Question

A 39.0 mL sample of 0.1950 M tellurous acid, H2TeO3, is titrated with 0.1950 M potassium hydroxide. What is the pH after 19.5 mL, 39.0 mL, 58.5 mL and 78.0 mL of KOH have been added? step mL KOH added pH 1 19.5 2 39.0 3 58.5 4 78.0 The first and second ionization constants for H2TeO3 are 5.4×10-7 and 3.7×10-9

Answer 1

pKa1 = -log Ka1 = 6.27

pKa2 = -log Ka2 = 8.43

millimoles of H2A = 39 x 0.195 = 7.60

1)   19.5 mL of KOH added :

millimoles of KOH = 19.5 x 0.195 = 3.80

H2A +   KOH -------------------> KHA + H2O

7.6           3.8                           0                0

3.8           0                              3.8

here H2A = KHA

pH = pKa1 + log [KHA / H2A]

pH = 6.27 + log (3.8 / 3.8 )

pH = 6.27

2) 39.0 mL KOH added :

it is first equivalence point

pH = 1/2 [pKa1 + pKa2 ]

      = 1/2 [6.27 + 8.43]

     = 7.35

pH = 7.35

3) 58.5 mL KOH added

it is second half equivalence point.

here pH = pKa2

pH = 8.43

4)   78 mL of KOH

it is second equivalece point.

only salt A2- left .

salt concentration = A2- = 39 x 0.1950 / (39 + 78)

                                     = 0.065 M

Kb1 = 2.70 x 10^-6

A2- + H2O --------------------> HA- + OH-

0.065                                      0            0

0.065-x                                   x            x

Kb = x^2 / 0.065 -x   = 2.70 x 10^-6

   x = 4.19 x 10^-4

[OH-] = 4.19 x 10^-4 M

pOH   = 3.38

pH + pOH = 14

pH = 10.62