Ionic Product of Water and pH

Spec Mapping — OCR H432 Module 5.1.3 — Acids, bases and buffers, sub-topic on the ionic product of water $K_w$Kw​, the definition of pH as $-\log_{10}[\text{H}^+]$−log10​[H+], the quantitative relationship between $[\text{H}^+]$[H+] and $[\text{OH}^-]$[OH−] in any aqueous solution at a given temperature, calculation of pH from $[\text{H}^+]$[H+] and the reverse, and the temperature dependence of $K_w$Kw​ as evidence that the auto-ionisation of water is an endothermic process (refer to the official OCR H432 specification document for exact wording). This is the gateway to every quantitative pH calculation in the rest of the module — strong acids (lesson 3), weak acids and $K_a$Ka​ (lessons 4–5), strong bases (lesson 6), buffers (lessons 7–8), titration curves (lesson 9) and indicators (lesson 10) all rely on $K_w$Kw​ as the bridge between $[\text{H}^+]$[H+] and $[\text{OH}^-]$[OH−].

The logical chain from Brønsted–Lowry theory (lesson 1) to quantitative pH (the rest of the module) passes through this single equilibrium constant. Water is amphoteric: one water molecule can donate a proton to another, producing a tiny but measurable concentration of H₃O⁺ and OH⁻ ions in pure water. The equilibrium constant for that self-ionisation is the ionic product of water, $K_w$Kw​. Once you have $K_w$Kw​, you can calculate the pH of pure water at any temperature, the pH of any aqueous solution from either $[\text{H}^+]$[H+] or $[\text{OH}^-]$[OH−], and you have the quantitative framework that the rest of the module fleshes out for specific acid and base types. This lesson is therefore short on chemistry but heavy on calculator routine — you must be able to switch fluently between $[\text{H}^+]$[H+], $[\text{OH}^-]$[OH−], pH and pOH for any aqueous solution at 298 K, and you must appreciate why the relationships shift at non-298 K temperatures.

Key Equation: $K_w = [\text{H}^+][\text{OH}^-] = 1.00 \times 10^{-14} \text{ mol}^2 \text{ dm}^{-6} \text{ at 298 K}$Kw​=[H+][OH−]=1.00×10−14 mol2 dm−6 at 298 K $\text{pH} = -\log_{10}[\text{H}^+] \qquad [\text{H}^+] = 10^{-\text{pH}}$pH=−log10​[H+][H+]=10−pH The units of $K_w$Kw​ follow directly from multiplying two concentrations together: $(\text{mol dm}^{-3})(\text{mol dm}^{-3}) = \text{mol}^2 \text{ dm}^{-6}$(mol dm−3)(mol dm−3)=mol2 dm−6. Mark schemes deduct for omitting these.

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Self-ionisation of water

Even chemically pure water is not entirely molecular. A vanishingly small fraction of water molecules transfer a proton to another water molecule at any instant:

$2\text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+\text{(aq)} + \text{OH}^-\text{(aq)}$2H2​O(l)⇌H3​O+(aq)+OH−(aq)

Or, using the abbreviated H⁺(aq) form:

$\text{H}_2\text{O(l)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)}$H2​O(l)⇌H+(aq)+OH−(aq)

The equilibrium lies very far to the left. At 298 K, roughly one water molecule in every $5 \times 10^8$5×108 is ionised at any instant — a tiny fraction, but enough to give pure water a measurable electrical conductivity and a definite pH of 7.00. The auto-ionisation is endothermic ($\Delta H \approx +56 \text{ kJ mol}^{-1}$ΔH≈+56 kJ mol−1 — breaking an O–H bond requires energy), which is why $K_w$Kw​ grows with temperature (see later).

The equilibrium constant expression

For the auto-ionisation written as a generic equilibrium, $K_c$Kc​ would be:

$K_c = \frac{[\text{H}^+][\text{OH}^-]}{[\text{H}_2\text{O}]}$Kc​=[H2​O][H+][OH−]​

But because water is the solvent, $[\text{H}_2\text{O}]$[H2​O] is essentially constant — even at extreme acidities or alkalinities, the concentration of water in a dilute aqueous solution is approximately $55.5 \text{ mol dm}^{-3}$55.5 mol dm−3 (1000 g of water in 1 dm³, divided by $M_r = 18.0$Mr​=18.0). We absorb that constant into the equilibrium constant and define a new constant:

