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
