Pure water looks like the most boring liquid in the world. It is not. At any instant a tiny handful of its molecules are busy passing protons to each other, and that quiet little equilibrium is what fixes 7 as the neutral point, lets you work out the pH of an alkali, and explains why hot water is slightly acidic without being acidic at all.
📘 What you need to know
Water ionises itself: H2O(l) ↔ H+(aq) + OH−(aq), or written in full, 2H2O(l) ↔ H3O+(aq) + OH−(aq).
Because [H2O] barely changes, it gets folded into the constant, giving Kw = [H+][OH−].
Kw = 1.00 × 10−14 at 298 K. This is on the data booklet, but learn it anyway.
In pure water the two ions are made in equal numbers, so [H+] = [OH−] = 1.00 × 10−7 and pH = 7.00.
The product is fixed, so push one concentration up and the other must come down.
Ionisation is endothermic, so heating shifts it right: Kw gets bigger and the pH of pure water drops.
Neutral means [H+] = [OH−], not pH 7. Hot pure water is neutral at pH 6.13.
Water passing protons to itself
Water is amphiprotic, as you saw two pages ago. So put two water molecules next to each other and one of them can act as the acid while the other acts as the base. One proton hops across.
Both products are conjugates of water: H3O+ is its conjugate acid and OH− is its conjugate base. Water is sitting in the middle of its own pair.
Where Kw comes from
It is an ordinary equilibrium, so start with an ordinary equilibrium constant.
Step 1: the normal expressionKc = [H+][OH−] ÷ [H2O]
Now think about how little water actually reacts. In one litre of water there are about 55 moles of molecules, and only around 0.0000001 mol of them split. That is such a small nibble that [H2O] is effectively constant. A constant divided into a constant is still a constant, so we multiply both sides by [H2O] and give the new number its own name.
Step 2: fold the water inKw = Kc × [H2O] = [H+][OH−] = 1.00 × 10−14 at 298 K
Notice that Kw has no denominator. If you write it as a fraction with [H2O] on the bottom you have written Kc, not Kw, and that costs a mark.
Why pure water is pH 7
In pure water there is nothing else around, so the only source of H+ and the only source of OH− is that one reaction. Every time it happens it makes one of each. So the two concentrations have to be identical.
Call them both x. Then:
Solving for pure water at 298 Kx × x = 1.00 × 10−14 x = √(1.00 × 10−14) = 1.00 × 10−7 mol dm−3
pH = −log10(1.00 × 10−7) = 7.00
So the number 7 is not a decision someone made. It falls out of the size of Kw.
The see-saw
Here is the most useful thing about Kw. The product of the two concentrations is fixed, so they behave like the two ends of a see-saw. Add acid and [H+] shoots up, which forces [OH−] down to keep the product at 10−14. The hydroxide ions never disappear entirely — there are just very few of them.
[H+] / mol dm−3
[OH−] / mol dm−3
pH
Type of solution
1.0 × 10−1
1.0 × 10−13
1
strongly acidic
1.0 × 10−3
1.0 × 10−11
3
acidic
1.0 × 10−5
1.0 × 10−9
5
weakly acidic
1.0 × 10−7
1.0 × 10−7
7
neutral
1.0 × 10−9
1.0 × 10−5
9
weakly alkaline
1.0 × 10−13
1.0 × 10−1
13
strongly alkaline
Spot the pattern. Read the powers in each row and add them together. They always come to 13, which with the two leading 1.0 values gives a product of exactly 10−14. Use that as a two-second check on any answer.
🧩 Getting the pH of a strong alkali
Write down [OH−]. For a strong base like NaOH it is simply the concentration of the base, since it fully dissociates.
Use [H+] = Kw ÷ [OH−] to get the hydrogen ion concentration.
Feed that into pH = −log10[H+].
Sanity check: an alkali must come out above 7. If it did not, you divided the wrong way round.
What heat does to Kw
Splitting a water molecule means breaking an O–H bond, and bond breaking takes energy in. So the forward reaction is endothermic.