$K_w = K_c \cdot [\text{H}_2\text{O}] = [\text{H}^+][\text{OH}^-]$Kw​=Kc​⋅[H2​O]=[H+][OH−]

This is the ionic product of water. At 298 K (25 °C):

$K_w = 1.00 \times 10^{-14} \text{ mol}^2 \text{ dm}^{-6}$Kw​=1.00×10−14 mol2 dm−6

Crucially, $K_w$Kw​ depends only on temperature, not on the identity or concentration of any solute. This is why it acts as a universal bridge between $[\text{H}^+]$[H+] and $[\text{OH}^-]$[OH−] in every aqueous solution at 298 K.

pH of pure water at 298 K

In pure water the self-ionisation generates equal numbers of H⁺ and OH⁻ ions — for every water molecule that gives up a proton, exactly one OH⁻ is produced. Let $[\text{H}^+] = [\text{OH}^-] = x$[H+]=[OH−]=x. Then:

$x^2 = K_w = 1.00 \times 10^{-14}$x2=Kw​=1.00×10−14 $x = \sqrt{1.00 \times 10^{-14}} = 1.00 \times 10^{-7} \text{ mol dm}^{-3}$x=1.00×10−14​=1.00×10−7 mol dm−3

So in pure water at 298 K, $[\text{H}^+] = [\text{OH}^-] = 1.00 \times 10^{-7} \text{ mol dm}^{-3}$[H+]=[OH−]=1.00×10−7 mol dm−3, and pH = 7.00.

A neutral solution is one in which $[\text{H}^+] = [\text{OH}^-]$[H+]=[OH−]. At 298 K this corresponds to pH = 7.00 — but only at 298 K. At other temperatures the neutral pH is different (because $K_w$Kw​ is different), yet the solution is still neutral. This is one of the most-confused points in OCR mark schemes; we return to it below.

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The definition of pH

pH is defined as:

$\text{pH} = -\log_{10}[\text{H}^+]$pH=−log10​[H+]

where $[\text{H}^+]$[H+] is the hydrogen ion concentration in mol dm⁻³. The minus sign converts the typically large negative exponent of $[\text{H}^+]$[H+] into a small positive number convenient for tabulation and discussion.

For pure water at 298 K:

$\text{pH} = -\log_{10}(1.00 \times 10^{-7}) = -(-7) = 7.00$pH=−log10​(1.00×10−7)=−(−7)=7.00

The Sørensen pH scale was introduced in 1909 by the Danish chemist Søren Sørensen, who needed a compact notation for the wide dynamic range of [H⁺] encountered in enzyme assays. The "p" stands for power — pH is literally the negative power of ten that gives [H⁺]. Every unit decrease in pH corresponds to a 10-fold increase in [H⁺].

Solution	$[\text{H}^+]$[H+] / mol dm⁻³	pH	Classification (298 K)
Concentrated HCl (~12 mol dm⁻³)	~12	~−1	Strongly acidic
1.00 mol dm⁻³ HCl	1.00	0.00	Strongly acidic
0.100 mol dm⁻³ HCl	0.100	1.00	Acidic
Stomach acid	~$3 \times 10^{-2}$3×10−2	~1.5	Strongly acidic
Vinegar	~$2 \times 10^{-3}$2×10−3	~2.7	Weakly acidic
Blood	$4 \times 10^{-8}$4×10−8	7.40	Slightly alkaline
Pure water	$1.00 \times 10^{-7}$1.00×10−7	7.00	Neutral
Seawater	~$5 \times 10^{-9}$5×10−9	~8.3	Mildly alkaline
Household ammonia	~$1 \times 10^{-11}$1×10−11	~11	Alkaline
0.100 mol dm⁻³ NaOH	$1.00 \times 10^{-13}$1.00×10−13	13.00	Strongly alkaline

The orders-of-magnitude span between vinegar (pH 2.7) and household ammonia (pH 11) is about $10^{8.3}$108.3, i.e. nearly nine orders of magnitude in $[\text{H}^+]$[H+]. The logarithmic scale is the only sensible way to tabulate this range.

flowchart TD
    A["Aqueous solution at 298 K"] --> B{"[H+] vs [OH-]"}
    B -- "[H+] > [OH-]" --> C["Acidic: pH less than 7"]
    B -- "[H+] = [OH-]" --> D["Neutral: pH = 7"]
    B -- "[H+] less than [OH-]" --> E["Alkaline: pH greater than 7"]
    C --> F["[H+] x [OH-] = 1e-14"]
    D --> F
    E --> F
    F --> G["Kw bridges [H+] and [OH-] in any solution"]

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Worked examples — pH from [H⁺]

Worked example 1 — direct pH

A solution has $[\text{H}^+] = 2.50 \times 10^{-3}$[H+]=2.50×10−3 mol dm⁻³. Calculate its pH.

