For the diprotic weak acid H2A, Ka1 = 2.1 × 10-6 and Ka2 = 7.0 ×...

For the diprotic weak acid H2A, Ka1 = 2.1 × 10-6 and Ka2 = 7.0 × 10-9. What is the pH of a 0.0700 M solution of H2A?

Expert Solution

Because Ka1 >> Ka2, then all of the H3O+ comes from the first hydrolysis reaction only.

Molarity               H2A + H2O <==> H3O+ + HA-
I                           0.0700 .                0            0
C                         -x .                        +x         +x

E                          0.0700-x               x             x

Ka1 = [H3O+][HA-] / [H2A]

= (x)(x) / (0.0700-x) = 2.1 x 10^-6

x^2 / (0.0700-x) = 2.1 x 10^-6 .

due to small values of Ka1 , 0.0700-x ≈0.0700

x^2 / 0.0700 = 2.1x 10^-6
x^2 = 1.74x 10^-7
x = 3.8 x 10^-4 M

3.8 x 10^-4 = [H3O+]

Now calculate the pH as follows:

pH = -log [H3O+]

= -log 3.8 x 10^-4

= 3.42

queen_honey_blossom answered 1 hour ago

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