Introduction to Kw and Water's Dissociation Revision Notes | A Level AQA Chemistry

1. Why Pure Water Conducts Electricity (Just About)

Pure water should not conduct electricity — it contains no dissolved ions. Yet if you measure the conductivity of very pure water, you detect a tiny but real current. Something must be producing ions, and that something is water itself.

This observation is the starting point for one of the most important equilibria in A-level chemistry. Water molecules can act as both an acid and a base simultaneously, transferring a proton from one molecule to another. The result is a tiny concentration of $\text{H}^+$ and $\text{OH}^-$ ions always present in any aqueous solution.

Understanding this self-ionisation, and the equilibrium constant that describes it, unlocks the entire topic of pH, buffers, and acid-base calculations at A-level. Everything that follows in this topic builds on what you are about to learn here.

2. The Self-Ionisation of Water

Picture two water molecules colliding. In a small fraction of collisions, one molecule donates a proton ($\text{H}^+$) to the other. The donor becomes a hydroxide ion ($\text{OH}^-$) and the acceptor becomes a hydronium ion ($\text{H}_3\text{O}^+$). For the purposes of AQA A-level, we write the proton as $\text{H}^+$ rather than $\text{H}_3\text{O}^+$, but both representations describe the same species.

The equilibrium for this process is written as:

The double arrow ($\rightleftharpoons$) is critical here — this is a reversible reaction at dynamic equilibrium. The forward reaction (ionisation) and the reverse reaction (recombination) occur at equal rates, so the tiny concentrations of $\text{H}^+$ and $\text{OH}^-$ remain constant at a given temperature.

3. Deriving Kw from the Equilibrium Constant

You already know how to write a $K_c$ expression from the equilibrium law: products over reactants, each raised to the power of its stoichiometric coefficient. Applying this to the self-ionisation of water gives:

Here is where the key simplification happens. The concentration of liquid water is effectively constant. One litre of water has a mass of approximately 1000 g, and the molar mass of water is 18 g mol$^{-1}$, giving a concentration of roughly 55.5 mol dm$^{-3}$. Because so little water ionises (the equilibrium lies far to the left), this value does not change measurably.

Since $[\text{H}_2\text{O}]$ is constant, it can be incorporated into the equilibrium constant itself. Multiplying both sides of the $K_c$ expression by $[\text{H}_2\text{O}]$ gives a new, simplified constant — the ionic product of water, $K_w$:

This is the expression you must know. $K_w = [\text{H}^+][\text{OH}^-]$ applies to every aqueous solution at a given temperature — acidic, alkaline, or neutral. The concentrations of $\text{H}^+$ and $\text{OH}^-$ will vary, but their product always equals $K_w$ at that temperature.

4. What Kw Tells Us About Pure Water

In pure water, every $\text{H}^+$ ion produced comes from the ionisation of a water molecule, and every such ionisation also produces exactly one $\text{OH}^-$ ion. The concentrations of $\text{H}^+$ and $\text{OH}^-$ must therefore be equal. This is the defining condition of a neutral solution.

At 298 K, $K_w = 1.00 \times 10^{-14}$ mol$^2$ dm$^{-6}$. Because $[\text{H}^+] = [\text{OH}^-]$ in pure water, we can write $[\text{H}^+]^2 = 1.00 \times 10^{-14}$, which gives $[\text{H}^+] = 1.00 \times 10^{-7}$ mol dm$^{-3}$. Taking $-\log_{10}$ of this value gives pH = 7.00 at 298 K — the familiar neutral pH.

Worked example: Calculating [H+] and [OH-] in Pure Water at 298 K

Given that Kw = 1.00 x 10-14 mol2 dm-6 at 298 K, calculate the concentration of H+ ions and OH- ions in pure water, and hence the pH.

1. $K_w = [\text{H}^+][\text{OH}^-] = 1.00 \times 10^{-14}$ mol$^2$ dm$^{-6}$ — This expression holds for all aqueous solutions at 298 K.
2. In pure water: $[\text{H}^+] = [\text{OH}^-]$, so $[\text{H}^+]^2 = 1.00 \times 10^{-14}$ — Each ionisation event produces one H+ and one OH-, so their concentrations must be equal in pure water.
3. $[\text{H}^+] = \sqrt{1.00 \times 10^{-14}} = 1.00 \times 10^{-7}$ mol dm$^{-3}$ — Taking the square root: $\sqrt{10^{-14}} = 10^{-7}$. Verify: $10^{-7} \times 10^{-7} = 10^{-14}$. Correct.
4. $[\text{OH}^-] = 1.00 \times 10^{-7}$ mol dm$^{-3}$ — Equal to [H+] in pure water, as expected.
5. $\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.00 \times 10^{-7}) = 7.00$ — $-\log_{10}(10^{-7}) = -(-7) = 7.00$. This confirms the neutral pH of pure water at 298 K.

