Why Not Every Hydrogen Atom in a Molecule Is Acidic

Ethanoic acid (acetic acid), \(\ce{CH3COOH}\), contains four hydrogen atoms. Yet when it reacts with sodium hydroxide in an ordinary aqueous neutralisation, one mole of acetic acid reacts with just one mole of \(\ce{OH-}\), not four.

Only the hydrogen in the carboxyl \(\ce{-OH}\) group is appreciably donated as a proton. The other three hydrogen atoms are attached to carbon and are not acidic under ordinary aqueous conditions.

The reason is that acidity isn’t determined by how many hydrogen atoms a molecule contains. It depends on whether a hydrogen can be transferred as a proton and how stable the molecule becomes after losing it.

Consider this question before reading on: if all four hydrogen atoms in acetic acid were acidic, how many moles of sodium hydroxide would one mole of acetic acid require?

Four. But experimental evidence gives a ratio of one mole of sodium hydroxide to one mole of acetic acid. Let’s work out why.

01An acidic hydrogen must be transferable as a proton

The Brønsted-Lowry definition describes an acid as a substance that donates a proton, \(\mathrm{H}^{+}\), to another substance.

This is different from simply containing hydrogen.

A hydrogen atom contains one proton and one electron. When an acid donates a proton, the hydrogen nucleus transfers to a base, while the bonding electrons remain with the original molecule.

For example, acetic acid transfers its O-H proton to water:

\[ \ce{CH3COOH + H2O <=> H3O+ + CH3COO-} \]

Water accepts the proton and becomes hydronium, \(\ce{H3O+}\). Acetic acid loses the proton and becomes the acetate ion, \(\ce{CH3COO-}\).

This gives us two conjugate acid-base pairs:

A conjugate acid-base pair differs by exactly one proton.

Notice what happened to the original acetic acid molecule. It did not lose all four hydrogen atoms. It lost one proton, and the remaining structure became an acetate ion.

Why is the O-H hydrogen acidic?

There are two important reasons.

1. The O-H bond is polar.

Oxygen is more electronegative than hydrogen. It attracts the shared electrons in the O-H bond towards itself.

This leaves the hydrogen partially positive, written as \(\delta+\).

The hydrogen can therefore be transferred as a proton relatively readily compared with the hydrogens attached to carbon.

2. The conjugate base is stabilised.

This is the more important part of the explanation.

When acetic acid loses its O-H proton, it forms the acetate ion. The resulting negative charge can be delocalised across the two oxygen atoms.

This is described using resonance:

\[ \ce{CH3-C(=O)-O- <-> CH3-C(-O-)=O} \]

These are two resonance contributors representing the same delocalised ion, not two separate molecules switching between structures.

The negative charge is spread across both oxygen atoms rather than being confined to just one location. This stabilises the conjugate base and makes proton donation more favourable.

What happens if one of the three C-H hydrogens is removed instead?

The resulting species is a carbon-based, enolate-like anion. Although some resonance stabilisation is possible, this is much less favourable than forming the acetate ion.

Water is not a sufficiently strong base to remove these carbon-bound hydrogens to any appreciable extent.

This is why the three C-H hydrogens are not counted as acidic protons in ordinary aqueous acid-base chemistry.

A useful comparison: acetic acid versus ethanol

Here is an important complication.

You might now predict that every hydrogen attached to oxygen must be strongly acidic. That isn’t correct either.

Compare acetic acid with ethanol, \(\ce{CH3CH2OH}\).

Both molecules contain an O-H bond, but acetic acid is much more acidic.

When ethanol loses its O-H proton, it forms the ethoxide ion, \(\ce{CH3CH2O-}\). Its negative charge is largely localised on one oxygen atom.

When acetic acid loses its O-H proton, it forms acetate, with negative charge delocalised across two oxygen atoms.

The acetate ion is more stabilised, so acetic acid donates its proton much more readily.

