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How do acids that can donate more than one proton ionise, and why does each successive ionisation get weaker?

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

A focused answer to the WACE Year 12 Chemistry dot point on polyprotic acids, their stepwise ionisation, why each successive Ka is smaller, and how this affects pH and titrations, with a worked example and common exam mistakes.

Generated by Claude Opus 4.88 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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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 H2SO4\text{H}_2\text{SO}_4 and carbonic acid H2CO3\text{H}_2\text{CO}_3) and triprotic acids (three protons, such as phosphoric acid H3PO4\text{H}_3\text{PO}_4).

Stepwise ionisation

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

H3PO4⇌H++H2PO4−Ka1\text{H}_3\text{PO}_4 \rightleftharpoons \text{H}^+ + \text{H}_2\text{PO}_4^- \quad K_{a1}

H2PO4−⇌H++HPO42−Ka2\text{H}_2\text{PO}_4^- \rightleftharpoons \text{H}^+ + \text{HPO}_4^{2-} \quad K_{a2}

HPO42−⇌H++PO43−Ka3\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 10510^5 each time.

Why successive constants decrease

After the first proton leaves, the remaining species carries a negative charge. Removing a second positive proton (H+\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

Ka1≫Ka2≫Ka3K_{a1} \gg K_{a2} \gg K_{a3}

Sulfuric acid is a special case

Sulfuric acid is strong in its first ionisation (H2SO4→H++HSO4−\text{H}_2\text{SO}_4 \rightarrow \text{H}^+ + \text{HSO}_4^-, essentially complete) but weak in its second (HSO4−⇌H++SO42−\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.

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 NaH2PO4\text{NaH}_2\text{PO}_4 (after 1 mol of base per mol of acid) and disodium hydrogen phosphate Na2HPO4\text{Na}_2\text{HPO}_4 (after 2 mol). The anions H2PO4−\text{H}_2\text{PO}_4^- and HPO42−\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 H2PO4−\text{H}_2\text{PO}_4^- and HPO42−\text{HPO}_4^{2-} buffers around pH 7.2\text{pH}\ 7.2, close to the value Ka2K_{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.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20216 marksCarbonic acid is diprotic with Ka1=4.3×10−7K_{a1} = 4.3 \times 10^{-7} and Ka2=4.8×10−11K_{a2} = 4.8 \times 10^{-11}. (a) Write the two ionisation equilibria. (b) For a 0.10 mol L−10.10\ \text{mol L}^{-1} solution, calculate the pH\text{pH}, justifying any approximation. (c) Explain why Ka2≪Ka1K_{a2} \ll K_{a1}.
Show worked answer →

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

(a) H2CO3⇌H++HCO3−\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^- (Ka1K_{a1}); HCO3−⇌H++CO32−\text{HCO}_3^- \rightleftharpoons \text{H}^+ + \text{CO}_3^{2-} (Ka2K_{a2}).

(b) Because Ka1≫Ka2K_{a1} \gg K_{a2}, essentially all the H+\text{H}^+ comes from the first step, so treat it as monoprotic. With x=[H+]x = [\text{H}^+] and the approximation 0.10−x≈0.100.10 - x \approx 0.10:

Ka1=x20.10⇒x2=(4.3×10−7)(0.10)=4.3×10−8,K_{a1} = \frac{x^2}{0.10} \Rightarrow x^2 = (4.3 \times 10^{-7})(0.10) = 4.3 \times 10^{-8},

x=2.07×10−4 mol L−1,pH=−log⁡10(2.07×10−4)=3.68.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 xx is far less than 5%5\% of 0.100.10.

(c) The second proton is removed from the negatively charged HCO3−\text{HCO}_3^- ion. The extra electrostatic attraction between the leaving H+\text{H}^+ and the increasingly negative ion makes the second ionisation far less favourable, so Ka2≪Ka1K_{a2} \ll K_{a1}.

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

WACE 20235 marks25.0 mL25.0\ \text{mL} of 0.100 mol L−10.100\ \text{mol L}^{-1} phosphoric acid H3PO4\text{H}_3\text{PO}_4 is titrated with 0.100 mol L−1 NaOH0.100\ \text{mol L}^{-1}\ \text{NaOH}. (a) Write the equation for complete neutralisation. (b) Calculate the volume of NaOH\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) H3PO4+3NaOH→Na3PO4+3H2O\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 L×0.100 mol L−1=2.50×10−3 mol= 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 NaOH=3×2.50×10−3=7.50×10−3 mol\text{NaOH} = 3 \times 2.50 \times 10^{-3} = 7.50 \times 10^{-3}\ \text{mol}. Volume =7.50×10−30.100=0.0750 L=75.0 mL= \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 (Ka1≫Ka2≫Ka3K_{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 mL75.0\ \text{mL}, and the stepwise ionisation explanation.

Openly AI-built from the official SCSA syllabus, past papers and marking guides. Not individually reviewed; always check SCSAand your school's materials for current rules.

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