Question

A 44.6 mL sample of 0.3060 M selenious acid, H₂SeO₃, is titrated with 0.3060 M sodium hydroxide.

What is the pH after 22.3 mL, 44.6 mL, 66.9 mL and 89.2 mL of NaOH have been added?

step	mL NaOH added	pH
1	22.3
2	44.6
3	66.9
4	89.2

The first and second ionization constants for H₂SeO₃ are 2.4×10^-3 and 4.8×10^-9

Answer 1

pKa1 = -logKa1

pKa1 = -log (2.4 x 10^-3) = 2.62

pKa2 = -log Ka2

pKa2 = 8.32

1) 22.3 mL

it is first half equivalece point . here pH = pKa1

pH = 2.62

2) 44.6

it is first equivalece point . here pH = pKa1 + pKa2 / 2

pH = 5.47

3) 66.9 mL

it is second half equivalece point . here pH = pKa2

pH = 8.32

4) 89.2 :

here only salt remains = salt concentration = 44.6 x 0.3060 / (44.6 + 89.2) = 0.102 M

A-2 + H2O -------------------> AH- + OH-

0.102 -x x x

2.08 x 10^-6 = x^2 / 0.102-x

x^2 + 2.08 x 10^-6 x - 2.125 x 10^-7 = 0

x = 4.60 x 10^-4

pH = 10.66