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:
([tex]pKa=-log(Ka)=-log(6,4*10^{-5})=4,19 [/tex])
Henderson-Hasselbach:
[tex]pH=pKa + log ( \frac{[A^{-}] }{[HA]} ) \\ \\ pH=4,19 + log ( \frac{0,05M}{0,05M}) \\ \\ pH=4,19[/tex]
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!
