Respuesta :

superman1987
when C6H5COO- is the conjugate base & the C6H5CO2H is the weak acid 

-and we have [C6H5COO-] = 0.15 M

and [C6H5CO2H] = 0.25 M 

- and we have Ka = 6.5 x 10^-5 so, we can use it and get Pka

∴ Pka = -㏒Ka

          = -㏒(6.5 x 10^-5)
          = 4.2

So, by using H-H equation:

PH = Pka + ㏒[C6H5COO-]/[C6H5CO2H]

     = 4.2 + (0.15 / 0.25)

     = 4.8