Question

Calculate the pH of 100 mL of a buffer solution containing 0.05 M benzoic acid (C6H5CO2H; Ka = 6.4 x 10-5) and 0.05 M sodium benzoate (NaC6H5CO2)

Answer 1

Hello!

The pH of a buffer solution of an acid and its conjugate base is easily calculated using the Henderson-Hasselbach's equation, with the value for the pKa for the benzoic acid:

(
`pKa=-log(Ka)=-log(6,4*10^(-5))=4,19`)

Henderson-Hasselbach:


`pH=pKa + log ( ([A^(-)] )/([HA]) ) \\ \\ pH=4,19 + log ( (0,05M)/(0,05M)) \\ \\ pH=4,19`

So, the pH for a 100-mL solution of 0,05 M Benzoic Acid and 0,05 M Sodium Benzoate is 4,19

Have a nice day!