Question

10 mL of 0.0100 M HCl are added to 25.0 mL of a buffer solution that is 0.010 M HC2H3O2 and 0.1 M C2H3O2-. What is the pH of the resulting solution?

Answer 1

mol of HCl added = 0.01M *10.0 mL = 0.1 mmol

CH3COONa will react with H+ to form CH3COOH

Before Reaction:

mol of CH3COONa = 0.1 M *25.0 mL

mol of CH3COONa = 2.5 mmol

mol of CH3COOH = 0.01 M *25.0 mL

mol of CH3COOH = 0.25 mmol

after reaction,

mol of CH3COONa = mol present initially - mol added

mmol of CH3COONa = (2.5 - 0.1) mmol

mol of CH3COONa = 2.4 mmol

mol of CH3COOH = mol present initially + mol added

mol of CH3COOH = (0.25 + 0.1) mmol

mol of CH3COOH = 0.35 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 {2.4/0.35}

= 5.58

Answer: 5.58