Question

You need to prepare an acetate buffer of pH 5.72 from a 0.876 M acetic acid solution and a 2.80 M KOH solution. If you have 625 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a bufer of pH 5.72? The pKa of acetic acid is 4.76.

Answer 1

before adding KOH
moles of CH3COOH = 0.876 M * 625.0 mL = 547.5 mmol
moles of CH3COOK = 0

let v mL of KOH is added
moles of KOH added= 2.80*v
after reaction,
moles of CH3COOH remaining = 547.5 - 2.80*v
moles of CH3COOK formed = 2.80*v

pka of CH3COOH = 4.76

They will form buffer
pH= pKa + log {[CH3COOK]/[CH3COOH]}
5.72= 4.76 + log ((2.80*v)/(547.5 - 2.80*v))
log ((2.80*v)/(547.5 - 2.80*v)) = 0.96
(2.80*v)/(547.5 - 2.80*v) = 9.12
2.80*v = 4993.2 - 25.54*v
v = 176.2 mL
Answer: 176.2 mL