A diprotic acid, H₂A, has Ka1 = 3.4 × 10⁻⁴ and Ka2 = 6.7 × 10⁻⁹. What is the pH of a 0.18 M solution of H₂A?

Following are the calculation to the pH:

For First ionization:

where

C = initial concentration of acid

For Second ionization:

Therefore, the pH is "2.15".

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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

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