The important lesson is that bond polarity helps predict acidity, but the stability of the conjugate base also matters.

A few examples help put this into perspective:

An important qualification: a C-H hydrogen is not impossible to remove under every conceivable condition. Very strong bases can remove certain carbon-bound hydrogens in specialised reactions. However, those conditions are different from the ordinary aqueous acid-base reactions considered here.

02How titration provides experimental evidence

We can test how many acidic protons a molecule donates by measuring how much base is needed to neutralise a known amount of acid.

This is the purpose of an acid-base titration.

Imagine setting up an experiment using acetic acid and sodium hydroxide.

What does the apparatus measure?

A measured volume of acetic acid is placed in a conical flask.

A burette above the flask contains sodium hydroxide of known concentration. The sodium hydroxide is added gradually, while the solution is stirred.

A pH probe measures the pH after each addition. The results can be plotted as pH against the volume of sodium hydroxide added.

The key measurement is the volume of sodium hydroxide required to reach the equivalence point, where the acid has been neutralised according to the reaction’s stoichiometric ratio.

What should we predict?

Suppose the flask contains:

• 25.00 mL of 0.1000 mol L\(^{-1}\) acetic acid.
• The burette contains 0.1000 mol L\(^{-1}\) sodium hydroxide.

If all four hydrogen atoms in acetic acid were acidic, four moles of sodium hydroxide would be needed per mole of acid.

If only one hydrogen is acidic, the ratio would be one mole of sodium hydroxide per mole of acid.

These predictions give very different equivalence volumes:

For acetic acid, the expected equivalence volume is approximately 25.00 mL.

The observation supports the conclusion that acetic acid is monoprotic in this aqueous neutralisation.

What are the limitations?

A titration measures the amount of base needed for neutralisation. It does not directly photograph a proton leaving a particular bond.

The molecular structure and the stability of the conjugate base explain which hydrogen is being transferred.

There is another complication with polyprotic acids. Their different protons can have very different acid strengths. Sometimes the successive neutralisation stages produce overlapping regions on the pH curve.

A single obvious jump does not automatically prove that a molecule has only one acidic proton.

A pH probe and a complete titration curve provide more information than relying on an indicator colour change alone. The sodium hydroxide concentration should also be standardised so that the calculated mole ratios are reliable.

The experiment therefore gives us a useful distinction: the number of hydrogen atoms in a formula is not necessarily the number of proton equivalents that an acid can donate.

04Strength, amount, concentration and number of acidic protons

These four ideas are related, but they describe different properties.

Confusing them is a common source of mistakes in acid-base questions.

Acid strength: how readily does the acid ionise?

For a general weak acid, \(\ce{HA}\), reacting with water:

\[ \ce{HA + H2O <=> H3O+ + A-} \]

The acid dissociation constant is:

\[ K_a=\frac{[\ce{H3O+}][\mathrm{A}^{-}]}{[\ce{HA}]} \]

A larger \(K_a\) indicates that the equilibrium lies further towards the products, meaning more acid molecules have ionised.

A smaller \(K_a\) indicates that less ionisation occurs.

For example:

• Hydrochloric acid is a strong acid and ionises essentially completely in water.
• Acetic acid is a weak acid and only partially ionises in water.

Strength is not determined by the number of hydrogen atoms in the formula.

For a polyprotic acid, each successive proton transfer has its own dissociation constant. The first proton can be much easier to remove than the second.

If you need to develop this distinction further, see How to Explain Acid Strength Using Equilibrium Rather Than pH Alone.

Amount and concentration: how much acid is present?

Amount is measured in moles.

Concentration describes how many moles are present per litre of solution.

The relationship is:

\[ n=cV \]

A solution containing 0.100 mol of acid in 1.00 L has a concentration of 0.100 mol L\(^{-1}\).

A solution containing the same 0.100 mol in 0.100 L has a concentration of 1.00 mol L\(^{-1}\).

The amount of acid is identical, but the concentrations are different.

