An acid that contains more than one acidic hydrogen is called a polyprotic acid. Acids like HNO₃, HCl, CH₃COOH, etc. are called monoprotic acids because they only contain one acidic hydrogen. Other acids such as H₂SO₄ and H₂S are diprotic acids because they contain two acidic hydrogens. A triprotic acid like H₃PO₄ has three acidic hydrogens. Polyprotic acids dissociate in a step-wise fashion with each step having its own acid dissociation constant, K_a1, K_a2, and so on. Oxalic acid is a diprotic acid with the following dissociations.

H₂C₂O₄(aq) + H₂O(l) ⇄ HC₂O₄^–(aq) + H₃O⁺(aq) \( K_{a1}=\frac{[HC_2O_4^-][H_3O^+]}{[H_2C_2O_4]}=5.6\times 10^{-2}\)
 
HC₂O₄^–(aq) + H₂O(l) ⇄ C₂O₄^2-(aq) + H₃O⁺(aq) \( K_{a2}=\frac{[C_2O_4^{2-}][H_3O^+]}{[HC_2O_4^-]}=6.4\times 10^{-5}\)

Note, the acid dissociation constants decrease, K_a1 > Ka₂ > Ka₃ and so on. It is much more difficult to remove a positively charged proton from a negatively charged species. Generally, the constants decrease by 10³ to 10⁷.

Example 1

The [H₃O⁺] concentration for a polyprotic acid comes from the first dissociation. Phosphoric acid, H₃PO₄, is a triprotic acid. Here we will calculate the concentrations of all species for a solution that is 3.5 M H₃PO₄. K_a1 = 7.5 x 10^-3, K_a2 = 6.2 x 10^-8, and K_a3 = 4.8 x 10^-13.

The equation and Ka for the first dissociation is below.

H₃PO₄(aq) + H₂O(l) ⇄ H₂PO₄^–(aq) + H₃O⁺(aq) \( K_{a1}=\frac{[H_2PO_4^-][H_3O^+]}{[H_3PO_4]}=7.5\times 10^{-3}\)

Set up an ICE table for the first dissociation.

Assume that 3.5 M – x ≅ 3.5 M.

\(\displaystyle \frac{x^2}{3.5}\;=\;7.5\times 10^{-3}\)
 
Solve for x.

\(\displaystyle \sqrt{3.50\times\;7.5\times 10^{-3}}\;=\;0.17\;M\)
 
Check the assumption. \(\displaystyle \frac{0.17\;M}{3.5\;M}\times\;100\;=\;4.9%\)
 
4.9% < 5%, therefore, the assumption is valid. [H₃PO₄]_eq = 3.3 M
[H₂PO₄^–] = [H₃O⁺]_eq = 0.17 M

The second dissociation is:

H₂PO₄^– (aq) + H₂O(l) ⇄ HPO₄^2-(aq) + H₃O⁺(aq) \(K_{a2}=\frac{[HPO_4^{2-}][H_3O^+]}{[H_2PO_4^-]}=6.2\times 10^{-8}\)

Set up an ICE table. This time we will use y rather than x.

Assume both 0.17 M + y and 0.17 M – y are approximately equal to 0.17 M.

\(\displaystyle \frac{y(0.17)}{0.17}\;=\;6.2\times 10^{-8}\)
 
We see x = 6.2 x 10^-8 M. This is the same as K_a2

Check the assumptions with the 5% rule:

\(\frac{6.2\times 10^{-8}\;M}{0.17\;M}\times\;100\;=\;3.6\times 10^{-5}\%\)

Our assumption was valid. The concentration of [HPO₄^2-]_eq = 6.2 x 10^-8 M.

For the third dissociation, write the balanced equation and then we can solve K_a3 for [PO₄^3-].

HPO₄^2- (aq) + H₂O (l) ⇄ PO₄^3- (aq) + H₃O⁺ (aq)

\(\displaystyle k_{a3}\;=\;\frac{[PO_4^{3-}][H_3O^+]}{[HPO_4^{2-}]}\;=\;4.8\times 10^{-13}\)
 
\(\displaystyle [PO_4^{3-}]\;=\;K_{a3}\times\frac{[HPO_4^{2-}]}{H_3O^+}\;=\;(4.8\times 10^{-13})\times\frac{6.2\times 10^{-8}\;M}{0.17\;M}\;=\;1.8\times 10^{-19}\;M\)

Again, the hydronium ion concentration is from the first dissociation and is negligible from the second and third dissociations.

Sulfuric acid is a triprotic acid and it is a strong acid. This means the first dissociation essentially goes to completion and we are left with hydrogen sulfate ion which does not dissociate much at all. Let’s do an example.

Example 2.

Determine the pH and the concentrations of all species in an aqueous 0.58 M H₂SO₄ solution. K_a2 = 1.2 x 10^-2

The first dissociation goes essentially to completion. The concentration of H₂SO₄ is 0.58 M. There is 0.58 M hydronium ion and 0.58 M hydrogen sulfate ion. The pH of the solution is 0.24.

H₂SO₄ (aq) + H₂O (l) → HSO₄^–(aq) + H₃O⁺(aq)

Next, we write the dissociation for HSOsub>4^–.

HSO₄^– (aq) + H₂O (l) ⇄ SO₄^2-(aq) + H₃O⁺(aq) \(K_{a2}\;=\;1.2\times 10^{-2}\)

Set up an ICE table.

Assume x is negligible — 0.58 M + x and 0.58 M – x is approximately equal to 0.58 M.

\(\displaystyle K_a\;=\;\frac{0.58\times\;x}{0.58}\;=\;1.2\times 10^{-2}\)
 
The sulfate ion concentration is 1.2 x 10^-2 M. The concentrations of all species are:

[H₃O⁺]_eq = 0.58 M + 1.2 x 10^-2 M = 0.59 M
[SO₄^2-]_eq = 1.2 x 10^-2 M
[HSO₄^–] = 0.58 M – 1.2 x 10^-2 M = 0.57 M

Watch the following video before attempting exercises

Exercises

Exercise 1.Calculate the pH of an aqueous 0.45 M selenious acid (H₂SeO₃) solution. K_a1 = 2.4 x 10^-3, K_a2 = 4.8 x 10^-9

Check Solution to Exercise 1

Exercise 2. Calculate the pH and the concentrations of all species in an aqueous solution of 0.25 M H₂S. K_a1 = 1.0 x 10^-7, K_a2 = 1.0 x 10^-19

Check Solution to Exercise 2

Exercise 3.Selenic acid, H₂SeO₄, is a strong acid. The first dissociation essentially goes to completion. Calculate the concentration of SeO₄^2- ion in a 2.5 M H₂SeO₄ aqueous solution. pK_a2 = 1.70

Check Solution to Exercise 3

Exercise 4. Tartaric acid, C₄H₆O₆, is a diprotic acid with pK_a1 = 2.89 and pK_a2 = 4.40. What is the pH of a 0.65 M aqueous solution of tartaric acid?

Check Solution to Exercise 4

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