Question

A 1.0 m solution of a compound with 2 ionizable groups (pka's = 6.2 and 9.5; 100 ml total) has a ph of 6.8. what are the concentrations of the relevant acid and conjugate base?

Answer 1

Answer:

[H₂A] = 0.2M

[HA⁻] = 0.8M

Explanation:

A compound with 2 ionizable groups has as equilbriums:

H₂A ⇄ HA⁻ + H⁺ ⇄ A²⁻ + H⁺

A pH of 6.8 is near to the first equilibrium:

H₂A ⇄ HA⁻ + H pKa = 6.2

To determine concentrations of the acid H₂A and conjugate base HA⁻ we use H-H equation:

pH = pKA + log [HA⁻] / [H₂A]

6.8 = 6.2 + log [HA⁻] / [H₂A]

0.6 = log [HA⁻] / [H₂A]

3.981 = [HA⁻] / [H₂A] (1)

As concentration of the buffer is 1.0M:

1M = [HA⁻] + [H₂A] (2)

Replacing (2) in (1):

3.981 = 1- [H₂A] / [H₂A]

3.981[H₂A] = 1 - [H₂A]

4.981 [H₂A] = 1

[H₂A] = 0.2M

And [HA] = 1M - 0.2M = 0.8M