Polyprotic acids: WACE Year 12 Chemistry

Syllabus dot point

“Describe the stepwise ionisation of polyprotic acids and explain why successive ionisation constants decrease”

8 min readUpdated 2 Jun 2026

Quick answer

A polyprotic acid can donate more than one proton, and it does so one step at a time. Each step has its own ionisation constant, and they decrease sharply ( $K_{a1} > K_{a2} > K_{a3}$ ) because removing a proton from an increasingly negative ion is progressively harder. This stepwise behaviour produces multiple end points in a titration.

What this dot point is asking

A monoprotic acid such as hydrochloric or ethanoic acid donates only one proton per molecule. A polyprotic acid can donate more than one: diprotic acids (two protons, such as sulfuric acid $\text{H}_2\text{SO}_4$ and carbonic acid $\text{H}_2\text{CO}_3$ ) and triprotic acids (three protons, such as phosphoric acid $\text{H}_3\text{PO}_4$ ).

Polyprotic acid

An acid capable of donating more than one proton per molecule. The protons are lost in successive steps, each with its own acid ionisation constant.

Stepwise ionisation

A polyprotic acid does not lose all its protons at once. Phosphoric acid ionises in three distinct steps:

\text{H}_3\text{PO}_4 \rightleftharpoons \text{H}^+ + \text{H}_2\text{PO}_4^- \quad K_{a1}

\text{H}_2\text{PO}_4^- \rightleftharpoons \text{H}^+ + \text{HPO}_4^{2-} \quad K_{a2}

\text{HPO}_4^{2-} \rightleftharpoons \text{H}^+ + \text{PO}_4^{3-} \quad K_{a3}

Each step has its own equilibrium constant, and the values fall steeply, typically by a factor of around $10^5$ each time.

Why successive constants decrease

After the first proton leaves, the remaining species carries a negative charge. Removing a second positive proton ( $\text{H}^+$ ) from an already negatively charged ion requires overcoming a stronger electrostatic attraction, so it is harder and less favourable. Removing a third proton from a doubly negative ion is harder still. This is why

K_{a1} \gg K_{a2} \gg K_{a3}

First ionisation dominates the pH

Because $K_{a1}$ is so much larger than $K_{a2}$ , almost all of the hydrogen ions in a polyprotic acid solution come from the first ionisation. For pH calculations of a weak polyprotic acid, you can usually treat it as monoprotic using $K_{a1}$ alone.

Sulfuric acid is a special case

Sulfuric acid is strong in its first ionisation ( $\text{H}_2\text{SO}_4 \rightarrow \text{H}^+ + \text{HSO}_4^-$ , essentially complete) but weak in its second ( $\text{HSO}_4^- \rightleftharpoons \text{H}^+ + \text{SO}_4^{2-}$ , an equilibrium). This mixed behaviour is worth remembering because it differs from a fully weak acid like phosphoric acid.

Effect on titrations

Because the protons come off in distinct steps, a titration of a polyprotic acid against a strong base can show more than one equivalence point, one for each proton being neutralised. A diprotic acid can show two end points and a triprotic acid up to three, provided the successive constants are far enough apart to resolve them.

Neutralising a diprotic acid

Question

How many moles of sodium hydroxide are needed to completely neutralise $0.10$ mol of phosphoric acid, $\text{H}_3\text{PO}_4$ ?

Solution

Phosphoric acid is triprotic, donating three protons per molecule. Complete neutralisation removes all three:

\text{H}_3\text{PO}_4 + 3\text{NaOH} \rightarrow \text{Na}_3\text{PO}_4 + 3\text{H}_2\text{O}

So each mole of acid needs 3 moles of base: $0.10 \times 3 = 0.30$ mol NaOH. Note that partial neutralisation (1 or 2 mol of base per mol of acid) would stop at an intermediate salt such as $\text{NaH}_2\text{PO}_4$ or $\text{Na}_2\text{HPO}_4$ .

Intermediate salts and amphiprotic ions

A practical consequence of stepwise ionisation is that polyprotic acids form a family of intermediate salts. Partially neutralising phosphoric acid gives, in turn, sodium dihydrogen phosphate $\text{NaH}_2\text{PO}_4$ (after 1 mol of base per mol of acid) and disodium hydrogen phosphate $\text{Na}_2\text{HPO}_4$ (after 2 mol). The anions $\text{H}_2\text{PO}_4^-$ and $\text{HPO}_4^{2-}$ are amphiprotic: each can donate the next proton or accept one back, which is precisely why phosphate buffers are so important in cells. A mixture of $\text{H}_2\text{PO}_4^-$ and $\text{HPO}_4^{2-}$ buffers around $\text{pH}\ 7.2$ , close to the value $K_{a2}$ would predict, because at that ratio the two species are present in comparable amounts.

