How to Distinguish Amphiprotic from Amphoteric in HSC Chemistry

Zinc oxide, \(\ce{ZnO}\), dissolves in both hydrochloric acid and sufficiently concentrated sodium hydroxide. A student sees this and calls it amphiprotic. Is that correct?

Not quite. Zinc oxide is amphoteric, but not amphiprotic. The difference comes down to proton transfer: an amphiprotic species must be able to both donate and accept a proton (\(\mathrm{H}^{+}\)), whereas an amphoteric substance can behave as both an acid and a base through broader types of acid-base reactions.

This distinction becomes much easier once you know exactly what to look for in a chemical equation.

01The distinction in one sentence

There are two definitions to remember:

• Amphiprotic: A species that can both donate and accept a proton.
• Amphoteric: A substance that can behave as both an acid and a base, without necessarily doing so through proton transfer in both reactions.

Amphiprotic is therefore the more specific term. Every amphiprotic species is amphoteric in the broad acid-base sense, but not every amphoteric substance is amphiprotic.

For example:

The important point is that reacting with both acids and bases does not automatically make something amphiprotic.

02Why the zinc oxide experiment is not enough

Imagine setting up a simple investigation to determine how zinc oxide behaves in different solutions.

Place equal amounts of white zinc oxide powder into three labelled test tubes:

• Tube A receives hydrochloric acid.
• Tube B receives excess, sufficiently concentrated sodium hydroxide.
• Tube C receives distilled water as a control.

Swirl each tube and observe whether the solid remains. A more quantitative investigation could filter, dry, and weigh the remaining solid.

What should happen?

The prediction is that zinc oxide will react with both the acid and the excess base, but not with water under these conditions.

The expected reactions are:

Tube A: Zinc oxide reacting with hydrochloric acid

\[
\ce{ZnO(s) + 2H3O+(aq) -> Zn^2+(aq) + 3H2O(l)}
\]

The oxide ions in zinc oxide accept protons from hydronium ions. Zinc oxide therefore behaves as a base in this reaction.

Tube B: Zinc oxide reacting with excess sodium hydroxide

\[
\ce{ZnO(s) + 2OH^-(aq) + H2O(l) -> [Zn(OH)4]^2-(aq)}
\]

Zinc oxide reacts with hydroxide ions to form the soluble complex ion tetrahydroxozincate.

The zinc centres accept electron pairs from hydroxide ions. This is an example of Lewis acid-base behaviour, rather than the donation of a proton by zinc oxide.

What does the experiment actually prove?

If zinc oxide dissolves in both acid and excess base, the observations support its classification as amphoteric.

However, they do not establish that zinc oxide is amphiprotic.

There is a problem with the original experimental design: measuring how much solid dissolves tells us whether a reaction occurs, but does not establish whether the solid can donate and accept protons.

In fact, the formula \(\ce{ZnO}\) contains no hydrogen atoms. It cannot donate a proton.

This is why we need to look beyond the visible reaction and examine what happens at the particle level.

As a small stoichiometric check, the acid reaction has a 1:2 mole ratio between zinc oxide and hydronium ions. Therefore, completely reacting \(0.00500\) mol of zinc oxide requires:

\[
n(\ce{H3O+})=2(0.00500)
\]

\[
=0.0100\text{ mol}
\]

This tells us how much acid the zinc oxide consumes. It still tells us nothing about proton donation by the oxide.

Why the definitions of acids and bases matter

This distinction makes more sense when you consider how acid-base models developed.

The Arrhenius model describes acids as substances that increase hydronium ion concentration in water and bases as substances that increase hydroxide ion concentration.

The Brønsted-Lowry model provides a more general explanation:

• An acid donates a proton.
• A base accepts a proton.

This is the model you need for identifying amphiprotic species.

A further extension is the Lewis model, which describes acids as electron-pair acceptors and bases as electron-pair donors. This helps explain reactions such as zinc oxide reacting with hydroxide ions.

The Lewis explanation is useful additional context, rather than a requirement for demonstrating the amphiprotic nature of the named salts in the 2017 HSC Chemistry syllabus.

03The proton-transfer test: how to identify an amphiprotic species

Forget about whether a substance dissolves in acid or has a particular pH for a moment.

Instead, ask two questions:

1. Can the species donate a proton?
2. Can the same species accept a proton?

If both answers are yes, it is amphiprotic.

The easiest way to understand this is through conjugate acid-base pairs.

A conjugate acid-base pair consists of two species that differ by exactly one proton.

