Why Water Can Act as Both an Acid and a Base

Add hydrochloric acid to water and the water molecules accept protons, forming hydronium ions, \(\ce{H3O+}\). Add ammonia to water and the water molecules donate protons, forming hydroxide ions, \(\mathrm{OH}^{-}\).

The same substance is acting as a base in one reaction and an acid in another. How is that possible?

The answer is that water can do two different things with a proton. Its oxygen atom has lone pairs of electrons that can accept a proton, while its O-H bonds allow it to donate a proton. This makes water amphiprotic: it can both donate and accept \(\mathrm{H}^{+}\).

The important detail is that water’s role depends on what it reacts with. It isn’t permanently an acid or permanently a base.

Before looking at the equations, predict what happens when ammonia reacts with water. Which substance loses a proton?

It is water. Ammonia accepts the proton, so water must donate it.

01Two reactions, two different roles for water

The easiest way to understand water’s behaviour is to compare two proton-transfer reactions.

1. Water acts as a base when it accepts a proton

Consider hydrochloric acid reacting with water:

\[
\ce{HCl(aq) + H2O(l) -> H3O+(aq) + Cl^-(aq)}
\]

Follow the proton:

• \(\ce{HCl}\) loses a proton, so it acts as the acid.
• \(\ce{H2O}\) accepts that proton, so it acts as the base.
• \(\ce{H3O+}\) forms when water gains a proton.
• \(\mathrm{Cl}^{-}\) forms when hydrochloric acid loses a proton.

Water’s oxygen uses a lone pair of electrons to bond with the incoming proton.

Notice that water has become positively charged. It has gained one positively charged proton without gaining an electron.

2. Water acts as an acid when it donates a proton

Now consider ammonia reacting with water:

\[
\ce{NH3(aq) + H2O(l) <=> NH4+(aq) + OH^-(aq)}
\]

This time, follow the proton in the opposite direction.

• \(\ce{NH3}\) accepts a proton, so it acts as the base.
• \(\ce{H2O}\) donates a proton, so it acts as the acid.
• \(\ce{NH4+}\) forms when ammonia gains a proton.
• \(\mathrm{OH}^{-}\) forms when water loses a proton.

Water’s O-H bond breaks, with the bonding electrons remaining with oxygen. The water molecule becomes a hydroxide ion.

The reaction uses an equilibrium arrow because ammonia is a weak base. Only a proportion of the ammonia molecules accept a proton at equilibrium.

How do the conjugate acid-base pairs fit in?

Every proton-transfer reaction produces two conjugate acid-base pairs.

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

Look at both reactions together:

Water appears in two different conjugate pairs:

• \(\ce{H3O+/H2O}\): water is the conjugate base of hydronium.
• \(\ce{H2O/OH^-}\): water is the conjugate acid of hydroxide.

This is the key to understanding its dual behaviour.

When water accepts a proton, it becomes \(\ce{H3O+}\), its conjugate acid. When water donates a proton, it becomes \(\mathrm{OH}^{-}\), its conjugate base.

A useful distinction: the two roles happen in separate proton-transfer events. A single water molecule does not need to donate and accept a proton simultaneously.

02Water can react with itself

There is an even more interesting example of water’s amphiprotic behaviour.

Water molecules can transfer protons to other water molecules. This process is called autoionisation or self-ionisation.

The reaction is:

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

Two water molecules participate.

• One water molecule donates a proton and becomes \(\mathrm{OH}^{-}\).
• The other water molecule accepts that proton and becomes \(\ce{H3O+}\).

One molecule acts as an acid, while the other acts as a base.

This is a particularly useful reaction because it shows that water’s ability to act as both an acid and a base isn’t just theoretical. Both behaviours are involved in the same overall equilibrium.

Does all the water ionise?

Not even close.

At 25°C, only a very small proportion of water molecules undergo autoionisation. Most remain as ordinary \(\ce{H2O}\) molecules.

The equilibrium is strongly favoured towards the reactants.

We describe this equilibrium using the ion-product constant of water, \(K_w\):

\[
K_w=[\ce{H3O+}][\mathrm{OH}^{-}]
\]

At 25°C:

\[
K_w=1.0\times10^{-14}
\]

The square brackets represent equilibrium concentrations in \(\mathrm{mol\,L^{-1}}\).

