100ml of 0.1M `H_(3)PO_(4)` is titrated with 0.05 M NaOH solution till 2nd equivalent point. Then, resultant solution was mixed with 10 ml of 0.5 MHCI solution
`K_(1)=10^(-3)M`
`K_(2)=10^(-8)M`
`K_(3)=10^(-13)M`
pH at 2nd equivalent point will be :
A. `5.5`
B. `10.5`
C. `8`
D. `7`
