To neutralize completely `20 mL` of `0.1M` aqueous solution of phosphorus `(H_(3)PO_(3))` acid the volume of `0.1M` aqueous `KOH` solution required is:

Options

A. `10 mL`
B. `20 mL`
C. `40 mL`
D. `60 mL`

Answer

3

Explanation

According to the law of equivalence
`Eqs of H_(3)PO_(3)=Eqs` of `KOH`
`N_(H_(3)PO_(3)) V_(H_(3)PO_(3))=N_(KOH)V_(KOH)`
Since
Normality `=n_("factor")` Morality
and
`n_("factor")` for `H_(3)PO_(3)` (a diprotic acid)=2
`n_("factor")` for KOH (a monoacidic base)=1
we have
`2MH_(3)PO_(3) VH_(3)PO_(3)=1 M_(KOH) V_(KOH)`
`(2) (0.1 M)(20 mL)=(1)(0.1 M)(V_(KOH))`
or `V_(KOH)=((2)(0.1 M)(20 mL))/((0.1 M))`
`=40 mL`