Equivalent weight of `H_(3)PO_(2)` in a reaction is found to be half of its molecular weight. It can be due to its

Options

A. oxidation to `H_(3)PO_(3)`
B. reaction of two `H^(+)` ions
C. oxidation to `H_(3)PO_(4)`
D. reduction to `PH_(3)`

Answer

A

Explanation

(`a`) It is monobasic acid, hence only one `H^(+)` can be neutralised `E=(M)/(1)`
(`b`) `underset(+1)underset(uarr)(H_(3)PO_(2)) rarr underset(+3)underset(uarr)(H_(3)PO_(3))`
`E=(M)/(2)`
(`c`) `underset(+1)underset(uarr)(H_(3)PO_(2)) rarr underset(+5)underset(uarr)(H_(3)PO_(4))`
`E=(M)/(4)`
(`d`) `underset(+1)underset(uarr)(H_(3)PO_(2)) rarr underset(-3)underset(uarr)(PH_(3))`
`E=(M)/(4)`