Question

A diprotic acid, H 2 A , has acid dissociation constants of K a1 = 2.27 × 10 − 4 and K a2 = 4.00 × 10 − 11 . Calculate the pH and molar concentrations of H 2 A , HA − , and A 2 − at equilibrium for each of the solutions. A 0.103 M solution of H 2 A .

Answer 1

Answer/Explanation:

What is meant by diprotic acid?

A diprotic acid is the type of acid that can undergo one or two dissociations depending on the pH.

A diprotic acid is represented by H2A.

Dissociation of a diprotic acid does not happen all at once.

It has two steps of dissociation and each dissociation step has its own Ka value. The step 1 and step 2 Ka values are designated Ka1 and Ka2 respectively.

Back to the question,

At equilibrium,

Step 1

pKa1 = -log(Ka1) = -log(2.27 × 10^−4) = 3.64

Step 2

pKa2 = -log(Ka2) = -log(4.00 x 10^-11) = 10.40

The pH of H2A is given by

-log(molar concentration)

pH = -log(0.103) = -(-0.987) = 0.987. Approximately equals to 1