The value of \(K_w\) is small because very little water ionises. It does not mean water is incapable of donating or accepting protons.

In pure water, the concentrations of hydronium and hydroxide ions are equal:

\[
[\ce{H3O+}]=[\mathrm{OH}^{-}]
\]

Why?

Every autoionisation event produces exactly one hydronium ion and one hydroxide ion. If no other substances are present, their concentrations must be equal.

03How experiments revealed that pure water contains ions

There is an interesting experimental problem behind the chemistry.

If water consists almost entirely of neutral molecules, why does even highly purified water conduct electricity at all?

Electrical current in an aqueous solution is carried by mobile ions. If water contained absolutely no ions, it would have essentially no ionic conductivity.

But measuring the conductivity of pure water is surprisingly difficult. Even tiny amounts of dissolved salts, acids, alkalis, or gases can change the result.

The historical investigation: measuring the conductivity of pure water

In the nineteenth century, physicist Friedrich Kohlrausch investigated the electrical conductivity of water.

Earlier measurements varied considerably. The problem was that supposedly pure water could pick up contaminants from the air, glass containers, and laboratory equipment.

Kohlrausch improved the purification process and used platinum equipment to reduce contamination. He then investigated whether dissolved gases were contributing to the conductivity readings.

The revised apparatus worked as follows:

1. Purified water was placed in a glass bulb connected by a tube to a small receiver containing platinum electrodes.
2. A branch tube connected the apparatus to a vacuum pump.
3. The water was evacuated and shaken to remove dissolved gases.
4. The bulb was warmed to approximately 30-40°C, while the receiver was cooled to between 0°C and −8°C.
5. Water vapour distilled from the bulb into the receiver, where the electrical resistance of the water could be measured using the platinum electrodes and a galvanometer.

The use of a cooled receiver helped limit the water’s contact with the glass surfaces during condensation.

What should the revised experiment show?

If dissolved gases were responsible for much of the conductivity, removing them should reduce the measured conductivity.

That is what happened.

A 1894 account in Nature, written by J. W. Rodger, reported the following measurements from Kohlrausch’s work:

These are historical relative readings from the report, not modern conductivity measurements in \(\mathrm{\mu S\,cm^{-1}}\).

The reduction can be calculated:

\[
\begin{aligned}
\text{Percentage decrease}
&=\frac{0.71-0.25}{0.71}\times100\\
&=64.8\%
\end{aligned}
\]

Removing dissolved gases reduced the conductivity reading by approximately 65%.

The experiment demonstrated that dissolved gases contributed significantly to the conductivity of water.

What did the experiment actually prove?

There is an important limitation.

A conductivity measurement detects the movement of ions. It does not directly identify which ions are present or show how they formed.

Even after dissolved gases were removed, the measured conductivity was not zero.

Modern chemistry explains the residual conductivity of sufficiently pure water through its autoionisation:

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

Both products are mobile ions that carry electrical charge.

However, traces of other contaminants can also contribute to a conductivity reading. The historical experiment alone could not prove that every remaining ion came from water autoionisation.

Modern measurements must control temperature and contamination carefully. For example, the theoretical conductivity of ultrapure water at 25°C is approximately \(0.055\ \mathrm{\mu S\,cm^{-1}}\). Even small amounts of dissolved ions can increase this value substantially.

Merck Millipore explains the relationship between ionic contamination, temperature, and water conductivity in its water quality monitoring guide.

The important chemical conclusion is that water undergoes a small amount of self-ionisation. Its ability to donate and accept protons explains the formation of both hydronium and hydroxide ions.

04Amphiprotic does not mean equally strong

Here is a misconception worth addressing before you move on.

If water can act as both an acid and a base, does that mean it is equally strong as an acid and a base?

No.

Being amphiprotic describes what a substance can do. It does not tell you how extensively either reaction occurs.

To understand this properly, you need to separate three different ideas.

These properties answer different questions.

1. Strength

Acid strength describes the tendency of an acid to donate protons.

Base strength describes the tendency of a base to accept protons.

For weak acids and bases, we use equilibrium constants:

• A larger \(K_a\) indicates a stronger acid.
• A larger \(K_b\) indicates a stronger base.

