Question

What is the equivalent weight of ${\mathrm{H}}_3\mathrm{PO}$ when it disproportionate into ${\mathrm{PH}}_3$ and ${\mathrm{H}}_3{\mathrm{PO}}_3$?

Answer 1

Equivalent weight :

• The law of equivalence is one equivalence of an element combined with one equivalence of others.
• The equivalent weight of an acid in an acid-base reaction is the portion of the weight of 1 mole of the acid that can furnish 1 mole of hydrogen ions.
• The equivalence of a substance is defined as the amount which combines with 1 mole of the hydrogen atoms or replaces the same number of hydrogen atoms in the reaction.

Calculation :

The disproportionate reaction $3{\mathrm{H}}_3{\mathrm{PO}}_2\left(\mathrm{aq}\right)\to {\mathrm{PH}}_3\left(\mathrm{aq}\right)+2{\mathrm{H}}_3{\mathrm{PO}}_3\left(\mathrm{aq}\right)$
$\mathrm{n}\mathrm{factor}=\frac{\mathrm{No}.\mathrm{electron}\mathrm{involved}}{\mathrm{No}.\mathrm{of}\mathrm{molecules}\mathrm{involved}}$

$\mathrm{n}\mathrm{factor}=\frac{4}{3}$

Equivalent weight will be :

$\mathrm{Equivalent}\mathrm{weight}=\frac{\mathrm{mol}\mathrm{weight}}{\mathrm{n}\mathrm{factor}}$

$\mathrm{Equivalent}\mathrm{weight}=\frac{\mathrm{M}}{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}$

$\mathrm{Equivalent}\mathrm{weight}=\frac{3\mathrm{M}}{4}$

Hence the equivalent weight of phosphorous is $\frac{3\mathrm{M}}{4}$.