Now apply Le Châtelier. Raise the temperature and the equilibrium shifts in the endothermic direction to absorb the extra heat, which here means shifting right. More water splits, so both [H+] and [OH−] go up, and their product — Kw — goes up with them.
More H+ means a lower pH. Water at 100 °C has a pH of about 6.13.
The curve is steep at the hot end because the number of split molecules roughly doubles for every 15 °C or so. It never reaches zero at the cold end — ice-cold water still ionises a little.
Temperature
Kw / 10−14
[H+] in pure water
pH of pure water
0 °C
0.114
3.4 × 10−8
7.47
10 °C
0.293
5.4 × 10−8
7.27
25 °C
1.00
1.0 × 10−7
7.00
50 °C
5.48
2.3 × 10−7
6.63
100 °C
56.2
7.5 × 10−7
6.13
Every one of those waters is neutral. The definition of neutral is that the two ion concentrations are equal, and they are equal in all five rows. “pH 7 means neutral” is a shortcut that only works at 298 K, which is why exam questions always give you the temperature.
Worked examples
WORKED EXAMPLE
Calculate the pH of 0.0500 mol dm−3 NaOH(aq) at 298 K. (Kw = 1.00 × 10−14)
Step 1: write down [OH−]
NaOH is a strong base and gives one OH− per formula unit.
[OH−] = 0.0500 mol dm−3Step 2: use Kw to get [H+][H+] = 1.00 × 10−14 ÷ 0.0500 = 2.00 × 10−13 mol dm−3Step 3: convert to pHpH = −log10(2.00 × 10−13) = 12.699…pH = 12.70it came out above 7, which is what an alkali must do — check ticks
WORKED EXAMPLE
A solution at 298 K has pH 9.60. Calculate its hydroxide ion concentration.
Step 1: pH gives you [H+] first[H+] = 10−9.60 = 2.512 × 10−10 mol dm−3Step 2: rearrange Kw for the other ion[OH−] = Kw ÷ [H+]Step 3: substitute[OH−] = 1.00 × 10−14 ÷ 2.512 × 10−10 = 3.98 × 10−5[OH−] = 3.98 × 10−5 mol dm−3the solution is alkaline, so [OH−] should beat [H+] — and it does
WORKED EXAMPLE
At 50 °C, Kw = 5.48 × 10−14. Find the pH of pure water at this temperature and state whether it is acidic.
Step 1: in pure water the two ions are equal[H+] × [H+] = 5.48 × 10−14Step 2: square root it[H+] = √(5.48 × 10−14) = 2.341 × 10−7 mol dm−3Step 3: convert to pHpH = −log10(2.341 × 10−7) = 6.63Step 4: answer the second half properly
[H+] still equals [OH−], so nothing is in excess.
pH = 6.63, and the water is still neutral, not acidicthe second half is where the marks hide — do not just stop at the number
💡 Exam tip
Write Kw = [H+][OH−] with no denominator. Water does not appear in the expression.
For strong base questions, the route is always [OH−] → [H+] → pH. Never try to log the hydroxide concentration and call it pH.
Check the direction of your final answer: acids below 7, alkalis above 7 (at 298 K). It catches division done the wrong way round.
If the temperature is not 298 K, do not assume Kw = 10−14. The question will give you the right value.
For “explain the effect of temperature”, you need three things: endothermic, shifts right, so Kw increases.
Remember that Kw changes only with temperature. Adding acid does not change it — it just moves the two concentrations along the see-saw.
⚠ Common mix-up
Saying hot water is acidic. Its pH is below 7 but [H+] = [OH−], so it is neutral. This is a favourite exam trap.
Thinking an acidic solution has no OH− at all. It has some — just very little, because the product is fixed.
Taking −log of [OH−] and writing “pH”. That is pOH, which is a different quantity.
Including [H2O] in the Kw expression. It was absorbed into the constant on purpose.
Assuming adding acid changes Kw. Only temperature does that.
Forgetting to square root. For pure water you need √Kw, not Kw itself.
Up next: Strong and Weak Acids and Bases. So far we have counted protons. Now we ask a different question — how willing is an acid to let its proton go, and why that has nothing to do with how concentrated it is.
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