Question

A 250.00 mL buffer solution is 0.250 M acetic acid (CH3COOH) and 0.100 M sodium acetate (NaCH3COO). The Ka of acetic acid is 1.8 • 10-5 at 25 o C. Calculate the pH of the resulting buffer solution when 10.00 mL of a 0.500 M aqueous HCl solution is added to this mixture.

Answer 1

mol of HCl added = 0.5M *10.0 mL = 5.0 mmol

CH3COO- will react with H+ to form CH3COOH

Before Reaction:

mol of CH3COO- = 0.1 M *250.0 mL

mol of CH3COO- = 25 mmol

mol of CH3COOH = 0.25 M *250.0 mL

mol of CH3COOH = 62.5 mmol

after reaction,

mol of CH3COO- = mol present initially - mol added

mmol of CH3COO- = (25 - 5.0) mmol

mol of CH3COO- = 20 mmol

mol of CH3COOH = mol present initially + mol added

mol of CH3COOH = (62.5 + 5.0) mmol

mol of CH3COOH = 67.5 mmol

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.7447

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

use:

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

= 4.7447+ log {20/67.5}

= 4.2165

pH is 4.22