When an acid donates a proton, it becomes its conjugate base. When a base accepts a proton, it becomes its conjugate acid.

Example: Hydrogen carbonate

Consider the hydrogen carbonate ion, \(\mathrm{HCO_{3}}^{-}\).

It sits between carbonic acid and carbonate:

\[
\ce{H2CO3 <=> HCO3^- <=> CO3^2-}
\]

Think of this as a proton-transfer ladder.

• Moving left means gaining one proton.
• Moving right means losing one proton.

Let’s test both directions.

1. Hydrogen carbonate acting as an acid

An acid donates a proton. Hydrogen carbonate can lose one \(\mathrm{H}^{+}\) to form carbonate:

\[
\ce{HCO3^-(aq) + H2O(l) <=> CO3^2-(aq) + H3O+(aq)}
\]

Hydrogen carbonate donates a proton to water.

The conjugate acid-base pairs are:

• \(\ce{HCO3^- / CO3^2-}\)
• \(\ce{H3O+ / H2O}\)

Water accepts the proton, so water acts as a Brønsted-Lowry base.

2. Hydrogen carbonate acting as a base

A base accepts a proton. Hydrogen carbonate can gain one \(\mathrm{H}^{+}\) to form carbonic acid:

\[
\ce{HCO3^-(aq) + H2O(l) <=> H2CO3(aq) + OH^-(aq)}
\]

This time, water donates a proton to hydrogen carbonate.

The conjugate acid-base pairs are:

• \(\ce{H2CO3 / HCO3^-}\)
• \(\ce{H2O / OH^-}\)

Water now acts as a Brønsted-Lowry acid.

Hydrogen carbonate can therefore perform both roles. It is amphiprotic.

Notice that the two reactions do not have to occur to the same extent. Amphiprotic describes what a species can do, not whether its acidic and basic behaviours are equally strong.

A quick decision method

When an exam question gives you an unfamiliar species, use this process.

Step 1: Remove one proton.

Ask what the species becomes after donating exactly one \(\mathrm{H}^{+}\).

If the resulting species is chemically valid, it may be able to act as an acid.

Step 2: Add one proton.

Ask what the species becomes after accepting exactly one \(\mathrm{H}^{+}\).

If the resulting species is chemically valid, it may be able to act as a base.

Step 3: Check both conjugate pairs.

Make sure each pair differs by exactly one proton and that the charges balance.

If both proton transfers are possible, the species is amphiprotic.

There is an important trap here: containing hydrogen does not automatically make a species amphiprotic.

For example:

• \(\ce{NH4+}\) can donate a proton to form \(\ce{NH3}\), but does not accept another proton in ordinary aqueous HSC acid-base chemistry. It behaves as an acid.
• \(\mathrm{CO_{3}}^{2-}\) can accept a proton to form \(\mathrm{HCO_{3}}^{-}\), but cannot donate a proton because it has no hydrogen. It behaves as a base.
• \(\ce{ZnO}\) can react with both acids and bases, but cannot donate a proton. It is amphoteric, not amphiprotic.

04Strength, amount and concentration: three different ideas

There is another common source of confusion.

Students sometimes assume that an amphiprotic substance must have equally strong acidic and basic properties, or that a concentrated acid must be a strong acid.

Neither assumption is correct.

These terms describe different chemical properties.

1. Strength

The strength of an acid describes its tendency to donate protons.

A strong acid ionises almost completely in water. A weak acid ionises only partially.

The acid dissociation constant, \(K_a\), measures the equilibrium position of the acid dissociation reaction.

• Larger \(K_a\): stronger acid.
• Smaller \(K_a\): weaker acid.

Similarly, \(K_b\) measures base strength.

For an amphiprotic species, its acidic and basic reactions can have very different equilibrium constants. One behaviour can dominate even though both are possible.

2. Amount

The amount of a substance is the number of moles present.

It is calculated using:

\[
n=cV
\]

where:

• \(n\) is amount in mol.
• \(c\) is concentration in mol L\(^{-1}\).
• \(V\) is volume in L.

Adding more of an amphiprotic substance increases its amount. It does not change the chemical identity of the species.

3. Concentration

Concentration describes the amount of substance dissolved in a particular volume.

For example, \(0.100\text{ mol L}^{-1}\) ethanoic acid is more concentrated than \(0.00100\text{ mol L}^{-1}\) ethanoic acid.

Diluting an acid decreases its concentration, but does not turn a strong acid into a weak acid.

At a fixed temperature and in the same solvent, the acid dissociation constant \(K_a\) remains unchanged when the solution is diluted, although the degree of ionisation and pH can change.

