Ionic Product of Water, Kw | A-Level Chemistry

Last updated: 18 August 2026

Diagnose → Learn → Master: Every study-note section includes 15 free MCQs — start with 5 Diagnostic MCQs to identify gaps in your understanding, then complete 10 Mastery MCQs near the end to check what you have learned.

Water is usually introduced as the solvent in acid–base chemistry. However, water is not chemically passive. Even a sample of pure water contains a very small concentration of hydrogen ions and hydroxide ions.

These ions are produced because water molecules can transfer protons between themselves. The resulting equilibrium gives us the ionic product of water, K_w, one of the most useful relationships in aqueous acid–base chemistry.

Once K_w is understood, several ideas that may initially seem separate become connected. We can calculate the pH of strong bases, relate pH to pOH, understand why neutral pH changes with temperature, and convert between the K_a of an acid and the K_b of its conjugate base.

Diagnostic Quiz

What Do You Already Know About K_w?

Try these five questions before reading the chapter. They check the acid–base ideas that make K_w much easier to understand.

5 quick questions
Focus on key acid–base foundations
Every answer option is explained

Which statement best describes a Brønsted–Lowry acid?

A species that accepts an electron pair

Accepting an electron pair describes a Lewis acid. The Brønsted–Lowry definition is based specifically on proton transfer.

A species that donates a proton

Correct. A Brønsted–Lowry acid donates H⁺ to another species.

A species that produces OH⁻ in every reaction

Producing OH⁻ is not the Brønsted–Lowry definition of an acid. Acids donate protons.

A species that must carry a positive charge

Acids do not have to be positively charged. Neutral species such as HCl and CH₃COOH can donate protons.

Which pair of ions is present in pure water because water undergoes auto-ionisation?

Na⁺ and OH⁻

Pure water contains no sodium source, so Na⁺ is not produced by auto-ionisation.

H⁺ and Cl⁻

Cl⁻ would require a chlorine-containing substance such as hydrochloric acid.

H⁺ and O²⁻

Oxide ions, O²⁻, are not the corresponding basic species present in ordinary aqueous water.

H⁺ and OH⁻

Correct. In the usual A-Level shorthand, auto-ionisation produces equal amounts of H⁺ and OH⁻.

Which relationship defines a neutral aqueous solution?

[H⁺] = [OH⁻]

Correct. Neutrality is defined by equal hydrogen-ion and hydroxide-ion concentrations.

[H⁺] = 0

Even pure water contains a small concentration of H⁺ because water auto-ionises.

[OH⁻] = 0

Neutral water contains OH⁻ as well as H⁺. Neither concentration is zero.

pH must equal 7 at every temperature

pH 7 corresponds to neutrality at 298 K, but the neutral pH changes when temperature changes.

Sodium hydroxide is a strong base. What happens to it when dissolved in water at ordinary A-Level concentrations?

It remains mainly as undissociated NaOH molecules

NaOH is treated as a strong base and strong electrolyte, so it dissociates essentially completely in aqueous solution.

It produces H⁺ as its main acid–base ion

NaOH is a base. Its dissociation supplies OH⁻ rather than H⁺.

It dissociates essentially completely to form Na⁺ and OH⁻

Correct. Therefore, for NaOH, its analytical concentration can normally be used directly to determine [OH⁻].

It reacts with water to form Na²⁺

Sodium forms Na⁺, not Na²⁺, in sodium hydroxide.

A sample of pure water has pH 7.00 at 298 K. Why is it neutral?

It contains no ions

Pure water does contain ions because a small amount auto-ionises.

Its H⁺ and OH⁻ concentrations are equal

Correct. Neutrality is determined by [H⁺] = [OH⁻]. At 298 K, each concentration is 1.00 × 10⁻⁷ mol dm⁻³.

Its H⁺ concentration is greater than its OH⁻ concentration

That would describe an acidic solution, not a neutral one.

Its OH⁻ concentration is greater than its H⁺ concentration

That would describe an alkaline solution.

The auto-ionisation of water

Water molecules can behave as both proton donors and proton acceptors. This becomes especially clear when two water molecules interact.

One water molecule can donate a proton to another water molecule:

2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)

One H₂O molecule has donated H⁺ and therefore acted as a Brønsted–Lowry acid. The second H₂O molecule has accepted H⁺ and therefore acted as a Brønsted–Lowry base.