The extent of proton transfer depends on the reacting species and the conditions.

Water can accept a proton from hydrochloric acid, but that does not mean it must accept protons equally readily from every other acid.

Likewise, water can donate a proton to ammonia, but this does not mean every base will accept one to the same extent.

2. Amount

Amount tells you how many moles of a substance are present.

For example, a container might hold:

\[
n(\ce{H2O})=5.00\ \mathrm{mol}
\]

Another container might hold:

\[
n(\ce{H2O})=10.0\ \mathrm{mol}
\]

The second container has twice as many moles of water.

Does that make the water a stronger acid or base?

No. It simply contains more water molecules.

3. Concentration

Concentration tells you how much substance is present per unit volume.

The relationship is:

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

where:

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

Consider two containers of ammonia solution, both with a concentration of \(0.0200\ \mathrm{mol\,L^{-1}}\).

One contains 100 mL, while the other contains 500 mL.

The larger container contains five times as many moles of ammonia, but both solutions have the same concentration.

They also contain the same chemical species, with the same intrinsic base strength at the same temperature.

The distinction is important: a larger amount does not automatically mean a higher concentration, and neither quantity determines whether a substance is a strong or weak acid or base.

A quick method for identifying water’s role

When you encounter a reaction involving water in an HSC question, follow the proton.

1. If water gains a proton, it acts as a base and becomes \(\ce{H3O+}\).
2. If water loses a proton, it acts as an acid and becomes \(\mathrm{OH}^{-}\).
3. Identify the conjugate pairs by finding species that differ by exactly one proton.
4. If asked about strength, examine the relevant equilibrium constant rather than assuming that both reactions occur equally.

Also remember that amphiprotic and amphoteric are related but not interchangeable terms. Amphiprotic specifically describes the ability to donate and accept protons. For the broader distinction, see How to Distinguish Amphiprotic from Amphoteric in HSC Chemistry.

05Questions and solutions

The following questions move from identifying water’s role to analysing experimental evidence, equilibrium, and some less obvious consequences of autoionisation.

Question 1

Hydrogen chloride and ammonia are separately dissolved in water.

(a) Write a balanced Brønsted-Lowry equation for each reaction.

(b) Identify the acid and base in each reaction.

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

(d) Explain why water acts as a base in one reaction and an acid in the other.

Solution 1

Water acts as a base when reacting with hydrogen chloride and as an acid when reacting with ammonia. Its role depends on the direction of proton transfer.

(a) Hydrogen chloride reacting with water:

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

(b) Hydrogen chloride donates a proton, so it is the acid. Water accepts the proton, so it is the base.

(c) The conjugate pairs are:

• \(\ce{HCl/Cl^-}\)
• \(\ce{H3O+/H2O}\)

(a) Ammonia reacting with water:

\[
\ce{NH3 + H2O <=> NH4+ + OH^-}
\]

(b) Ammonia accepts a proton, so it is the base. Water donates the proton, so it is the acid.

(c) The conjugate pairs are:

• \(\ce{NH4+/NH3}\)
• \(\ce{H2O/OH^-}\)

(d) Water has both the ability to accept a proton through an oxygen lone pair and the ability to donate a proton from an O-H bond.

In the first reaction, hydrochloric acid donates a proton to water. In the second, ammonia accepts a proton from water.

The two reactions demonstrate why water is classified as amphiprotic.

Question 2

The hydrogen carbonate ion, \(\mathrm{HCO_{3}}^{-}\), is another amphiprotic species.

It can react with water in two different ways:

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

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

At 25°C, the following equilibrium constants are provided:

\[
K_w=1.0\times10^{-14}
\]

\[
K_{a1}(\ce{H2CO3})=4.3\times10^{-7}
\]

\[
K_{a2}(\mathrm{HCO_{3}}^{-})=4.7\times10^{-11}
\]

(a) Identify the role of water in each reaction.

(b) Identify the conjugate acid-base pairs in both reactions.

(c) Calculate the base dissociation constant, \(K_b\), for hydrogen carbonate acting as a base.

(d) Determine which of the two proton-transfer reactions is more favoured and explain what this reveals about water’s role.

