Upon titrating `50mL` of `1.96%(w//v) H_(2)SO_(4)` with a `KOH` solution (containing `11.2g KOH` per litre of solution on adding `50mL KOH` solution.

Options

A. Equivalent point has been just reached.
B. Equivalent point has been crossed
C. `pH` of resulting solution `=1`
D. `pH` of resulting solution `=7`

Explanation

`[H_(2)SO_(4)]=(19.6)/(98)=0.2 M` and `KOH=(11.2)/(56)=0.2M`
so for equivalent point, `(0.2xx50xx2)/(0.2)=100mL` of `KOH`solution should be added.
`[H^(+)]=(0.2xx2-0.2)/(2)=0.1M` so `pH=1`.