This reversible process is called the auto-ionisation of water. Only a very small proportion of the water molecules are ionised at any instant, but the ions are always present.

Pure water does not mean ion-free water.

Pure water contains mostly H₂O molecules together with a very small equilibrium concentration of H₃O⁺ and OH⁻ ions.

Why do A-Level equations often use H⁺ instead of H₃O⁺?

A proton is not present as an isolated bare H⁺ ion in liquid water. It is strongly associated with surrounding water molecules, and H₃O⁺ is a more explicit representation of protonated water.

In acid–base calculations, however, H⁺(aq) is conventionally used as shorthand. We can therefore represent the equilibrium more simply as:

H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)

When we write [H⁺] in the calculations that follow, we are using this conventional aqueous shorthand.

Defining the ionic product of water, K_w

The concentrations of H⁺ and OH⁻ in water are not independent. Because they arise from an equilibrium, their equilibrium concentrations are linked.

The activity of liquid water is effectively constant and is incorporated into the equilibrium constant. The resulting constant is called the ionic product of water.

K_w = [H⁺][OH⁻]

At 298 K:

K_w = 1.00 × 10⁻¹⁴ mol² dm⁻⁶

This is not merely a statement about pure water. At a given temperature, the relationship applies to aqueous solutions generally.

If [H⁺] increases, [OH⁻] must decrease so that their product remains equal to K_w. Similarly, increasing [OH⁻] causes [H⁺] to decrease.

K_w is constant when temperature is constant.

Adding acid or alkali changes [H⁺] and [OH⁻], but it does not change K_w if the temperature remains unchanged.

Rearranging the K_w expression

If [OH⁻] is known, we can calculate [H⁺]:

[H⁺] = K_w ÷ [OH⁻]

If [H⁺] is known, we can calculate [OH⁻]:

[OH⁻] = K_w ÷ [H⁺]

This is why K_w is so useful in strong-base calculations. A strong base gives us [OH⁻] directly from its dissociation stoichiometry, while K_w allows us to convert that concentration into [H⁺] and hence pH.

Neutral, acidic and alkaline solutions

A solution is classified as acidic, neutral or alkaline by comparing its equilibrium concentrations of H⁺ and OH⁻.

Solution	Relationship	Meaning
Acidic	[H⁺] > [OH⁻]	Hydrogen-ion concentration is greater.
Neutral	[H⁺] = [OH⁻]	The two ion concentrations are equal.
Alkaline	[H⁺] < [OH⁻]	Hydroxide-ion concentration is greater.

Notice that the fundamental definition of neutral is [H⁺] = [OH⁻], not simply “pH = 7”.

Finding [H⁺] and [OH⁻] in pure water at 298 K

In pure water, every auto-ionisation event produces H⁺ and OH⁻ in equal amounts. Therefore:

[H⁺] = [OH⁻]

At 298 K:

K_w = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴

Since the two concentrations are equal:

K_w = [H⁺]²

Hence:

[H⁺] = √K_w

[H⁺] = √(1.00 × 10⁻¹⁴) = 1.00 × 10⁻⁷ mol dm⁻³

Since the solution is neutral:

[OH⁻] = 1.00 × 10⁻⁷ mol dm⁻³

The pH is therefore:

pH = −log₁₀(1.00 × 10⁻⁷) = 7.00

At 298 K: neutral water has pH 7.00 because [H⁺] = [OH⁻] = 1.00 × 10⁻⁷ mol dm⁻³.

At another temperature: neutral water still has [H⁺] = [OH⁻], but those concentrations may no longer be 1.00 × 10⁻⁷ mol dm⁻³.

The relationship between pH, pOH and pK_w

pH expresses hydrogen-ion concentration on a logarithmic scale:

pH = −log₁₀[H⁺]

In exactly the same way, pOH is defined using hydroxide-ion concentration:

pOH = −log₁₀[OH⁻]

We can also define pK_w:

pK_w = −log₁₀K_w

Starting from:

K_w = [H⁺][OH⁻]

taking −log₁₀ of both sides gives:

pK_w = pH + pOH

This is the general relationship.