05Questions and solutions

The following questions move from straightforward classification to more demanding problems involving equilibrium, experimental evidence and competing proton-transfer reactions.

Question 1

Classify each of the following species as:

• Amphiprotic.
• Amphoteric but not amphiprotic.
• Neither amphiprotic nor amphoteric.

Assume ordinary aqueous HSC acid-base conditions.

\[
\mathrm{HCO_{3}}^{-},\quad
\mathrm{CO_{3}}^{2-},\quad
\ce{ZnO},\quad
\ce{NH4+}
\]

Justify each classification by referring to the species’ ability to donate or accept a proton, or its broader acid-base behaviour.

Solution 1

Only \(\mathrm{HCO_{3}}^{-}\) is amphiprotic, while \(\ce{ZnO}\) is amphoteric but not amphiprotic.

The classifications are:

The important distinction is that being able to react with an acid or a base individually does not establish amphiprotic behaviour.

For that classification, you must demonstrate both directions of proton transfer.

Question 2

The dihydrogen phosphate ion, \(\mathrm{H_{2}PO_{4}}^{-}\), is an example of an amphiprotic species.

(a) Write the ionic equation showing \(\mathrm{H_{2}PO_{4}}^{-}\) acting as an acid in water.

(b) Write the ionic equation showing \(\mathrm{H_{2}PO_{4}}^{-}\) acting as a base in water.

(c) Identify the conjugate acid-base pairs in each reaction.

(d) Explain why water changes its role between the two reactions.

Solution 2

The dihydrogen phosphate ion is amphiprotic because it can donate a proton to form \(\mathrm{HPO_{4}}^{2-}\) and accept a proton to form \(\ce{H3PO4}\).

(a) Acting as an acid

The ion donates one proton to water:

\[
\ce{H2PO4^-(aq) + H2O(l) <=> HPO4^2-(aq) + H3O+(aq)}
\]

The conjugate acid-base pairs are:

• \(\ce{H2PO4^- / HPO4^2-}\)
• \(\ce{H3O+ / H2O}\)

(b) Acting as a base

The ion accepts one proton from water:

\[
\ce{H2PO4^-(aq) + H2O(l) <=> H3PO4(aq) + OH^-(aq)}
\]

The conjugate acid-base pairs are:

• \(\ce{H3PO4 / H2PO4^-}\)
• \(\ce{H2O / OH^-}\)

(c) Conjugate pairs

In each reaction, the conjugate acid and base differ by exactly one proton.

This is a useful way to check your equations. If the two species in a proposed conjugate pair differ by more than one proton, they are not a conjugate acid-base pair.

(d) Water changes its role because the direction of proton transfer changes.

In the first reaction, water accepts a proton, so it acts as a base.

In the second reaction, water donates a proton, so it acts as an acid.

This demonstrates that acid and base are descriptions of a species’ behaviour in a particular reaction, rather than permanent labels attached to the species.

Question 3

A student prepares \(25.0\text{ mL}\) of a \(0.200\text{ mol L}^{-1}\) sodium hydrogen carbonate solution.

The student transfers the entire sample into a \(250.0\text{ mL}\) volumetric flask and adds distilled water to the calibration mark.

(a) Calculate the amount of hydrogen carbonate ions initially present.

(b) Calculate the final concentration of hydrogen carbonate ions.

(c) Explain whether dilution changes the amphiprotic nature or intrinsic acid strength of hydrogen carbonate.

Solution 3

The final solution contains \(0.00500\text{ mol}\) of hydrogen carbonate ions at a concentration of \(0.0200\text{ mol L}^{-1}\). The species remains amphiprotic.

(a) Calculate the amount of hydrogen carbonate ions.

First, convert the initial volume into litres:

\[
V=25.0\text{ mL}=0.0250\text{ L}
\]

Use:

\[
n=cV
\]

Substitute:

\[
n=0.200\times0.0250
\]

\[
\boxed{n=0.00500\text{ mol}}
\]

The sodium hydrogen carbonate dissociates into sodium ions and hydrogen carbonate ions in solution. Therefore, the amount of \(\mathrm{HCO_{3}}^{-}\) is also \(0.00500\text{ mol}\).

(b) Calculate the final concentration.

The final volume is:

\[
V=250.0\text{ mL}=0.2500\text{ L}
\]

Dilution does not remove any hydrogen carbonate ions, so the amount remains unchanged.

Using:

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

\[
c=\frac{0.00500}{0.2500}
\]

\[
\boxed{c=0.0200\text{ mol L}^{-1}}
\]

The concentration has decreased by a factor of ten.

