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
