Question

Equivalent mass of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ in a reaction, ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}{\text{ + NaOH}} \to {\text{N}}{{\text{a}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$ if the molar mass of the compound is M,
(A)- $\dfrac{{\text{M}}}{{{\text{10}}}}$
(B)- $\dfrac{{\text{M}}}{5}$
(C)- $\dfrac{{\text{M}}}{3}$
(D)- $\dfrac{{\text{M}}}{4}$

Answer 1

Hint Equivalent mass of any compound is calculated by dividing the molecular mass of the given compound by the n - factor i.e.
Equivalent weight = Molar mass / n – factor

Complete step by step solution:
For finding the equivalent weight of the Phosphoric acid ( ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ ) which takes part in the above given reaction we have to divide its molar mass which is given M by the n – factor.
-Value of n – factor for an acid is the number of hydrogen atoms replaced by that acid.
-For Phosphoric acid ( ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ ) value of n- factor is 3, because it replaces three hydrogen atoms from itself on the reactant side and converts into ${\text{N}}{{\text{a}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ on the product side by the replacement of three hydrogen atoms through three sodium (${\text{Na}}$ ) atoms.
-Equivalent weight of phosphoric acid ( ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ ) = $\dfrac{{{\text{Molar mass}}}}{{{\text{n - factor}}}}$
Equivalent weight of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}{\text{ = }}\dfrac{{\text{M}}}{{\text{3}}}$ .

Hence, equivalent mass of phosphoric acid ( ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ ) is $\dfrac{{\text{M}}}{3}$ i.e. option (C) is correct.

Additional information:
Equivalent weight of any compound is also calculated by dividing the molar mass of that compound by the cationic or anionic charge, by the acidity or basicity, by the exchanged number of electrons.

Note: Here some of you may do wrong calculation if you consider the cationic charge of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ is one as hydrogen carry $+ 1$ charge, but this consideration will be wrong because we have to take overall cationic charge of the compound not of the single cation.