(c) Explain the effect of dilution.

Dilution does not change the identity or amphiprotic nature of hydrogen carbonate.

It can still donate a proton to form carbonate and accept a proton to form carbonic acid.

Its intrinsic acid dissociation constant, \(K_a\), also remains unchanged at the same temperature and in the same solvent.

However, the equilibrium concentrations and pH can change because the solution has been diluted.

This question tests an important distinction: changing concentration does not change whether a species is amphiprotic.

Question 4

A chemistry student prepares two aqueous solutions at the same temperature:

• \(0.10\text{ mol L}^{-1}\) sodium hydrogen carbonate.
• \(0.10\text{ mol L}^{-1}\) potassium dihydrogen phosphate.

The student measures the following pH values:

These are illustrative experimental readings.

The student concludes:

“Hydrogen carbonate is amphiprotic because its solution is basic. Dihydrogen phosphate cannot be amphiprotic because its solution is acidic.”

Evaluate this conclusion. Explain what the pH measurements establish and what additional chemical evidence is required to demonstrate amphiprotic behaviour.

Solution 4

The student’s conclusion is incorrect. Both hydrogen carbonate and dihydrogen phosphate are amphiprotic, despite producing solutions with different pH values.

The pH measurements indicate the net acidic or basic behaviour of each solution. They do not establish whether the dissolved species can both donate and accept protons.

For sodium hydrogen carbonate, \(\mathrm{HCO_{3}}^{-}\) can:

• Donate a proton to form \(\mathrm{CO_{3}}^{2-}\).
• Accept a proton to form \(\ce{H2CO3}\).

Its basic behaviour is more significant in water, so the solution is mildly basic.

For potassium dihydrogen phosphate, \(\mathrm{H_{2}PO_{4}}^{-}\) can:

• Donate a proton to form \(\mathrm{HPO_{4}}^{2-}\).
• Accept a proton to form \(\ce{H3PO4}\).

Its acidic behaviour dominates in water, so the solution is acidic.

The experimental readings are consistent with these different equilibrium behaviours.

However, pH alone cannot demonstrate that both proton transfers are possible.

To establish amphiprotic behaviour, you must identify the two possible proton transfers and their corresponding conjugate acid-base pairs.

The crucial distinction is:

pH tells you about the net acidity or basicity of a solution. Amphiprotic describes the proton-transfer capabilities of a species.

These are related, but they are not the same thing.

Question 5

A solution initially contains \(0.0250\text{ mol}\) of hydrogen carbonate ions in \(250.0\text{ mL}\) of water.

A chemist performs two additions in sequence:

1. Adds \(0.0100\text{ mol}\) of hydrochloric acid.
2. Adds \(0.0150\text{ mol}\) of sodium hydroxide.

Assume:

• Hydrochloric acid and sodium hydroxide dissociate completely.
• The relevant proton-transfer reactions proceed to completion for stoichiometric bookkeeping.
• No carbon dioxide escapes from the solution.
• Volume changes are ignored.
• All amounts refer to the dissolved carbon-containing species.

At \(25^\circ\text{C}\), use the approximate values:

\[
\mathrm{p}K_{a1}(\ce{H2CO3})=6.35
\]

\[
\mathrm{p}K_{a2}(\mathrm{HCO_{3}}^{-})=10.33
\]

(a) Determine the amounts of \(\ce{H2CO3}\), \(\mathrm{HCO_{3}}^{-}\) and \(\mathrm{CO_{3}}^{2-}\) immediately after the hydrochloric acid addition.

(b) Determine the final amounts of all three species after the sodium hydroxide addition.

(c) Explain why the first hydroxide ions added react with carbonic acid rather than immediately converting all the hydrogen carbonate ions into carbonate.

(d) Explain how the sequence demonstrates both the acidic and basic behaviour of hydrogen carbonate.

Solution 5

The final solution contains:

\[
\boxed{0.0200\text{ mol }\mathrm{HCO_{3}}^{-}}
\]

\[
\boxed{0.00500\text{ mol }\mathrm{CO_{3}}^{2-}}
\]

\[
\boxed{0\text{ mol }\ce{H2CO3}}
\]

The key is to recognise that hydroxide reacts with the stronger acid in the mixture first. Only after the carbonic acid has been consumed does the excess hydroxide react with hydrogen carbonate.

(a) Determine the amounts after hydrochloric acid addition.