Solution 2

The reaction in which hydrogen carbonate accepts a proton is more favoured. In this reaction, water acts as an acid and donates a proton.

(a) Consider each equation separately.

In the first reaction:

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

Hydrogen carbonate accepts a proton, so it acts as a base. Water donates a proton, so it acts as an acid.

In the second reaction:

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

Hydrogen carbonate donates a proton, so it acts as an acid. Water accepts a proton, so it acts as a base.

(b) The conjugate pairs are:

First reaction:

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

Second reaction:

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

Each pair differs by one proton.

(c) The base dissociation constant of hydrogen carbonate is related to the acid dissociation constant of its conjugate acid, carbonic acid.

Use:

\[
K_b=\frac{K_w}{K_{a1}}
\]

Substitute:

\[
K_b=\frac{1.0\times10^{-14}}{4.3\times10^{-7}}
\]

\[
\boxed{K_b=2.3\times10^{-8}}
\]

(d) Compare the two equilibrium constants:

\[
K_b=2.3\times10^{-8}
\]

\[
K_{a2}=4.7\times10^{-11}
\]

The base dissociation constant is approximately 490 times larger than the acid dissociation constant.

Therefore, hydrogen carbonate has a greater tendency to accept a proton than to donate one under these conditions.

The first reaction is more favoured, meaning water predominantly acts as an acid in this particular comparison.

The second reaction is still possible. Its smaller equilibrium constant simply indicates that it is less favoured.

This is an important distinction: a species can have two possible proton-transfer behaviours without those behaviours being equally strong.

Question 3

A historical investigation into the conductivity of purified water produced the following measurements at 18°C:

(a) Calculate the percentage decrease in conductivity after dissolved gases were removed.

(b) Explain why removing dissolved gases was an important improvement to the experimental design.

(c) The second sample still had a non-zero conductivity. Explain how water’s autoionisation accounts for this observation.

(d) Explain why the conductivity result alone cannot prove that the only ions present are hydronium and hydroxide.

Solution 3

The conductivity decreased by approximately 65%, showing that dissolved gases contributed significantly to the original reading. However, the remaining conductivity cannot be attributed exclusively to water autoionisation without further evidence.

(a) Calculate the percentage decrease:

\[
\text{Percentage decrease}
=\frac{\text{Initial}-\text{Final}}{\text{Initial}}
\times100
\]

Substitute:

\[
=\frac{0.71-0.25}{0.71}\times100
\]

\[
\boxed{64.8\%}
\]

The conductivity decreased by approximately 65%.

(b) Water distilled in contact with air can absorb dissolved gases, particularly carbon dioxide.

Carbon dioxide reacts with water to form carbonic acid, which can produce additional ions in solution.

These ions increase electrical conductivity, potentially making the water appear to contain more ions from its own autoionisation than it actually does.

Removing dissolved gases therefore helps isolate the contribution from the water itself.

(c) Water undergoes autoionisation:

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

One water molecule donates a proton and becomes hydroxide. Another accepts the proton and becomes hydronium.

Both products are charged particles and can carry electrical current.

Therefore, even highly purified water can have a small, non-zero conductivity.

(d) Conductivity measures the overall ability of mobile ions to carry electrical charge.

It does not identify individual ions.

Trace contamination from salts, acids, alkalis, or the apparatus itself could also contribute to the reading.

The experiment provides evidence that dissolved gases affect conductivity and that highly purified water still has a small conductivity. The chemical model of autoionisation explains why water itself can produce mobile ions.

It is the combination of experimental evidence and chemical reasoning that supports the explanation.

Question 4

A student prepares two separate hydrochloric acid solutions.

Both solutions receive exactly \(1.00\times10^{-5}\ \mathrm{mol}\) of HCl.

• Solution A contains \(100.0\ \mathrm{mL}\) of water.
• Solution B contains \(1.00\ \mathrm{L}\) of water.

Assume that HCl ionises completely, the final solution volumes are approximately equal to the initial water volumes, and the contribution from water’s autoionisation is negligible.

(a) Calculate the hydronium ion concentration in each solution.

(b) Calculate the pH of each solution.

(c) Explain why the solutions have different pH values even though they contain the same amount of hydrochloric acid.

(d) Identify the role of water in both reactions and explain whether the strength of hydrochloric acid has changed.