At 298 K:

pK_w = −log₁₀(1.00 × 10⁻¹⁴) = 14.00

Therefore:

pH + pOH = 14.00 at 298 K

Do not memorise pH + pOH = 14 as a universal law.

The general equation is pH + pOH = pK_w. The value 14.00 applies at 298 K because pK_w is 14.00 at that temperature.

Finding the neutral pH from pK_w

In a neutral solution:

[H⁺] = [OH⁻]

Therefore:

pH = pOH

Since:

pH + pOH = pK_w

the neutral pH must be:

Neutral pH = ½pK_w

At 298 K, pK_w = 14.00, so:

Neutral pH = ½(14.00) = 7.00

How temperature affects K_w

Like other equilibrium constants, K_w depends on temperature.

The forward auto-ionisation of water is endothermic:

2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq) ΔH > 0

If temperature increases, the equilibrium position shifts towards the ions. Consequently, pure water contains greater concentrations of both H⁺ and OH⁻.

Since:

K_w = [H⁺][OH⁻]

a simultaneous increase in both concentrations means that K_w increases as temperature increases.

Change	K_w	pK_w	Neutral pH
Temperature increases	Increases	Decreases	Decreases
Temperature decreases	Decreases	Increases	Increases

Why can neutral water have a pH below 7?

Suppose the temperature is increased. More water molecules ionise, so [H⁺] increases. Since pH depends on [H⁺], the pH decreases.

However, [OH⁻] increases by exactly the same amount because auto-ionisation produces the two ions together.

[H⁺] = [OH⁻]

The water therefore remains neutral, even though its pH may now be below 7.

Incorrect reasoning: “The pH is below 7, so the water must be acidic.”

Correct reasoning: Compare [H⁺] with [OH⁻]. If they are equal, the solution is neutral. The neutral pH itself changes with temperature.

Does adding acid or alkali change K_w?

Not if the temperature remains unchanged.

Suppose acid is added. The acid increases [H⁺]. The auto-ionisation equilibrium shifts to the left, reducing [OH⁻].

Higher [H⁺] × lower [OH⁻] = same K_w

If a base is added, [OH⁻] increases and [H⁺] decreases instead.

Concentration changes and temperature changes do different things.

Adding acid or base changes [H⁺] and [OH⁻] while K_w remains constant at fixed temperature. Changing the temperature changes K_w itself.

The relationship between K_a, K_b and K_w

K_w provides an important connection between an acid and its conjugate base.

Consider the conjugate pair HA and A⁻.

HA acts as an acid:

HA(aq) ⇌ H⁺(aq) + A⁻(aq)

K_a = [H⁺][A⁻] ÷ [HA]

A⁻ can act as a base by accepting a proton from water:

A⁻(aq) + H₂O(l) ⇌ HA(aq) + OH⁻(aq)

K_b = [HA][OH⁻] ÷ [A⁻]

Multiply the two equilibrium expressions:

K_aK_b = ([H⁺][A⁻] ÷ [HA])([HA][OH⁻] ÷ [A⁻])

[HA] and [A⁻] cancel:

K_aK_b = [H⁺][OH⁻]

Since [H⁺][OH⁻] = K_w:

K_aK_b = K_w

Therefore:

K_a = K_w ÷ K_b

K_b = K_w ÷ K_a

The corresponding pK relationship

Taking −log₁₀ of:

K_aK_b = K_w

gives:

pK_a + pK_b = pK_w

At 298 K:

pK_a + pK_b = 14.00

A strong acid has a weak conjugate base.

At a fixed temperature, K_aK_b must remain equal to K_w. A large K_a must therefore correspond to a small K_b, and vice versa.

Worked example: finding K_a from K_b

At 298 K, a weak base has K_b = 2.00 × 10⁻⁵. Calculate K_a for its conjugate acid.

K_aK_b = K_w

K_a = K_w ÷ K_b

K_a = (1.00 × 10⁻¹⁴) ÷ (2.00 × 10⁻⁵)

K_a = 5.00 × 10⁻¹⁰

Answer: K_a = 5.00 × 10⁻¹⁰.

Using K_w in calculations

Most K_w calculations become straightforward if you separate them into two stages.

Work out either [H⁺] or [OH⁻] from the chemical information.
Use K_w to connect that concentration to the other ion.

