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Question: The equivalent weight of \({{H}_{3}}P{{O}_{2}}\) , when it disproportionate into \(P{{H}_{3}}\text{ ...

The equivalent weight of H3PO2{{H}_{3}}P{{O}_{2}} , when it disproportionate into PH3 and H3PO4P{{H}_{3}}\text{ and }{{\text{H}}_{3}}P{{O}_{4}} is,
(A) 82
(B) 61.5
(C) 33
(D) 49.5

Explanation

Solution

As we know that in a chemical reaction the equivalence of a substance is explained as the amount of which combines with 1 mole of the hydrogen atoms or replaces the same number of hydrogen atoms in the reaction. Thus equivalent weight in grams is the weight of 1 grams.

Complete step by step solution:
In this question we have to find the H3PO2{{H}_{3}}P{{O}_{2}} disproportionate and form PH3P{{H}_{3}}. We can write the equation involved as follow:
H3PO2→PH3 P++4e−→P3− \begin{aligned} & {{H}_{3}}P{{O}_{2}}\to P{{H}_{3}} \\\ & {{P}^{+}}+4{{e}^{-}}\to {{P}^{3-}} \\\ \end{aligned}
The equivalent weight of H3PO2{{H}_{3}}P{{O}_{2}} can be calculated by dividing the ratio of molecular weight to the valence factor =M4\dfrac{M}{4}
Again H3PO2{{H}_{3}}P{{O}_{2}}dissociates and form H3PO3{{H}_{3}}P{{O}_{3}}this reaction is mentioned below:
H3PO2→H3PO3 P+−2e−→P3+ \begin{aligned} & {{H}_{3}}P{{O}_{2}}\to {{H}_{3}}P{{O}_{3}} \\\ & {{P}^{+}}-2{{e}^{-}}\to {{P}^{3+}} \\\ \end{aligned}
The equivalent weight of H3PO2{{H}_{3}}P{{O}_{2}} can be calculated by dividing the ratio of molecular weight to the valence factor=M2\dfrac{M}{2}
The molar mass of H3PO2{{H}_{3}}P{{O}_{2}} is 66g/mol
Now the equivalent weight will be M4+M2=16.5+33\dfrac{M}{4}+\dfrac{M}{2}=16.5+33= 49.5

Hence the correct answer is option (D).

Additional information:
We know that the law of equivalence is one equivalence of an element combined with one equivalence of others. Equivalent weight of an acid in an acid base reaction is the portion of weight of 1 mole of the acid that can furnish 1 mole of hydrogen ions.

Note: As we know that a redox reaction that reactant is transformed into products by simultaneous reduction reaction as well as the oxidation reaction is termed as the disproportionation reaction. For s reaction say that hydrogen peroxide transformed into water and oxygen. In this reaction hydrogen peroxide oxidizes to form oxygen and it reduces and forms water.