Question

Calculate the pH of a solution of 0.25 M Acetic acid and 0.34M Sodium acetate before and after 0.036 M NaOH is added to the solution. Ka of HAc = 1.8 x 10 -5

Answer 1

Pka   = -logka

       = -log1.8*10^-5

       = 7.45

PH = Pka + log[Ch3COOH]/[CH3COOH]

      = 4.75 + log0.34/0.25

      = 4.75+ 0.1335 = 4.8835

by the addition of 0.036M NaOH

[CH3COONa}   = 0.34+ 0.036   = 0.376M

[CH3COOH]   = 0.25-0.036    = 0.214 M

   PH   = 4.75 + log0.376/0.214

           = 4.75 + 0.2447 = 4.997