Only then convert the relevant ion concentration into pH or pOH if required.

Example 1: pH of a strong base

Calculate the pH of 0.0200 mol dm⁻³ NaOH at 298 K.

Sodium hydroxide dissociates essentially completely:

NaOH(aq) → Na⁺(aq) + OH⁻(aq)

The mole ratio between NaOH and OH⁻ is 1 : 1, so:

[OH⁻] = 0.0200 mol dm⁻³

Use K_w:

[H⁺] = K_w ÷ [OH⁻]

[H⁺] = (1.00 × 10⁻¹⁴) ÷ 0.0200 = 5.00 × 10⁻¹³ mol dm⁻³

Therefore:

pH = −log₁₀(5.00 × 10⁻¹³) = 12.30

Answer: pH = 12.30.

Example 2: do not forget the OH⁻ stoichiometry

Calculate the pH of 0.0250 mol dm⁻³ Ba(OH)₂ at 298 K.

Barium hydroxide dissociates according to:

Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻(aq)

One mole of Ba(OH)₂ therefore supplies two moles of OH⁻.

[OH⁻] = 2(0.0250) = 0.0500 mol dm⁻³

We can find pOH:

pOH = −log₁₀(0.0500) = 1.301

At 298 K:

pH = 14.00 − 1.301 = 12.699

Answer: pH = 12.70.

Incorrect: [OH⁻] = 0.0250 mol dm⁻³ simply because the Ba(OH)₂ concentration is 0.0250 mol dm⁻³.

Correct: Use the dissociation stoichiometry first. Each Ba(OH)₂ produces two OH⁻, so [OH⁻] = 0.0500 mol dm⁻³.

Example 3: finding [OH⁻] from pH

A solution has pH 4.50 at 298 K. Calculate [OH⁻].

First convert pH into [H⁺]:

[H⁺] = 10⁻⁴·⁵⁰ = 3.16 × 10⁻⁵ mol dm⁻³

Now use K_w:

[OH⁻] = K_w ÷ [H⁺]

[OH⁻] = (1.00 × 10⁻¹⁴) ÷ (3.16 × 10⁻⁵) = 3.16 × 10⁻¹⁰ mol dm⁻³

Answer: [OH⁻] = 3.16 × 10⁻¹⁰ mol dm⁻³.

Example 4: neutral water at a different temperature

At a certain temperature, K_w = 4.00 × 10⁻¹⁴ mol² dm⁻⁶. Calculate the pH of pure water.

Pure water is neutral, so:

[H⁺] = [OH⁻]

Therefore:

[H⁺] = √K_w

[H⁺] = √(4.00 × 10⁻¹⁴) = 2.00 × 10⁻⁷ mol dm⁻³

Hence:

pH = −log₁₀(2.00 × 10⁻⁷) = 6.699

Answer: pH = 6.70. The water is still neutral because [H⁺] = [OH⁻].

Use the K_w appropriate to the stated temperature.

Do not automatically substitute 1.00 × 10⁻¹⁴ or assume that neutral pH is 7.00 if the question gives K_w at another temperature.

Common mistakes with the ionic product of water

Common mistake	Why it is wrong	Correct approach
Thinking pure water contains no ions	Water undergoes a small amount of auto-ionisation	Remember that pure water contains equal concentrations of H⁺ and OH⁻
Writing K_w = [H⁺] + [OH⁻]	K_w is defined using the product of the two concentrations	Use K_w = [H⁺][OH⁻]
Assuming neutral always means pH 7	The value of K_w, and therefore neutral pH, changes with temperature	Define neutrality using [H⁺] = [OH⁻]
Assuming pH + pOH is always 14	The general relationship is pH + pOH = pK_w	Use 14.00 only when pK_w = 14.00
Thinking that adding acid changes K_w	At fixed temperature, K_w remains constant	Increasing [H⁺] causes [OH⁻] to decrease
Using the concentration of Ba(OH)₂ directly as [OH⁻]	Each Ba(OH)₂ produces two OH⁻ ions	Use the dissociation equation to determine the correct mole ratio first
Writing K_a + K_b = K_w	The relationship involves multiplication	Use K_aK_b = K_w
Calling a solution acidic solely because its pH is below 7	The neutral pH may itself be below 7 at temperatures above 298 K	Compare [H⁺] with [OH⁻], or compare pH with ½pK_w

K_w checklist

Know the equilibrium

Water undergoes auto-ionisation.
Water can donate or accept a proton.
K_w = [H⁺][OH⁻].
At 298 K, K_w = 1.00 × 10⁻¹⁴ mol² dm⁻⁶.

