. 32.67, 49, 98

Hint: To determine the equivalent weight we need to find the molecular weight of the compound and the basicity value of the compound. The molecular weight is divided by the basicity value to get the equivalent weight.

Complete answer:Equivalent weight is also known as gram equivalent as the mass is in terms of gram.
The formula to calculate the equivalent weight is shown below.
$E.W = \dfrac{M}{B}$
Where,
E.W is the equivalent weight.
M is the molecular weight
B is the basicity of the compound.
The basicity of the acid is defined as the number of hydrogen ions present which can be replaced.
The reaction 1 is shown below.
${H_3}P{O_4} + O{H^ - } \to {H_2}PO_4^{2 - } + 2{H_2}O$
In this reaction, one hydrogen atom of ${H_3}P{O_4}$ is replaced by $O{H^ - }$ to form ${H_2}PO_4^{2 - }$ and two mole of ${H_2}O$. Therefore, the value of basicity in this reaction is 1.
The reaction 2 is shown below.
${H_3}P{O_4} + 2O{H^ - } \to HPO_4^{2 - } + 2{H_2}O$
In this reaction, two hydrogen atom of ${H_3}P{O_4}$ is replaced by two mole of $O{H^ - }$ to form $HPO_4^{2 - }$ and two mole of ${H_2}O$. Therefore, the value of basicity in this reaction is 2.

The reaction 3 is shown below.
${H_3}P{O_4} + 3O{H^ - } \to PO_4^{3 - } + 3{H_2}O$
In this reaction, three hydrogen atom of ${H_3}P{O_4}$ is replaced by three mole of $O{H^ - }$ to form $PO_4^{3 - }$ and three mole of ${H_2}O$. Therefore, the value of basicity in this reaction is 3.
The molecular weight of phosphoric acid ${H_3}P{O_4}$ is 98 g/mol.
To calculate the equivalent weight of ${H_3}P{O_4}$in reaction 1, substitute the values in the equation.
$E.W = \dfrac{{98}}{1}$
Thus, the equivalent weight is 98.

To calculate the equivalent weight of ${H_3}P{O_4}$in reaction 2, substitute the values in the equation.
$E.W = \dfrac{{98}}{2}$
Thus, the equivalent weight is 49.
To calculate the equivalent weight of ${H_3}P{O_4}$in reaction 3, substitute the values in the equation.
$E.W = \dfrac{{98}}{3}$
Thus, the equivalent weight is 32.67.
Thus, the equivalent weight of ${H_3}P{O_4}$ in each of the reactions will be 98, 49, 32.67.
Therefore, the correct option is A.

Note:Acidity is used for the base compound where replaceable hydroxide ion is counted and basicity is used for acidic compounds. In this question the hydrogen ion is replaced by the hydroxide ion.