Question

Calculate the ph of a buffer solution that contains 0.25 m benzoic acid (c6h5co2h) and 0.15m sodium benzoate (c6h5coona). [ka = 6.5 x 10–5 for benzoic acid]

Answer 1

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