Question

Consider the titration of 24.0 mL of 0.100 M HC₂H₃O₂ with 0.120 M NaOH. Determine the pH after adding 4.00 mL of base. I have found that the volume of the base needed to reach the equivalence point is 20.0 mL.

Show your work and be as clear as possible. Thank you in advance!

Answer 1

This will form a buffer, since conjugate will be present (acetate ion) and the acid as well so

pH = pKa + log(conjugate / acid)

conjugate formed = 0 + M*V (of base)

conjugate formed = 0 + 4*0.12 = 0.48 mmol of conjugate is formed (acetate ion)

for acid, there is a neutralization so

acid left = acid original - base = 24*0.1 - 4*0.12= 1.92 mmol of acid left

then

pKa = -log(Ka)

KA = 1.8*10^-5

pKa = -log(1.8*10^-5) = 4.74472

then

pH = 4.74472 + log(0.48 /1.92) = 4.142

pH = 4.142