Solution 4

Solution A has a pH of 4.00, while Solution B has a pH of 5.00. The solutions contain the same amount of hydrochloric acid, but their different volumes produce different concentrations.

(a) Start with the proton-transfer reaction:

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

HCl is a strong acid and ionises essentially completely under these conditions.

Therefore, the amount of hydronium produced is approximately equal to the amount of HCl added.

For Solution A:

\[
[\ce{H3O+}]=\frac{n}{V}
\]

\[
=\frac{1.00\times10^{-5}}{0.1000}
\]

\[
\boxed{[\ce{H3O+}]=1.00\times10^{-4}\ \mathrm{mol\,L^{-1}}}
\]

For Solution B:

\[
[\ce{H3O+}]=\frac{1.00\times10^{-5}}{1.00}
\]

\[
\boxed{[\ce{H3O+}]=1.00\times10^{-5}\ \mathrm{mol\,L^{-1}}}
\]

(b) Use:

\[
\mathrm{pH}=-\log_{10}[\ce{H3O+}]
\]

Solution A:

\[
\mathrm{pH}=-\log_{10}(1.00\times10^{-4})
\]

\[
\boxed{\mathrm{pH}=4.00}
\]

Solution B:

\[
\mathrm{pH}=-\log_{10}(1.00\times10^{-5})
\]

\[
\boxed{\mathrm{pH}=5.00}
\]

(c) Both solutions contain the same amount of hydrochloric acid:

\[
n=1.00\times10^{-5}\ \mathrm{mol}
\]

However, Solution A has one-tenth the volume of Solution B.

Consequently, its hydronium concentration is ten times greater.

This produces a pH difference of one unit.

The important point is that the amount of acid is identical, but the concentration is different.

(d) Water acts as a base in both reactions because it accepts a proton from hydrochloric acid.

The conjugate acid-base pairs are:

• \(\ce{HCl/Cl^-}\)
• \(\ce{H3O+/H2O}\)

The strength of hydrochloric acid has not changed. It remains a strong acid in both solutions.

Only the concentration has changed.

A common mistake is to interpret the lower pH as evidence that the acid itself has become stronger. The lower pH is caused by the greater hydronium concentration, not a change in acid strength.

Question 5 – Extension challenge

A chemist prepares three samples of high-purity water.

The samples are sealed to prevent contamination and allowed to reach equilibrium.

The following values are provided:

\[
K_w(25^\circ\mathrm{C})=1.0\times10^{-14}
\]

\[
K_w(60^\circ\mathrm{C})=9.6\times10^{-14}
\]

Assume that the concentration approximation for \(K_w\) is valid and that no other substances contribute ions.

A student makes the following claims:

• Sample B is the most acidic because it contains the greatest total amount of hydronium ions.
• Samples A and B are acidic because their pH values are below 7.
• Sample C is neutral because its pH is exactly 7.
• Water becomes a stronger acid at higher temperatures because more hydronium ions are produced.

(a) Calculate the hydronium and hydroxide concentrations in each sample.

(b) Calculate the pH of each sample.

(c) Calculate the amount, in moles, of hydronium ions in each sample.

(d) Evaluate all four student claims. Explain the difference between amount, concentration, neutrality, and acid strength in this situation.

Solution 5

All three samples are neutral. Sample B contains ten times as many moles of hydronium ions as Sample A, but their hydronium concentrations and pH values are identical. Samples A and B also have a lower pH than Sample C without being acidic.

This is a useful example of why pH, total amount of ions, and acid-base behaviour must be considered separately.

(a) Start with the ion-product expression:

\[
K_w=[\ce{H3O+}][\mathrm{OH}^{-}]
\]

Because the samples contain only pure water, their hydronium and hydroxide concentrations must be equal.

