`25 mL` of `0.107 M H_(3)PO_(4)` was titrated with `0.115 M` solution of `NaOH` to the end point identified by indicator bromocresol green.This required `23.1 mL`. The titration was repeated using phenolphthalein as indicator. This time `25 mL` of `0.107 M H_(3)PO_(4)` reuired `46.2 mL` of the `0.115 M NaOH`. What is the coefficient of `n` in this equation for each reaction?
`H_(3)PO_(4)+n OH^(-)rarr nH_(2)O+[H_(3-n)PO_(4)]^(n-)`

Explanation

Number of milliequivalents of `H_(3)PO_(4)`
`=25xx0.107xxn`
`=2.675xxn`
In first titration: Number of milliequivalents of `OH^(-)` used
`=23.1xx0.115xx1=2.66`
{Acidity of NaOH=1)
In second titration: Number of milliequivalents of `OH^(-)` used
`=46.2xx0.115xx1=5.313`
`therefore` In first titration: `2.675xxn=2.66`
i.e., `" "` n=1
`therefore` In second titration : `2.675xxn =5.313`
n=2