Question

A 1.0 0Liter buffer solution is .760 M acetic acid and .350 M in sodium acetate. Calculate the pH of the solution after the addition of 100 mL of 0.500 M HCl. The Ka for acetic acid is 1.8 * 10^-5

Answer 1

mol of HCl added = 0.5M *100.0 mL = 50.0 mmol

CH3COO- will react with H+ to form CH3COOH

Before Reaction:

mol of CH3COO- = 0.35 M *1000.0 mL

mol of CH3COO- = 350 mmol

mol of CH3COOH = 0.76 M *1000.0 mL

mol of CH3COOH = 760 mmol

after reaction,

mol of CH3COO- = mol present initially - mol added

mmol of CH3COO- = (350 - 50.0) mmol

mol of CH3COO- = 300 mmol

mol of CH3COOH = mol present initially + mol added

mol of CH3COOH = (760 + 50.0) mmol

mol of CH3COOH = 810 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

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

= 4.745+ log {3*10^2/8.1*10^2}

= 4.313

Answer: 4.31