Question

a) What is the pH of a solution prepared by adding 25.0 mL of 0.10 M acetic acid (Ka=1.8 x 10^-5) and 20.0 mL of 0.10 M sodium acetate?

b) What is the pH after the addition of 1.0 mL of 0.10 M HCl to 20 mL of the solution above?

c) What is the pH after the addition of 1.0 mL of 0.10 M NaOH to 20 mL of the original solution?

Answer 1

a) using Hendersen-Hasselbalck equation,

pH = pKa + log(base/acid)

     = 4.75 + log[(0.1 x 20/45)/(0.1 x 25/45)]

     = 4.653

b) after HCl = 0.1 M x 1 ml = 0.1 mmol added

pH = pKa + log(base/acid)

     = 4.75 + log[(0.1 x 20 - 0.1/45)/(0.1 x 25 + 0.1/45)]

     = 4.604

c) after NaOH = 0.1 M x 1 ml = 0.1 mmol added

pH = pKa + log(base/acid)

     = 4.75 + log[(0.1 x 20 + 0.1/45)/(0.1 x 25 - 0.1/45)]

     = 4.682