Answer: [H+] = [OH-] = 1.00 x 10-7 mol dm-3; pH = 7.00 at 298 K

Now consider what happens when you dissolve an acid in water. Adding $\text{H}^+$ ions increases $[\text{H}^+]$. To maintain the constant value of $K_w$, the equilibrium shifts to the left, reducing $[\text{OH}^-]$. The product $[\text{H}^+][\text{OH}^-]$ always equals $K_w$ — this is a powerful constraint you can use in calculations.

5. Kw and Temperature — A Critical Relationship

Like all equilibrium constants, $K_w$ changes only when temperature changes. The self-ionisation of water is endothermic overall — the energy required to break bonds exceeds the energy released in forming new bonds, so the forward reaction absorbs energy. By Le Chatelier's principle, increasing temperature shifts the equilibrium to the right, producing more $\text{H}^+$ and $\text{OH}^-$ ions and increasing $K_w$.

The table reveals something that catches many students out. At 37 °C, the pH of pure water is 6.80, not 7.00. This does not mean the water has become acidic. Neutral means $[\text{H}^+] = [\text{OH}^-]$, and that condition still holds — both concentrations have increased equally. The pH scale shifts with temperature because pH 7 is only the neutral point at 298 K.

6. Using Kw in Calculations

The real power of $K_w$ is that it lets you find $[\text{OH}^-]$ from $[\text{H}^+]$, or vice versa, in any aqueous solution. This is essential for calculating the pH of strong alkalis, where you are given $[\text{OH}^-]$ rather than $[\text{H}^+]$ directly.

Worked example: Calculating the pH of a Strong Alkali Using Kw

Calculate the pH of a 0.0200 mol dm-3 solution of sodium hydroxide at 298 K. Kw = 1.00 x 10-14 mol2 dm-6.

1. NaOH is a strong alkali and fully dissociates: $\text{NaOH}(s) \rightarrow \text{Na}^+(aq) + \text{OH}^-(aq)$, so $[\text{OH}^-] = 0.0200$ mol dm$^{-3}$ — One mole of NaOH gives one mole of OH-(aq) upon complete dissociation.
2. $K_w = [\text{H}^+(aq)][\text{OH}^-(aq)]$, so $[\text{H}^+(aq)] = \frac{K_w}{[\text{OH}^-(aq)]}$ — Rearranging Kw = [H+(aq)][OH-(aq)] to make [H+(aq)] the subject.
3. $[\text{H}^+(aq)] = \frac{1.00 \times 10^{-14}}{0.0200}$ — Substituting Kw = 1.00 x 10-14 mol2 dm-6 and [OH-(aq)] = 0.0200 mol dm-3.
4. $[\text{H}^+(aq)] = \frac{1.00 \times 10^{-14}}{2.00 \times 10^{-2}} = 5.00 \times 10^{-13}$ mol dm$^{-3}$ — Dividing: $\frac{10^{-14}}{10^{-2}} = 10^{-12}$, then $\frac{1.00}{2.00} = 0.500$, giving $5.00 \times 10^{-13}$. Verify: $5.00 \times 10^{-13} \times 0.0200 = 1.00 \times 10^{-14}$. Correct.
5. $\text{pH} = -\log_{10}(5.00 \times 10^{-13}) = 12.30$ — $-\log_{10}(5.00 \times 10^{-13}) = -(\log_{10} 5.00 + \log_{10} 10^{-13}) = -(0.699 + (-13)) = -(0.699 - 13) = 12.30$ (to 2 d.p.).

Answer: pH = 12.30 at 298 K

Notice the sanity check built into step 4: multiplying $[\text{H}^+]$ back by $[\text{OH}^-]$ should always return $K_w$. This is a reliable way to verify your arithmetic before calculating pH. A pH above 7 at 298 K confirms the solution is alkaline, which is consistent with a sodium hydroxide solution.

7. Exam Technique and Common Mistakes

AQA questions on $K_w$ tend to test three things: writing the correct expression, using it to find $[\text{H}^+]$ or $[\text{OH}^-]$, and explaining the effect of temperature on $K_w$ and pH. Each of these has a specific trap that costs marks.

Every spec point in this lesson connects back to one central idea: $K_w = [\text{H}^+][\text{OH}^-]$ is a temperature-dependent constant that governs the relationship between hydrogen and hydroxide ion concentrations in all aqueous solutions. Mastering this expression — where it comes from, what it means, and how to use it — is the foundation for every pH and buffer calculation that follows.