Question

A buffer soln contains 0.1M CH₃COOH and 0.1M sodium acetate buffer. Calculate the pH after the addition of 4.0 mL of 0.025 M HCl to 10 mL of the buffer. The pKa for acetic acid is 4.74/

Answer 1

mol of HCl added = 0.025M *4.0 mL = 0.1 mmol

CH3COONa will react with H+ to form CH3COOH

Before Reaction:

mol of CH3COONa = 0.1 M *10.0 mL

mol of CH3COONa = 1 mmol

mol of CH3COOH = 0.1 M *10.0 mL

mol of CH3COOH = 1 mmol

after reaction,

mol of CH3COONa = mol present initially - mol added

mmol of CH3COONa = (1 - 0.1) mmol

mol of CH3COONa = 0.9 mmol

mol of CH3COOH = mol present initially + mol added

mol of CH3COOH = (1 + 0.1) mmol

mol of CH3COOH = 1.1 mmol

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.74+ log {0.9/1.1}

= 4.65

pH is 4.65