Neither amount nor concentration tells you, by itself, how strongly the acid ionises.

Proticity: how many protons can the acid donate?

Proticity describes the number of acidic protons available for transfer per molecule.

• Monoprotic acids can donate one proton per molecule.
• Diprotic acids can donate two protons per molecule.
• Triprotic acids can donate three protons per molecule.

For example, sulfuric acid, \(\ce{H2SO4}\), is diprotic.

Its two protons can be transferred in successive stages:

\[ \ce{H2SO4 + H2O -> H3O+ + HSO4-} \]

\[ \ce{HSO4- + H2O <=> H3O+ + SO4^2-} \]

The first dissociation is essentially complete in water, while the second is an equilibrium reaction.

Both protons can be neutralised by a sufficiently strong base, but they do not have equal acid strength.

This is a particularly useful distinction: an acid can be diprotic without both protons being equally strong.

06Questions and solutions

The following questions move from identifying acidic hydrogens to analysing equilibrium, titration data, and polyprotic acids.

Question 1

Ethanoic acid has the molecular formula \(\ce{CH3COOH}\).

(a) State how many hydrogen atoms are present in one molecule.

(b) Identify the hydrogen atom that is donated when ethanoic acid reacts with water.

(c) Write the Brønsted-Lowry reaction with water and identify both conjugate acid-base pairs.

Solution 1

Only one of the four hydrogen atoms is appreciably acidic in ordinary aqueous conditions: the hydrogen in the O-H group.

(a) There are four hydrogen atoms in total: three attached to carbon and one attached to oxygen.

(b) The O-H hydrogen is donated as a proton.

The O-H bond is polar, and removing this proton produces the resonance-stabilised acetate ion.

(c) The reaction is:

\[ \ce{CH3COOH + H2O <=> H3O+ + CH3COO-} \]

The conjugate pairs are:

• \(\ce{CH3COOH}\) and \(\ce{CH3COO-}\).
• \(\ce{H2O}\) and \(\ce{H3O+}\).

Each pair differs by one proton.

The three C-H hydrogens are not appreciably transferred to water because removing one would produce a much less favourable carbon-based anion.

Question 2

A 20.00 mL sample of propanoic acid, \(\ce{CH3CH2COOH}\), is titrated with 0.1250 mol L\(^{-1}\) sodium hydroxide.

The equivalence point is reached after 16.00 mL of sodium hydroxide has been added.

(a) Calculate the number of moles of sodium hydroxide used.

(b) Determine the concentration of propanoic acid.

(c) Propanoic acid contains six hydrogen atoms. Explain why only one mole of sodium hydroxide is required per mole of propanoic acid.

Solution 2

The propanoic acid concentration is \(0.1000\text{ mol L}^{-1}\), and it donates one acidic proton per molecule in this aqueous neutralisation.

(a) Calculate the moles of sodium hydroxide.

Use:

\[ n=cV \]

Convert the volume into litres:

\[ 16.00\text{ mL}=0.01600\text{ L} \]

Substitute:

\[ \begin{aligned} n(\ce{NaOH}) &=0.1250\times0.01600\\ &=0.002000\text{ mol} \end{aligned} \]

(b) Calculate the propanoic acid concentration.

The neutralisation equation is:

\[ \ce{CH3CH2COOH + OH- -> CH3CH2COO- + H2O} \]

The mole ratio is 1:1.

Therefore:

\[ n(\ce{CH3CH2COOH})=0.002000\text{ mol} \]

The initial acid volume is 0.02000 L.

Using \(c=n/V\):

\[ \begin{aligned} c&=\frac{0.002000}{0.02000}\\ &=0.1000\text{ mol L}^{-1} \end{aligned} \]

(c) Explain the number of acidic protons.

Propanoic acid has the structure \(\ce{CH3CH2COOH}\).

