Question

Consider the titration of 50.0 mL of a 0.0200 M solution of butanoic acid with 0.100 M NaOH. First, calculate the equivalence point (Ve). Then, calculate the pH at the following points along the titration curve. Assume fractions are exact numbers and ignore activities for all calculations in this problem.

A) VNaOH= 0.000 mL

B) VNaOH= 1/2Ve

C) VNaOH= 4/5Ve

D) VNaOH= Ve

E) VNaOH= 3/2Ve

Answer 1

A)

Ka of butanoic acid = 1.5×10^–5

pKa = 4.82

pH = 1/2 (pKa - log C)

      = 1/2 (4.82 - log 0.0200)

pH = 3.26

B)

at equivalence point :

moles of acid = moles of base

50 x 0.0200 = 0.100 x V

V = 10 mL

equivalence point volume = 10 mL

volume of NaOH Ve = 5 mL

pH = 4.82

C)

VNaOH = 4/5 x 10

V_NaOH = 8.0 mL

mmoles of acid = 50 x 0.0200 = 1

mmoles of NaOH = 8 x 0.1 = 0.8

C3H7COOH   + NaOH   ---------------> C3H7COO- + H2O

    1                      0.8                                 0                     0

   0.2                      0                                  0.8

pH = 4.82 + log [0.8 / 0.2]

pH = 5.422

D)

VNaOH= Ve

V_NaOH= 10.0 mL

concentration of salt = 1 / 10 + 50 = 0.0167 M

pH = 7 + 1/2 (pKa + log C)

     = 7 + 1/2 (4.82 + log 0.0167 )

pH = 8.52

E)

VNaOH = 3/2 x 10

V_NaOH = 15 mL

pH = 11.89