Judge acidity correctly

Acidic means [H⁺] > [OH⁻].
Neutral means [H⁺] = [OH⁻].
Alkaline means [H⁺] < [OH⁻].
Do not define neutral simply as pH 7.

Connect the p quantities

pH = −log₁₀[H⁺].
pOH = −log₁₀[OH⁻].
pH + pOH = pK_w.
At 298 K, pH + pOH = 14.00.

Solve calculations carefully

Use stoichiometry to obtain [H⁺] or [OH⁻] first.
For neutral water, [H⁺] = √K_w.
Use the K_w given for the stated temperature.
For conjugate pairs, K_aK_b = K_w.

Mastery Quiz

The Ionic Product of Water Mastery Quiz

Now apply K_w to calculations and unfamiliar situations. These questions test whether you can connect equilibrium, pH, temperature and conjugate acid–base relationships rather than simply recall definitions.

10 application-focused questions
Includes calculations and misconception testing
Every answer option is explained

At 298 K, an aqueous solution has [H⁺] = 2.00 × 10⁻⁵ mol dm⁻³. What is [OH⁻]?

K_w = 1.00 × 10⁻¹⁴ mol² dm⁻⁶

2.00 × 10⁻⁵ mol dm⁻³

This would make [H⁺] and [OH⁻] equal, but the solution is not stated to be neutral. Use K_w = [H⁺][OH⁻].

2.00 × 10⁻⁹ mol dm⁻³

Multiplying 2.00 × 10⁻⁵ by this value would not give 1.00 × 10⁻¹⁴.

5.00 × 10⁻¹⁰ mol dm⁻³

Correct. [OH⁻] = (1.00 × 10⁻¹⁴) ÷ (2.00 × 10⁻⁵) = 5.00 × 10⁻¹⁰ mol dm⁻³.

5.00 × 10⁻²⁰ mol dm⁻³

This results from handling the powers of ten incorrectly. The exponents are divided as 10⁻¹⁴ ÷ 10⁻⁵ = 10⁻⁹ before the numerical factors are considered.

What is the pH of 0.0100 mol dm⁻³ Ca(OH)₂ at 298 K, assuming complete dissociation?

Ca(OH)₂(aq) → Ca²⁺(aq) + 2OH⁻(aq)

2.00

This treats the hydroxide concentration as though it were a hydrogen-ion concentration and does not account for the basic nature of the solution.

12.30

Correct. [OH⁻] = 2(0.0100) = 0.0200 mol dm⁻³. pOH = 1.699, so pH = 14.00 − 1.699 = 12.30.

12.00

This would correspond approximately to [OH⁻] = 0.0100 mol dm⁻³. Each Ca(OH)₂ supplies two OH⁻ ions, so [OH⁻] is twice this value.

7.00

Calcium hydroxide supplies excess OH⁻, so the solution is alkaline rather than neutral.

At a certain temperature, K_w = 4.00 × 10⁻¹⁴ mol² dm⁻⁶. What is the pH of pure water at this temperature?

6.70

Correct. Pure water has [H⁺] = [OH⁻], so [H⁺] = √(4.00 × 10⁻¹⁴) = 2.00 × 10⁻⁷ mol dm⁻³. Therefore pH = 6.70.

7.00

Neutral pH is 7.00 only when K_w = 1.00 × 10⁻¹⁴. Here K_w is larger.

13.40

This effectively uses pK_w itself rather than half of pK_w for a neutral solution.

3.35

The concentration of H⁺ is √K_w. The logarithm should then be taken once; dividing the final pH by two again is incorrect.

Hydrochloric acid is added to water while the temperature is kept constant. Which statement is correct after equilibrium is re-established?

[H⁺] increases and K_w increases

[H⁺] does increase, but K_w does not change because the temperature has not changed.

[H⁺] and [OH⁻] both increase by the same factor

If both increased, their product would increase. At fixed temperature, K_w must remain constant.

