Question

Answer the following question about the titration of 80mL of .100 M H2SO3 with .100 M NaOH.

1. What is the inital pH of the solution before any NaOH is added?

2. What is the pH of the solution at the first equivalence point?

3. What is the pH of the solution at the second equivalence point?

4. What is the relationship between the first equivalence and the second equivalence points?

Answer 1

millimoles of H2SO3 = 80 x 0.1 = 8

H2SO3 , Ka1 = 1.54 x 10^-2 , pKa1 = 1.81

Ka2 = 6.3 x 10^-8 , pKa2 = 7.20

1 ) before additon of any NaOH :

H2SO3 --------------------------> HSO3 - + H+

0.100 0 0

0.1 - x x x

Ka1 = x^2 / 0.1-x

1.54 x 10^-2 = x^2 / 0.1-x

x^2 + 1.54 x 10^-2 x - 1.54 x 10^-3 = 0

x = 0.03229

x = [H+] = 0.03229 M

pH = -log [H+]

pH = 1.49

2) at first equivalence point :

pH = (pKa 1 + pKa2 )/ 2

pH = (1.81 + 7.20) /2

pH = 4.50

3 ) at second equivalece point:

80 mL of base needed

only salt SO3^-2 will remains = its molarity = 80 x 0.1 / (80 +80) = 0.05 M

SO3-2 + H2O ----------------------> HSO3- + OH-

0.05 -x x x

Kb2 = x^2 / 0.05 -x

Kw / Ka2 = x^2 / 0.05 -x

1.59 x 10^-7 = x^2 / 0.05 -x

x^2 + 1.59 x 10^-7 x - 7.94 x 10^-9 = 0

x = [OH-] = 8.90 x 10^-5

pOH = 4.05

pH = 9.95

4)

at second equivalece point pH nearly twice of first equivalece point