question archive Consider the titration of a 20
Subject:ChemistryPrice:2.87 Bought7
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:
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