[H⁺] remains unchanged but [OH⁻] decreases

Adding a strong acid directly increases [H⁺], so [H⁺] cannot remain unchanged.

[H⁺] increases, [OH⁻] decreases and K_w remains constant

Correct. At fixed temperature, [H⁺][OH⁻] must continue to equal the same K_w.

At 298 K, a weak base B has K_b = 1.80 × 10⁻⁵. What is K_a for BH⁺?

K_w = 1.00 × 10⁻¹⁴

1.80 × 10⁻¹⁹

This comes from multiplying K_b by K_w. The relationship requires division to obtain K_a.

1.80 × 10⁹

This is far too large and results from reversing the ratio incorrectly. The conjugate acid of a weak base such as this is itself weak.

5.56 × 10⁻¹⁰

Correct. K_a = K_w ÷ K_b = (1.00 × 10⁻¹⁴) ÷ (1.80 × 10⁻⁵) = 5.56 × 10⁻¹⁰.

5.56 × 10¹⁰

The numerical coefficient is similar to the correct calculation, but the power of ten has been handled incorrectly.

A solution has pH 9.25 at 298 K. What is [OH⁻]?

5.62 × 10⁻¹⁰ mol dm⁻³

This is [H⁺], obtained from 10⁻⁹·²⁵. The question asks for [OH⁻].

9.25 mol dm⁻³

pH is a logarithmic quantity and is not itself an ion concentration.

1.78 × 10⁻⁵ mol dm⁻³

Correct. pOH = 14.00 − 9.25 = 4.75, so [OH⁻] = 10⁻⁴·⁷⁵ = 1.78 × 10⁻⁵ mol dm⁻³.

4.75 mol dm⁻³

4.75 is the pOH, not the hydroxide-ion concentration. The inverse logarithm must still be taken.

The temperature of pure water is increased. Which combination of changes is correct?

K_w decreases, pK_w increases and neutral pH increases

This is the opposite of what occurs when the endothermic auto-ionisation is favoured by increasing temperature.

K_w increases, pK_w decreases and neutral pH decreases

Correct. Greater ionisation increases K_w. Since pK_w = −log K_w, pK_w falls, and neutral pH = ½pK_w also falls.

K_w increases, pK_w increases and neutral pH remains 7

Increasing K_w means pK_w decreases, not increases. Neutral pH is also temperature-dependent.

K_w remains unchanged because equilibrium constants never depend on temperature

Equilibrium constants do depend on temperature. K_w is no exception.

At a particular temperature, pK_w = 13.20. An aqueous solution has pH 6.60. What can be concluded?

The solution is acidic because its pH is below 7

The neutral pH at this temperature is not 7. It is ½pK_w = 6.60.

The solution is neutral

Correct. Neutral pH = ½(13.20) = 6.60, so [H⁺] = [OH⁻].

The solution is alkaline because pK_w is below 14

pK_w determines the neutral reference point. Its value alone does not make a particular solution alkaline.

There is insufficient information because neutral pH is always 7

Neutral pH is not always 7. With pK_w known, the neutral pH can be calculated directly.

At 298 K, an acid HA has pK_a = 4.76. What is pK_b for its conjugate base A⁻?

4.76

Conjugate acids and bases do not generally have equal pK values.

18.76

This results from adding 14.00 rather than using pK_a + pK_b = 14.00.

2.94

This does not satisfy the required sum of 14.00 at 298 K.

9.24

Correct. pK_b = 14.00 − 4.76 = 9.24.

Acid HX has a larger K_a than acid HY at the same temperature. Which statement about their conjugate bases is correct?

X⁻ has a smaller K_b than Y⁻

Correct. HX is the stronger acid because it has the larger K_a. Since K_aK_b = K_w, its conjugate base X⁻ must have the smaller K_b.

X⁻ has a larger K_b than Y⁻

A stronger acid has a weaker conjugate base, not a stronger one.

X⁻ and Y⁻ must have identical K_b values

Different K_a values require different corresponding K_b values because each product must equal K_w.

No comparison can be made because K_a and K_b are unrelated

They are directly related for a conjugate pair through K_aK_b = K_w.

Frequently asked questions

Questions students often ask about K_w

These questions connect the basic K_w equation to ideas that become important later in acid–base equilibria, buffers and titration calculations.