Why this matters

Polyprotic acids appear throughout chemistry and biology: carbonic acid buffers blood and oceans, phosphoric acid buffers cells, and sulfuric acid is the most produced industrial chemical. Understanding stepwise ionisation explains their buffering, their titration curves and their pH.

Treating all protons as equally available

A frequent error is assuming a polyprotic acid releases all its protons fully and at once, giving far too low a pH. In reality only the first ionisation is significant for most weak polyprotic acids. Another slip is forgetting the stoichiometric ratio in neutralisation, for example using a 1:1 ratio for phosphoric acid instead of 1:3.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2021-style6 marks

Carbonic acid is diprotic with

K_{a1} = 4.3 \times 10^{-7}

and

K_{a2} = 4.8 \times 10^{-11}

. (a) Write the two ionisation equilibria. (b) For a

0.10\ \text{mol L}^{-1}

solution, calculate the

\text{pH}

, justifying any approximation. (c) Explain why

K_{a2} \ll K_{a1}

Show worked answer →

A 6 mark question rewards the equilibria, the justified pH, and the electrostatic explanation.

(a) $\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-$ ( $K_{a1}$ ); $\text{HCO}_3^- \rightleftharpoons \text{H}^+ + \text{CO}_3^{2-}$ ( $K_{a2}$ ).

(b) Because $K_{a1} \gg K_{a2}$ , essentially all the $\text{H}^+$ comes from the first step, so treat it as monoprotic. With $x = [\text{H}^+]$ and the approximation $0.10 - x \approx 0.10$ :

K_{a1} = \frac{x^2}{0.10} \Rightarrow x^2 = (4.3 \times 10^{-7})(0.10) = 4.3 \times 10^{-8},

x = 2.07 \times 10^{-4}\ \text{mol L}^{-1},\qquad \text{pH} = -\log_{10}(2.07 \times 10^{-4}) = 3.68.

The approximation is valid because

x

is far less than

5\%

0.10

(c) The second proton is removed from the negatively charged $\text{HCO}_3^-$ ion. The extra electrostatic attraction between the leaving $\text{H}^+$ and the increasingly negative ion makes the second ionisation far less favourable, so $K_{a2} \ll K_{a1}$ .

Markers reward both equilibria, $\text{pH} = 3.68$ with the single-step justification, and the electrostatic reasoning.

WACE 2023-style5 marks

25.0\ \text{mL}

0.100\ \text{mol L}^{-1}

phosphoric acid

\text{H}_3\text{PO}_4

is titrated with

0.100\ \text{mol L}^{-1}\ \text{NaOH}

. (a) Write the equation for complete neutralisation. (b) Calculate the volume of

\text{NaOH}

needed to reach the final equivalence point. (c) Explain why the titration curve shows distinct steps.

Show worked answer →

A 5 mark question rewards the equation, the stoichiometric volume, and the stepwise reasoning.

(a) $\text{H}_3\text{PO}_4 + 3\text{NaOH} \rightarrow \text{Na}_3\text{PO}_4 + 3\text{H}_2\text{O}$ .

(b) Moles of acid $= 0.0250\ \text{L} \times 0.100\ \text{mol L}^{-1} = 2.50 \times 10^{-3}\ \text{mol}$ . Complete neutralisation needs 3 mol base per mol acid, so moles $\text{NaOH} = 3 \times 2.50 \times 10^{-3} = 7.50 \times 10^{-3}\ \text{mol}$ . Volume $= \dfrac{7.50 \times 10^{-3}}{0.100} = 0.0750\ \text{L} = 75.0\ \text{mL}$ .

(c) Phosphoric acid loses its three protons in steps with widely separated ionisation constants ( $K_{a1} \gg K_{a2} \gg K_{a3}$ ). Each proton is neutralised in turn, so the pH rises sharply at each successive equivalence point, producing distinct steps in the curve (here at 25.0 and 50.0 mL before the final point at 75.0 mL).

Markers reward the 1:3 equation, $75.0\ \text{mL}$ , and the stepwise ionisation explanation.

Practise this

Sources & how we know this

Chemistry ATAR course: syllabus and support materials — SCSA
Chemistry: past ATAR course examinations — SCSA