Question

Calculate the pH and concentration of all species of malonic acid (C3H4O4) in each of the follwing solutions.

K1=1.42*10^-3, K2=2.01*10^-6

0.100 M NaC3H3O4 solution

0.100 M Na2C3H2O4 solution

Answer 1

Ka1=1.42*10^-3

Kb1 = 1 x 10^-14 / 1.42*10^-3 = 7.04 x 10^-12

NaC3H3O4 + H2O -------------------> C3H4O4 + OH-

    0.1                                                       0             0

0.1 - x                                                      x              x

Kb1 = x^2 / 0.1 - x

7.04 x 10^-12 = x^2 / 0.1 - x

x = 8.39 x 10^-7

[OH-] = 8.39 x 10^-7 M

[C3H4O4] = 8.39 x 10^-7 M

[NaC3H3O4] = 0.1 - x = 0.0999 M

pOH = -log [OH-] = -log [8.39 x 10^-7]

         = 6.08

pH = 7.92

b)

Ka2 = 2.01 x 10^-6

Kb2 = 1 x 10^-14 / 2.01 x 10^-6 = 4.98 x 10^-9

C3H2O4-2 + H2O -------------------> C3H3O4- + OH-

    0.1                                                       0             0

0.1 - x                                                      x              x

Kb2 = x^2 / 0.1 - x

4.98 x 10^-9 = x^2 / 0.1 - x

x = 2.23 x 10^-5

[OH-] = 2.23 x 10^-5 M

pOH = -log [OH-] = -log ( 2.23 x 10^-5)

         = 4.65

pH = 9.35