Why does pure water contain H⁺ and OH⁻ if no acid or alkali has been added?

Water itself can transfer protons between water molecules. One molecule acts as a Brønsted–Lowry acid while another acts as a Brønsted–Lowry base, producing H₃O⁺ and OH⁻. This reversible process is the auto-ionisation of water.

Is H⁺ really present as a free proton in aqueous solution?

No. A proton is strongly associated with water molecules. H₃O⁺ is a more explicit representation, although H⁺(aq) is the conventional shorthand used throughout A-Level acid–base calculations.

Why is H₂O not included in the K_w expression?

Liquid water is the solvent and its activity is effectively constant. That constant contribution is incorporated into K_w, leaving the useful expression K_w = [H⁺][OH⁻].

Does adding a strong acid increase K_w?

No, provided the temperature remains unchanged. Adding acid increases [H⁺], but [OH⁻] decreases so that [H⁺][OH⁻] continues to equal the same K_w.

Is pH + pOH always equal to 14?

No. The general relationship is pH + pOH = pK_w. At 298 K, pK_w = 14.00, so pH + pOH = 14.00 at that temperature. At another temperature, pK_w may have a different value.

Can neutral water have a pH below 7?

Yes. Increasing temperature increases K_w, so both [H⁺] and [OH⁻] in pure water increase. The pH therefore decreases, but the water remains neutral because [H⁺] still equals [OH⁻].

How do I decide whether a solution is acidic at a temperature other than 298 K?

Compare [H⁺] with [OH⁻]. If [H⁺] > [OH⁻], the solution is acidic. Alternatively, calculate the neutral pH as ½pK_w and compare the solution's pH with that value.

Why does a stronger acid have a weaker conjugate base?

For a conjugate acid–base pair, K_aK_b = K_w. At a fixed temperature, K_w is constant. A larger K_a must therefore be accompanied by a smaller K_b.

Should I find the pH of a strong base using K_w or using pOH?

Either route is valid. You may calculate [H⁺] from K_w = [H⁺][OH⁻] and then find pH, or calculate pOH from [OH⁻] and use pH + pOH = pK_w. At the same temperature, both methods give the same result.

Why must I check the formula of a strong base before using its concentration as [OH⁻]?

The base concentration and hydroxide-ion concentration are not always equal. NaOH gives one OH⁻ per formula unit, whereas Ba(OH)₂ gives two. You must first use the dissociation stoichiometry to determine [OH⁻].

The ionic product of water turns the small auto-ionisation of H₂O into a powerful quantitative relationship. At a fixed temperature, K_w links [H⁺] and [OH⁻]; pK_w links pH and pOH; and K_w also connects the strengths of conjugate acids and bases through K_aK_b = K_w. Most importantly, remember that neutral means [H⁺] = [OH⁻], not automatically pH 7. With these ideas secure, calculations involving weak acids, weak bases, buffers and acid–base equilibria become much easier to organise.

The auto-ionisation of water

Why do A-Level equations often use H⁺ instead of H₃O⁺?

Defining the ionic product of water, Kw

Rearranging the Kw expression

Neutral, acidic and alkaline solutions

Finding [H⁺] and [OH⁻] in pure water at 298 K

The relationship between pH, pOH and pKw

Finding the neutral pH from pKw

How temperature affects Kw

Why can neutral water have a pH below 7?

Does adding acid or alkali change Kw?

The relationship between Ka, Kb and Kw

The corresponding pK relationship

Worked example: finding Ka from Kb

Using Kw in calculations

Example 1: pH of a strong base

Example 2: do not forget the OH⁻ stoichiometry

Example 3: finding [OH⁻] from pH

Example 4: neutral water at a different temperature

Common mistakes with the ionic product of water

Kw checklist

Know the equilibrium

Judge acidity correctly

Connect the p quantities

Solve calculations carefully

Questions students often ask about Kw

Defining the ionic product of water, K_w

Rearranging the K_w expression

The relationship between pH, pOH and pK_w

Finding the neutral pH from pK_w

How temperature affects K_w

Does adding acid or alkali change K_w?

The relationship between K_a, K_b and K_w

Worked example: finding K_a from K_b

Using K_w in calculations

K_w checklist

Questions students often ask about K_w