Question

In a driven reaction between 1 mole of NaOH and 1 mole of H2C3H2O4 identify the species of interest present in the solution as we add NaOH and state how to handle that solution to determine pH.

a. Before adding NaOH: H2C3H2O4 is the only thing present which will then undergo an equilibrium process and we can use an ice table to determine pH. Since it is a diprotic acid, we might need two ice tables.

b. After adding 0.5 mole of NaOH:

c. After adding 1.0 mole of NaOH:

d. After adding 1.5 mole of NaOH:

e. After adding 2.0 mole of NaOH:

Answer 1

malonic acid pKa1 = 2.83 , pKa2 = 5.69

a)

before adding NaOH :

species present = HC3H2O4-, H2C3H2O4 , C3H2O4^2-

H2C3H2O4   ------------> H+ +   HC3H2O4-

1 mol                                0                   0

1 - x                                  x                     x

Ka1 = x^2 / 1 - x

1.48 x 10^-3 = x^2 / 1 - x

x = 0.0377 M

[H+]= 0.0377 M

pH = 1.42

b) After adding 0.5 mole of NaOH:

this is first half equivalence point. so

here pH = pKa1

pH = 2.83

species present = HC3H2O4-, H2C3H2O4

c) After adding 1.0 mole of NaOH:

this is equivalence point. here

pH = pKa1 + pKa2 / 2

    = 2.83 + 5.69 / 2

pH = 4.26

species present = HC3H2O4-

d) After adding 1.5 mole of NaOH:

this is second half equivalence point . here

pH = pKa2

pH = 5.69

species present = HC3H2O4- , C3H2O4^2-

e). After adding 2.0 mole of NaOH:

species present = C3H2O4^2-