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:

$pH=pKa + log ( \frac{[A^{-}] }{[HA]} ) \\ \\ pH=4,19 + log ( \frac{0,05M}{0,05M}) \\ \\ pH=4,19$

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!