$\text{pH} = -\log_{10}(2.50 \times 10^{-3}) = 2.60 \quad (2 \text{ dp})$pH=−log10​(2.50×10−3)=2.60(2 dp)

Worked example 2 — slightly alkaline solution

A solution has $[\text{H}^+] = 4.80 \times 10^{-9}$[H+]=4.80×10−9 mol dm⁻³. Calculate its pH and state whether it is acidic, neutral or alkaline.

$\text{pH} = -\log_{10}(4.80 \times 10^{-9}) = 8.32 \quad (2 \text{ dp})$pH=−log10​(4.80×10−9)=8.32(2 dp)

The pH is greater than 7, so the solution is alkaline at 298 K.

Significant-figures convention: OCR expects pH to 2 decimal places. The reasoning is the "log decimal-place rule" — the number of decimal places in a logarithm corresponds to the number of significant figures in the original value. So $[\text{H}^+] = 2.50 \times 10^{-3}$[H+]=2.50×10−3 (3 s.f.) gives pH = 2.602 → quoted as 2.60 (2 d.p., reflecting that the mantissa 2.50 has 3 s.f.).

Worked examples — [H⁺] from pH

Taking antilogs: $[\text{H}^+] = 10^{-\text{pH}}$[H+]=10−pH.

Worked example 3 — moderately acidic

A solution has pH = 4.72. Calculate $[\text{H}^+]$[H+].

$[\text{H}^+] = 10^{-4.72} = 1.91 \times 10^{-5} \text{ mol dm}^{-3}$[H+]=10−4.72=1.91×10−5 mol dm−3

Worked example 4 — alkaline solution

A solution has pH = 10.40. Calculate $[\text{H}^+]$[H+] and $[\text{OH}^-]$[OH−] at 298 K.

$[\text{H}^+] = 10^{-10.40} = 3.98 \times 10^{-11} \text{ mol dm}^{-3}$[H+]=10−10.40=3.98×10−11 mol dm−3 $[\text{OH}^-] = \frac{K_w}{[\text{H}^+]} = \frac{1.00 \times 10^{-14}}{3.98 \times 10^{-11}} = 2.51 \times 10^{-4} \text{ mol dm}^{-3}$[OH−]=[H+]Kw​​=3.98×10−111.00×10−14​=2.51×10−4 mol dm−3

So a solution of pH 10.40 has roughly six orders of magnitude more OH⁻ than H⁺ — strongly alkaline territory.

Worked examples — using $K_w$Kw​ for alkaline solutions

For an alkaline solution we typically know $[\text{OH}^-]$[OH−] (from the base concentration and stoichiometry) and want pH. We use $K_w$Kw​ to convert:

$[\text{H}^+] = \frac{K_w}{[\text{OH}^-]}$[H+]=[OH−]Kw​​

Worked example 5 — 0.0200 mol dm⁻³ NaOH

NaOH is a strong base, fully dissociated, so $[\text{OH}^-] = 0.0200$[OH−]=0.0200 mol dm⁻³.

$[\text{H}^+] = \frac{1.00 \times 10^{-14}}{0.0200} = 5.00 \times 10^{-13} \text{ mol dm}^{-3}$[H+]=0.02001.00×10−14​=5.00×10−13 mol dm−3 $\text{pH} = -\log_{10}(5.00 \times 10^{-13}) = 12.30 \quad (2 \text{ dp})$pH=−log10​(5.00×10−13)=12.30(2 dp)

Worked example 6 — 1.25 × 10⁻³ mol dm⁻³ KOH

$[\text{H}^+] = \frac{1.00 \times 10^{-14}}{1.25 \times 10^{-3}} = 8.00 \times 10^{-12} \text{ mol dm}^{-3}$[H+]=1.25×10−31.00×10−14​=8.00×10−12 mol dm−3 $\text{pH} = -\log_{10}(8.00 \times 10^{-12}) = 11.10 \quad (2 \text{ dp})$pH=−log10​(8.00×10−12)=11.10(2 dp)

Strong-base pH is taken up in detail in lesson 6.

