Question

How much 5.50 M NaOH must be added to 530.0 mL of a buffer that is 0.0200 M acetic acid and 0.0230 M sodium acetate to raise the pH to 5.75?

Answer 1

before adding NaOH,

mol of CH3COOH = M*V = 0.0200 M * 530.0 mL = 10.6 mmol

mol of CH3COONa = M*V = 0.0230 M * 530.0 mL = 12.19 mmol

Let V mL of NaOH added

mol of NaOH added = 5.50*V mmol

after reaction,

mol of CH3COOH = (10.6 - 5.50*V) mmol

mol of CH3COONa = (12.19 + 5.50*V) mmol

pKa of CH3COOH = 4.75

use:

pH = pKa + log {[CH3COONa]/[CH3COOH]}

5.75 = 4.75 + log ( (12.19 + 5.50*V) /(10.6 - 5.50*V))

log ( (12.19 + 5.50*V) /(10.6 - 5.50*V))= 1

( (12.19 + 5.50*V) /(10.6 - 5.50*V))= 10

12.19 + 5.50*V = 106 - 55.0*V

60.5*V = 93.81

V = 1.55 mL

Answer: 1.55 mL