Answer:
pH = 2.10
Explanation:
We name an acid as diprotic because it can release two protons:
H₂A + H₂O ⇄ H₃O⁺ + HA⁻ Ka₁
HA⁻ + H₂O ⇄ H₃O⁺ + A⁻² Ka₂
We propose the mass balance:
Analytical concentration = [H₂A] + [HA⁻] + [A⁻²]
As Ka₂ is so small, we avoid the [A⁻²] so:
0.18 M = [H₂A] + [HA⁻]
But we can not avoid the HA⁻, because the Ka₁. Ka₁'s expression is:
Ka₁ = [H₃O⁺] . [HA⁻] / [H₂A]
We propose the charge balance:
[H₃O⁺] = [HA⁻] + [A⁻²] + [OH⁻]
As we did not consider the A⁻², we can miss the term and if
Kw = H⁺ . OH⁻
We replace Kw/H⁺ = OH⁻. So the new equation is:
[H₃O⁺] = [HA⁻] + Kw / [H₃O⁺]
The acid is so concentrated, so we can avoid the term with the Kw, so:
[H₃O⁺] = [HA⁻]
In the mass balance we would have:
0.18 M = [H₂A]
We replace at Ka₁
Ka₁ = [H₃O⁺] . [HA⁻] / [H₂A]
Ka1 . 0.18 / [H₃O⁺] = [HA⁻]
We replace at the charge balance:
[H₃O⁺] = Ka1 . 0.18 / [H₃O⁺]
[H₃O⁺]² = 3.4×10⁻⁴ . 0.18
[H₃O⁺] = √(3.4×10⁻⁴ . 0.18)
[H₃O⁺] = 7.82×10⁻³
- log [H₃O⁺] = pH → - log 7.82×10⁻³
pH = 2.10