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Relationship between pH and pOH

If we define $\text{pOH} = -\log_{10}[\text{OH}^-]$pOH=−log10​[OH−] and $\text{p}K_w = -\log_{10}(K_w)$pKw​=−log10​(Kw​), then taking $-\log_{10}$−log10​ of both sides of $K_w = [\text{H}^+][\text{OH}^-]$Kw​=[H+][OH−] gives:

$\text{p}H + \text{pOH} = \text{p}K_w$pH+pOH=pKw​

At 298 K, $\text{p}K_w = 14.00$pKw​=14.00, so:

$\text{p}H + \text{pOH} = 14.00 \text{ (at 298 K only!)}$pH+pOH=14.00 (at 298 K only!)

OCR focuses on pH rather than pOH, but the relationship is a powerful shortcut. A solution of pH 12.30 has pOH = 1.70 at 298 K, i.e. $[\text{OH}^-] = 10^{-1.70} = 2.00 \times 10^{-2}$[OH−]=10−1.70=2.00×10−2 mol dm⁻³ — which matches the 0.0200 mol dm⁻³ NaOH of worked example 5.

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Variation of $K_w$Kw​ with temperature

The auto-ionisation of water is endothermic:

$2\text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+\text{(aq)} + \text{OH}^-\text{(aq)}, \quad \Delta H \approx +56 \text{ kJ mol}^{-1}$2H2​O(l)⇌H3​O+(aq)+OH−(aq),ΔH≈+56 kJ mol−1

By Le Chatelier's principle, raising the temperature shifts the equilibrium to the right, increasing both $[\text{H}^+]$[H+] and $[\text{OH}^-]$[OH−] and therefore increasing $K_w$Kw​.

Temperature / °C	$K_w$Kw​ / mol² dm⁻⁶	Neutral pH	$[\text{H}^+]$[H+] / mol dm⁻³
0	$1.14 \times 10^{-15}$1.14×10−15	7.47	$3.38 \times 10^{-8}$3.38×10−8
10	$2.92 \times 10^{-15}$2.92×10−15	7.27	$5.40 \times 10^{-8}$5.40×10−8
25	$1.00 \times 10^{-14}$1.00×10−14	7.00	$1.00 \times 10^{-7}$1.00×10−7
37 (body temp)	$2.51 \times 10^{-14}$2.51×10−14	6.80	$1.58 \times 10^{-7}$1.58×10−7
50	$5.48 \times 10^{-14}$5.48×10−14	6.63	$2.34 \times 10^{-7}$2.34×10−7
100	$5.13 \times 10^{-13}$5.13×10−13	6.14	$7.16 \times 10^{-7}$7.16×10−7

At 100 °C the pH of pure water is 6.14 — but the water is still neutral, because $[\text{H}^+] = [\text{OH}^-]$[H+]=[OH−]. This is the subtlety: "neutral" is defined by equality of [H⁺] and [OH⁻], NOT by pH = 7. The pH = 7 shortcut only applies at 298 K.

A student who writes "pure water at 50 °C has pH = 6.63 so it is acidic" has misunderstood the definition of neutrality. Always say: "neutral when [H⁺] = [OH⁻]."

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Worked example 7 — neutral water at 50 °C

At 50 °C, $K_w = 5.48 \times 10^{-14}$Kw​=5.48×10−14 mol² dm⁻⁶. Calculate the pH of pure water and state whether it is acidic, neutral or alkaline.

In pure water $[\text{H}^+] = [\text{OH}^-] = \sqrt{K_w}$[H+]=[OH−]=Kw​​:

$[\text{H}^+] = \sqrt{5.48 \times 10^{-14}} = 2.34 \times 10^{-7} \text{ mol dm}^{-3}$[H+]=5.48×10−14​=2.34×10−7 mol dm−3 $\text{pH} = -\log_{10}(2.34 \times 10^{-7}) = 6.63 \quad (2 \text{ dp})$pH=−log10​(2.34×10−7)=6.63(2 dp)

The water is still neutral because $[\text{H}^+] = [\text{OH}^-]$[H+]=[OH−]. The pH is below 7 only because $K_w$Kw​ has increased.

Worked example 8 — pH 12.50 to [OH⁻]

A solution has pH = 12.50 at 298 K. Calculate $[\text{OH}^-]$[OH−].