Therefore:

\[
[\ce{H3O+}]=[\mathrm{OH}^{-}]=x
\]

Rearrange:

\[
x=\sqrt{K_w}
\]

For Samples A and B at 60°C:

\[
x=\sqrt{9.6\times10^{-14}}
\]

\[
x=3.10\times10^{-7}\ \mathrm{mol\,L^{-1}}
\]

Therefore:

\[
\boxed{
[\ce{H3O+}]=[\mathrm{OH}^{-}]
=3.10\times10^{-7}\ \mathrm{mol\,L^{-1}}
}
\]

For Sample C at 25°C:

\[
x=\sqrt{1.0\times10^{-14}}
\]

\[
x=1.00\times10^{-7}\ \mathrm{mol\,L^{-1}}
\]

Therefore:

\[
\boxed{
[\ce{H3O+}]=[\mathrm{OH}^{-}]
=1.00\times10^{-7}\ \mathrm{mol\,L^{-1}}
}
\]

(b) Calculate the pH of each sample.

For Samples A and B:

\[
\mathrm{pH}=-\log_{10}(3.10\times10^{-7})
\]

\[
\boxed{\mathrm{pH}=6.51}
\]

For Sample C:

\[
\mathrm{pH}=-\log_{10}(1.00\times10^{-7})
\]

\[
\boxed{\mathrm{pH}=7.00}
\]

(c) Calculate the amount of hydronium ions using:

\[
n=cV
\]

For Sample A:

\[
n=3.10\times10^{-7}\times0.500
\]

\[
\boxed{n_A=1.55\times10^{-7}\ \mathrm{mol}}
\]

For Sample B:

\[
n=3.10\times10^{-7}\times5.00
\]

\[
\boxed{n_B=1.55\times10^{-6}\ \mathrm{mol}}
\]

For Sample C:

\[
n=1.00\times10^{-7}\times0.500
\]

\[
\boxed{n_C=5.00\times10^{-8}\ \mathrm{mol}}
\]

The comparison is:

The hydroxide concentrations and amounts are identical to the corresponding hydronium values in every sample.

(d) Now consider each claim.

Claim 1: Sample B is the most acidic because it contains the greatest total amount of hydronium ions.

Incorrect.

Sample B contains ten times as many moles of hydronium as Sample A, but it also has ten times the volume.

Their concentrations are identical:

\[
[\ce{H3O+}]_A=[\ce{H3O+}]_B
\]

They also have identical pH values.

Having more total hydronium ions does not automatically make a solution more acidic. The concentration and the corresponding hydroxide concentration must also be considered.

Claim 2: Samples A and B are acidic because their pH values are below 7.

Incorrect.

A solution is neutral when:

\[
[\ce{H3O+}]=[\mathrm{OH}^{-}]
\]

It is acidic when:

\[
[\ce{H3O+}]>[\mathrm{OH}^{-}]
\]

It is basic when:

\[
[\ce{H3O+}]<[\mathrm{OH}^{-}]
\]

Both Samples A and B have equal hydronium and hydroxide concentrations.

They are therefore neutral, despite having a pH of 6.51.

The assumption that neutral water must always have a pH of 7 is only correct at 25°C.

Claim 3: Sample C is neutral because its pH is exactly 7.

Correct, but the reasoning needs care.

Sample C is neutral because its hydronium and hydroxide concentrations are equal.

Its pH is 7.00 because the temperature is 25°C and the ion-product constant has the corresponding value of \(1.0\times10^{-14}\).

The pH value alone is not the fundamental test of neutrality at every temperature.

Claim 4: Water becomes a stronger acid at higher temperatures because more hydronium ions are produced.

This conclusion is not justified.

Increasing the temperature increases \(K_w\), meaning the autoionisation equilibrium becomes more product-favoured.

More hydronium ions are produced, but an equal amount of hydroxide ions is also produced.

Remember the reaction:

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

One water molecule donates a proton, while another accepts it.

The increased ion concentrations reflect greater autoionisation, not an excess of acidic behaviour over basic behaviour.

The higher-temperature samples remain neutral.

The deeper lesson is that three different measurements describe three different things:

• Concentration tells you the amount of ions per litre.
• Amount tells you the total number of moles of ions in the sample.
• Acid-base strength describes the tendency of a species to donate or accept protons at equilibrium.

You cannot use one of these properties as a substitute for another.

The next useful step is to apply the same reasoning to other amphiprotic species, particularly hydrogen carbonate and dihydrogen phosphate. By comparing their \(K_a\) and \(K_b\) values, you can predict whether they behave predominantly as acids or bases in water, rather than assuming that both possible reactions occur equally.