Its six hydrogen atoms consist of:

• Three C-H hydrogens in the \(\ce{CH3}\) group.
• Two C-H hydrogens in the \(\ce{CH2}\) group.
• One O-H hydrogen in the carboxyl group.

Only the O-H hydrogen is appreciably donated in ordinary aqueous acid-base reactions.

The carboxylate ion formed after proton transfer is stabilised by resonance.

Therefore, one mole of propanoic acid requires one mole of sodium hydroxide for neutralisation.

Question 3

Two solutions are prepared at 25°C.

Solution A:

• 100.0 mL of 0.0100 mol L\(^{-1}\) hydrochloric acid.

Solution B:

• 10.0 mL of 0.100 mol L\(^{-1}\) acetic acid.

The \(K_a\) of acetic acid is \(1.8\times10^{-5}\).

Assume hydrochloric acid ionises completely and the solutions behave ideally.

(a) Calculate the amount of acid in each solution.

(b) Calculate the initial hydronium concentration and approximate pH of each solution.

(c) Which solution has the greater acid concentration?

(d) Which acid is stronger, and why does the more concentrated solution not have the lower pH?

Solution 3

Both solutions contain \(0.00100\) mol of acid, but hydrochloric acid produces a higher hydronium concentration and therefore has the lower pH.

(a) Calculate the amount of acid in each solution.

For solution A:

\[ \begin{aligned} n_A&=cV\\ &=0.0100\times0.1000\\ &=0.00100\text{ mol} \end{aligned} \]

For solution B:

\[ \begin{aligned} n_B&=cV\\ &=0.100\times0.0100\\ &=0.00100\text{ mol} \end{aligned} \]

Both solutions contain the same amount of acid.

(b) Calculate the hydronium concentrations and pH values.

For solution A, hydrochloric acid ionises completely.

Therefore:

\[ [\ce{H3O+}]=0.0100\text{ mol L}^{-1} \]

\[ \begin{aligned} \text{pH} &=-\log_{10}(0.0100)\\ &=2.00 \end{aligned} \]

For solution B, acetic acid only partially ionises.

Let \(x\) represent the hydronium concentration produced by ionisation.

Using the equilibrium expression:

\[ 1.8\times10^{-5} =\frac{x^2}{0.100-x} \]

Since \(x\) is small compared with 0.100:

\[ \begin{aligned} x&\approx\sqrt{(1.8\times10^{-5})(0.100)}\\ &\approx1.34\times10^{-3}\text{ mol L}^{-1} \end{aligned} \]

Therefore:

\[ \begin{aligned} \text{pH} &=-\log_{10}(1.34\times10^{-3})\\ &\approx2.87 \end{aligned} \]

(c) Compare the concentrations.

Solution B is more concentrated.

Its acid concentration is 0.100 mol L\(^{-1}\), compared with 0.0100 mol L\(^{-1}\) for solution A.

Solution B is ten times more concentrated.

(d) Compare acid strengths.

Hydrochloric acid is stronger because it ionises essentially completely in water.

Acetic acid is weak because most of its molecules remain undissociated.

Consequently, even though solution B is more concentrated, solution A has a higher hydronium concentration and a lower pH.

This demonstrates why concentration and strength must be considered separately.

Question 4

A 10.00 mL sample of 0.150 mol L\(^{-1}\) sulfuric acid is titrated with 0.100 mol L\(^{-1}\) sodium hydroxide until complete neutralisation.

(a) Calculate the volume of sodium hydroxide required.

(b) Explain why two moles of sodium hydroxide are required per mole of sulfuric acid, even though its second dissociation is less complete than its first.

Solution 4

A volume of 30.0 mL of sodium hydroxide is required because sulfuric acid can donate two protons per molecule during complete neutralisation.

(a) Calculate the required volume.

First, calculate the moles of sulfuric acid:

\[ \begin{aligned} n(\ce{H2SO4}) &=cV\\ &=0.150\times0.01000\\ &=0.00150\text{ mol} \end{aligned} \]

Sulfuric acid is diprotic.

