Question

0.1 M NaOH is titrated with 25 mL acetic acid. The equivalence point was reached at 23.25 mL.

a) Based on your experimentally observed equivalence point of 23.25 mL, calculate the original concentration of the acetic acid that came from the stock bottle.

b) Based on your experimentally observed half equivalence point of 11.625 mL pH= 4.8, calculate the Ka of acetic acid.

c) Based on your answers to 1& 2 calculate the expected pH at the equivalence point

Answer 1

a) The balanced equation for the acetic acid and NaOH titration is -

            CH₃COOH (aq) + NaOH (Aq) -------------> CH₃COONa (aq) + H₂O (l)

The Molarity of acetic acid , therefore = (0.1 M)(25 mL) / (23.25 mL) = 0.107 M

b) At half equivalence point , pH = pKa

                     pH = 4.8 = pKa

c) The number of mols of sodium acetate formed at equivalence point = (0.107 M)(0.02325 L) = 0.00248 mol

Therefore, concentration of sodium acetate = 0.00248 mol / (0.02325+0.025) L = 0.051 M

Thus -

              AcO^- (aq) + H₂O (l) <----------> AcOH (aq) + OH^- (aq)

I(mol) 0.051          -                               0                  0

C             -x                                             +x                 +x

Eq        (0.051-x)                                  x                  x

                      K_b = K_w / K_a = 6.30*`10^-10 = (x)(x) / (0.051-x)

By solving, x = 5.66*10^-6 M

Hence, [OH^-] = 5.66*10^-6 M

Thus, pOH = -log (5.66*10^-6 ) = 5.24

Thus , pH at equivalence point = 14 - 5.24 = 8.75