Question

Consider the titration of a 20.0mL sample of 0.105M HC2H3O2 with 0.125M NaOH. Determine each of the following. a) Initial pH b) the volume of added base required to reach the equivelence point c) the pH at 5.0 mL of added base d) the pH at one-half of the equivelence point e) the pH at the equivelence point f) ph after adding 5.0ml of base beyond the equivalence point

Answer 1

Titration

a) initial pH

only acid present

HC2H3O2 <==> C2H3O2- + H+

let x amount has dissociated

Ka = [C2H3CO2-][H+]/[HC2H3O2]

1.8 x 10^-5 = x^2/0.105

x = [H+] = 1.37 x 10^-3 M

pH = -log[H+] = 2.86

b) volume of base required to reach equivalence point = (0.105 M x 20 ml)/0.125 M = 16.8 ml of NaOH

c) pH after 5 ml of NaOH added

moles of acid = 0.105 M x 20 ml = 2.1 mmol

moles of base added = 0.125 M x 5 ml = 0.625 mmol

excess [HC2H3O2] = 1.475 mmol/25 ml = 0.059 M

formed [C2H3O2Na] = 0.625 mmol/25 ml = 0.025 M

pH = pKa + log(base/acid)

      = 4.74 + log(0.025/0.059)

      = 4.37

d) pH at half equivalence point

concentration of HC2H3O2 = C2H3O2Na

pH = pKa = 4.74

e) pH at equivalence point

all of acid is neutralised

salt formed [C2H3O2Na] = 0.105 M x 20 ml/36.8 ml = 0.057 M

salt hydrolyses

C2H3O2- + H2O <==> HC2H3O2 + OH-

let x amount has hydrolysed

Kb = Kw/Ka = [HC2H3O2][OH-]/[C2H3O2-]

1 x 10-14/1.8 x 10^-5 = x^2/0.057

x = [OH-] = 5.63 x 10^-6 M

pOH = -log[OH-] = 5.25

pH = 14 - pOH = 8.78

f) after 5 ml excess base is added

[OH-] = 0.125 M x 5 ml/41.8 ml = 0.015 M

pOH = -log[OH-] = 1.82

pH = 14 - pOH = 12.18