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/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

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.