The overall neutralisation equation is:

\[ \ce{H2SO4 + 2OH- -> SO4^2- + 2H2O} \]

The mole ratio is therefore 1:2.

Calculate the required moles of sodium hydroxide:

\[ \begin{aligned} n(\ce{NaOH}) &=2\times0.00150\\ &=0.00300\text{ mol} \end{aligned} \]

Now calculate the volume:

\[ V=\frac{n}{c} \]

\[ \begin{aligned} V&=\frac{0.00300}{0.100}\\ &=0.0300\text{ L}\\ &=30.0\text{ mL} \end{aligned} \]

(b) Explain the different dissociation strengths.

Sulfuric acid undergoes two successive proton-transfer reactions.

The first dissociation is essentially complete:

\[ \ce{H2SO4 + H2O -> H3O+ + HSO4-} \]

The second dissociation is an equilibrium:

\[ \ce{HSO4- + H2O <=> H3O+ + SO4^2-} \]

The second proton is less readily donated than the first.

However, this does not mean that it cannot be neutralised.

When sodium hydroxide is added, hydroxide ions react with hydronium ions and remove them from the solution. This favours further proton transfer from hydrogensulfate.

The overall reaction therefore consumes two moles of hydroxide per mole of sulfuric acid.

The important distinction is that the two protons have different strengths, but both contribute to the acid’s total neutralisation capacity.

Question 5

Consider a hypothetical diprotic acid, \(\ce{H2A}\), containing two O-H hydrogens.

Its first dissociation is effectively complete in water, while its second dissociation has:

\[ K_{a2}=1.0\times10^{-7} \]

A 25.0 mL sample of 0.100 mol L\(^{-1}\) \(\ce{H2A}\) is prepared at 25°C.

Its initial pH is approximately 1.00.

The sample is titrated with 0.100 mol L\(^{-1}\) sodium hydroxide. The final stoichiometric neutralisation endpoint occurs after 50.0 mL of sodium hydroxide has been added.

A student argues:

“The initial pH shows that the acid produces approximately 0.100 mol L\(^{-1}\) hydronium ions. The second dissociation is negligible, so only one hydrogen is actually acidic. The second hydrogen should not count.”

Assess this argument.

In your response:

(a) Explain why the second dissociation contributes almost nothing to the initial hydronium concentration.

(b) Calculate the volume of sodium hydroxide required to neutralise the first proton equivalent.

(c) Explain why the second proton can still be neutralised, despite contributing very little to the initial pH.

(d) Determine the total number of acidic protons per molecule and explain the apparent contradiction between the initial pH and the titration result.

Solution 5

The student’s argument is incorrect. The acid is diprotic, even though its second proton contributes almost nothing to the initial hydronium concentration.

The mistake is treating the initial pH as a measurement of the total number of protons that can eventually be donated.

(a) Explain the initial hydronium concentration.

The first dissociation is effectively complete:

\[ \ce{H2A + H2O -> H3O+ + HA-} \]

The initial acid concentration is 0.100 mol L\(^{-1}\).

Therefore, the first dissociation produces approximately:

\[ [\ce{H3O+}]\approx0.100\text{ mol L}^{-1} \]

This gives:

\[ \text{pH}\approx-\log_{10}(0.100)=1.00 \]

The second dissociation is:

\[ \ce{HA- + H2O <=> H3O+ + A^2-} \]

Its equilibrium constant is:

\[ K_{a2}= \frac{[\ce{H3O+}][\mathrm{A}^{2-}]} {[\ce{HA-}]} \]

Rearrange:

\[ \frac{[\mathrm{A}^{2-}]}{[\ce{HA-}]} = \frac{K_{a2}}{[\ce{H3O+}]} \]

Substitute the initial values:

\[ \begin{aligned} \frac{[\mathrm{A}^{2-}]}{[\ce{HA-}]} &=\frac{1.0\times10^{-7}}{0.100}\\ &=1.0\times10^{-6} \end{aligned} \]

This means the concentration of \(\mathrm{A}^{2-}\) is extremely small compared with \(\ce{HA-}\) initially.

