Question

A 100. mL buffer solution is 0.300 M in HAc and 0.150 M in NaAc. Calculate the pH of the solution after the addition of 35.0 mL of 0.20 M HCl. The Ka for HAc is 1.8 × 10-5.

Answer 1

mol of HCl added = 0.2M *35.0 mL = 7.0 mmol

Ac- will react with H+ to form HAc

Before Reaction:

mol of Ac- = 0.15 M *100.0 mL

mol of Ac- = 15 mmol

mol of HAc = 0.3 M *100.0 mL

mol of HAc = 30 mmol

after reaction,

mol of Ac- = mol present initially - mol added

mmol of Ac- = (15 - 7.0) mmol

mol of Ac- = 8 mmol

mol of HAc = mol present initially + mol added

mol of HAc = (30 + 7.0) mmol

mol of HAc = 37 mmol

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

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

use:

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

= 4.745+ log {8/37}

= 4.08

Answer: 4.08