The initial amount of hydrogen carbonate is:

\[
n(\mathrm{HCO_{3}}^{-})=0.0250\text{ mol}
\]

Hydrochloric acid supplies hydronium ions:

\[
\ce{HCl + H2O -> H3O+ + Cl-}
\]

The hydronium ions react with hydrogen carbonate:

\[
\ce{HCO3^- + H3O+ -> H2CO3 + H2O}
\]

The mole ratio is 1:1.

Therefore, \(0.0100\text{ mol}\) of hydrogen carbonate reacts with \(0.0100\text{ mol}\) of hydronium ions.

The remaining hydrogen carbonate is:

\[
0.0250-0.0100=0.0150\text{ mol}
\]

The amount of carbonic acid formed is:

\[
0.0100\text{ mol}
\]

Immediately after the acid addition:

Hydrogen carbonate has acted as a base because it accepted protons from hydronium ions.

(b) Determine the final amounts after sodium hydroxide addition.

The solution now contains \(0.0100\text{ mol}\) of carbonic acid.

The added sodium hydroxide supplies:

\[
n(\mathrm{OH}^{-})=0.0150\text{ mol}
\]

Hydroxide ions react with carbonic acid:

\[
\ce{H2CO3 + OH^- -> HCO3^- + H2O}
\]

The reaction consumes \(0.0100\text{ mol}\) of hydroxide ions.

The remaining hydroxide is:

\[
0.0150-0.0100
=0.00500\text{ mol}
\]

At this point, all the carbonic acid has been consumed, and the amount of hydrogen carbonate has returned to:

\[
0.0150+0.0100=0.0250\text{ mol}
\]

There is still \(0.00500\text{ mol}\) of hydroxide remaining.

This excess hydroxide can now react with hydrogen carbonate:

\[
\ce{HCO3^- + OH^- -> CO3^2- + H2O}
\]

The reaction has a 1:1 mole ratio.

Therefore, \(0.00500\text{ mol}\) of hydrogen carbonate is converted into carbonate.

The final hydrogen carbonate amount is:

\[
0.0250-0.00500
=0.0200\text{ mol}
\]

The final carbonate amount is:

\[
0.00500\text{ mol}
\]

The final amounts are:

(c) Explain why hydroxide reacts with carbonic acid first.

This is the most important part of the question.

A tempting approach is to assume that all hydroxide ions immediately react with hydrogen carbonate.

However, the solution contains two possible proton donors:

• Carbonic acid, \(\ce{H2CO3}\).
• Hydrogen carbonate, \(\mathrm{HCO_{3}}^{-}\).

Their acid strengths are different.

The supplied values show:

\[
\mathrm{p}K_{a1}=6.35
\]

\[
\mathrm{p}K_{a2}=10.33
\]

A lower \(pK_a\) indicates a stronger acid.

Therefore, carbonic acid is the stronger proton donor.

Hydroxide reacts with the available carbonic acid first. Only after the carbonic acid has been consumed can the remaining hydroxide react with hydrogen carbonate.

This is why simply identifying the possible reactions is not enough. You must also consider which reactants are present and their relative acid strengths.

(d) Explain how both behaviours are demonstrated.

Hydrogen carbonate initially acts as a base:

\[
\ce{HCO3^- + H3O+ -> H2CO3 + H2O}
\]

It accepts a proton to form carbonic acid.

Later, after the carbonic acid has been consumed, the remaining hydroxide reacts with hydrogen carbonate:

\[
\ce{HCO3^- + OH^- -> CO3^2- + H2O}
\]

Hydrogen carbonate donates a proton to form carbonate.

Therefore, hydrogen carbonate demonstrates both proton-accepting and proton-donating behaviour.

It is amphiprotic.

Notice that the final solution contains more hydrogen carbonate than carbonate. This does not mean that hydrogen carbonate failed to act as an acid. Some of it did donate protons, even though most remains in its original form.

The final amounts reflect the quantities of acid and base added, the order of the reactions, and the relative strengths of the proton donors.

A useful independent check is conservation of the carbon-containing species:

\[
0.0200+0.00500=0.0250\text{ mol}
\]

The total number of moles of carbon-containing species is conserved, as expected under the stated assumption that no carbon dioxide escapes.

The next useful step is learning to compare the two possible equilibria of an amphiprotic species using \(K_a\), \(K_b\), and pH. This lets you predict which behaviour dominates in water, rather than simply identifying which behaviours are possible.

For a fuller walkthrough of the hydrogen carbonate and dihydrogen phosphate ionic equations, see Amphiprotic Salts and Ionic Equations | HSC Chemistry.