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Question: The molecular mass of \[{{\rm{H}}_3}{\rm{P}}{{\rm{O}}_3}\] is 82. Its equivalent mass, if it is comp...

The molecular mass of H3PO3{{\rm{H}}_3}{\rm{P}}{{\rm{O}}_3} is 82. Its equivalent mass, if it is completely neutralised is
A. 82
B. 27.327.3
C. 41
D. 246

Explanation

Solution

Equivalent mass is the mass of a given substance that will combine with or displace a fixed quantity of another substance. The equivalent mass of a compound can be calculated from the equivalent mass formula [given below]. We shall substitute the appropriate values in the formula to find the equivalent mass.
Formula used:
Equivalentweight=molecularweightofcompoundnumberofelectronstransferred{\rm{Equivalent weight = }}\dfrac{{{\rm{molecular weight of compound}}}}{{{\rm{number of electrons transferred}}}}

Complete step by step answer:
Equivalent mass is the mass of a given substance that will combine with or displace a fixed quantity of another substance. It is used to compare chemically different elements. For acids and bases, it is the mass of the compound required to supply 1 mole of hydrogen ions. In the case of redox reactions, it is the weight of each reactant required to react with 1 mole of electrons.
As we know that, phosphorous acid [H3PO3{{\rm{H}}_3}{\rm{P}}{{\rm{O}}_3} ] is a dibasic acid. It means that it can donate two protons or hydrogen ions per molecule in a solution. H3PO3{{\rm{H}}_3}{\rm{P}}{{\rm{O}}_3} in a solution will dissociate into:
H3PO32H++HPO32{{\rm{H}}_3}{\rm{P}}{{\rm{O}}_3} \to {\rm{2}}{{\rm{H}}^ + } + {\rm{HPO}}_3^{2 - }
From here, we can see that the number of electrons being transferred is 2. So, by putting these values in the formula we get:
Equivalentweight=822{\rm{Equivalent weight}} = \dfrac{{82}}{2}
Equivalentweight=41\Rightarrow {\rm{Equivalent weight}} = 41
So, the equivalent weight of H3PO3{{\rm{H}}_3}{\rm{P}}{{\rm{O}}_3} in a completely neutralised solution will be 41 u.

The correct option in Option C, i.e. 41 u.

Note:
Molar mass is used to find the molarity of a compound which is used in volumetric analysis. Equivalent mass can also be used in volumetric analysis. It is used to find the normality of the solution, which is analogous to molarity.