The large hydronium concentration suppresses the second dissociation because hydronium is already a product of that equilibrium.

This is an example of the common-ion effect.

(b) Calculate the first proton equivalent.

First, calculate the initial moles of acid:

\[ \begin{aligned} n(\ce{H2A}) &=cV\\ &=0.100\times0.0250\\ &=0.00250\text{ mol} \end{aligned} \]

The first dissociation transfers one proton per molecule.

Therefore, the first neutralisation stage requires:

\[ n(\ce{OH-})=0.00250\text{ mol} \]

Using:

\[ V=\frac{n}{c} \]

\[ \begin{aligned} V&=\frac{0.00250}{0.100}\\ &=0.0250\text{ L}\\ &=25.0\text{ mL} \end{aligned} \]

The first stoichiometric equivalence volume is 25.0 mL.

(c) Explain why the second proton can still be neutralised.

As sodium hydroxide is added, hydroxide ions react with hydronium ions:

\[ \ce{H3O+ + OH- -> 2H2O} \]

This removes hydronium from the solution.

But hydronium is a product of the second dissociation equilibrium:

\[ \ce{HA- + H2O <=> H3O+ + A^2-} \]

Removing a product shifts this equilibrium towards the right.

Consequently, more \(\ce{HA-}\) dissociates, releasing additional protons.

The hydroxide ions can also react directly with hydrogensulfate-like species \(\ce{HA-}\):

\[ \ce{HA- + OH- -> A^2- + H2O} \]

The second proton is therefore progressively neutralised as more sodium hydroxide is added.

It was not readily released in the original acidic solution, but removing hydronium changes the equilibrium conditions.

(d) Determine the total number of acidic protons.

The acid has two acidic protons per molecule.

The first neutralisation stage requires:

\[ 0.00250\text{ mol OH-} \]

The second stage requires the same additional amount:

\[ 0.00250\text{ mol OH-} \]

Therefore, the total hydroxide requirement is:

\[ \begin{aligned} n(\ce{OH-})_{\text{total}} &=2\times0.00250\\ &=0.00500\text{ mol} \end{aligned} \]

The total volume of sodium hydroxide is:

\[ \begin{aligned} V&=\frac{0.00500}{0.100}\\ &=0.0500\text{ L}\\ &=50.0\text{ mL} \end{aligned} \]

This agrees with the experimental endpoint.

The overall neutralisation reaction is:

\[ \ce{H2A + 2OH- -> A^2- + 2H2O} \]

The apparent contradiction disappears once we distinguish between two different measurements:

• Initial pH measures the hydronium concentration present at that particular moment.
• Titration measures the total amount of base needed to neutralise the acid’s transferable protons.

The second proton is strongly suppressed in the initial solution, but it becomes available as hydronium is removed.

This is why a pH measurement alone cannot establish how many acidic protons a molecule can donate.

07What to remember when analysing an acid’s structure

When you encounter a molecule containing several hydrogen atoms, don’t simply count them.

Instead, identify which hydrogen can transfer as a proton, then consider the stability of the conjugate base left behind.

For acetic acid, the O-H proton is appreciably acidic because its removal produces a resonance-stabilised acetate ion. The three C-H hydrogens are not appreciably donated in ordinary aqueous conditions.

Keep four questions separate:

1. How many protons can the molecule donate? This determines its proticity.
2. How readily does each proton dissociate? This determines acid strength.
3. How many moles of acid are present? This determines the amount.
4. How many moles are present per litre? This determines concentration.

Once you can identify the transferable protons, the next step is to write their conjugate acid-base pairs and analyse each dissociation equilibrium. That connects molecular structure directly to \(K_a\), pH, and the different stages of a polyprotic acid titration.