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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.
Expert Answer
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
