Chemistry · Acid-Base

Ion Product of Water

The ion product of water links the hydronium and hydroxide concentrations of every aqueous solution: at 25 °C, $Kw =\frac{ 10^{-14} mol^{2}}{L^{2}}$ , from which $pH + pOH = 14$ follows.

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Formula

K_{W} = [H_{3} O^{+}] \cdot [OH^{-}]

Kw in mol²/L² (10⁻¹⁴ at 25 °C) · [H₃O⁺] in mol/L · [OH⁻] in mol/L

Variables & units

Variables & units
Symbol	Meaning	Unit
$Kw$	Ion product of water (temperature-dependent)	mol²/L²
[H₃O⁺]	Hydronium-ion concentration	mol/L
[OH⁻]	Hydroxide-ion concentration	mol/L

LaTeX

K_W = [\text{H}_3\text{O}^+] \cdot [\text{OH}^-]

QUANTA STUDYquanta-study.de

Diagram with a logarithmic concentration axis: the line [H₃O⁺] falls from 10⁰ at pH 0 to 10⁻¹⁴ at pH 14, the line [OH⁻] rises as its mirror image; they cross, highlighted, at pH 7 and 10⁻⁷ mol/L. — As the pH rises, [H₃O⁺] falls while [OH⁻] rises by the same factor, because their product stays Kw = 10⁻¹⁴ mol²/L²; at pH 7 both equal 10⁻⁷ mol/L and the solution is neutral (25 °C).

Derivation & background

The basis is the autoprotolysis 2 H₂O ⇌ H₃O⁺ + OH⁻. Kw is strongly temperature-dependent: $pKw = 14.0$ at 25 °C, ≈ 13.6 at 37 °C (body temperature), ≈ 12.3 at 100 °C. Neutral means $[H_{3}O^{+}] = [OH^{-}]$ ; at 100 °C the neutral point therefore lies at $pH \approx{} 6.1$ . The "14" is thus no law of nature but the 25 °C value.

Exam blueprint

Validity range

Applies to all aqueous solutions; the numerical value $\frac{10^{-14} mol^{2}}{L^{2}}$ holds only at 25 °C, because Kw is strongly temperature-dependent.

Derivation steps

The law of mass action of the autoprotolysis is combined with the practically constant water concentration.

12 H₂O ⇌ H₃O⁺ + OH⁻ is an equilibrium with a very small constant.
2The constant water concentration is absorbed: $Kw = [H_{3}O^{+}]\cdot{}[OH^{-}]$ .

Rearrangements

Hydronium concentration from [OH⁻]

[\text{H}_3\text{O}^+] = \frac{K_W}{[\text{OH}^-]}

The route from the pOH of a base to the pH.

Logarithmic form

pH + pOH = pK_W = 14

The value 14 holds only at 25 °C; otherwise use the appropriate pKw.

Task variant

$[OH^{-}] =\frac{ 10^{-3} mol}{L}$ : what pH results at 25 °C?

$[H_{3}O^{+}] =\frac{ 10^{-14}}{10^{-3}} =\frac{ 10^{-11} mol}{L}$ , so $pH = 11 (pOH = 3)$ .

Why does neutral water at 100 °C not have pH 7?

Kw rises with temperature (pKw ≈ 12.3 at 100 °C).
Neutral means $[H_{3}O^{+}] = [OH^{-}] = \sqrt{Kw} \approx{}\frac{ 7\times{}10^{-7} mol}{L}$ , so $pH \approx{} 6.1$ ; it is still neutral.

Common mistakes

Treating pH 7 as a universal neutral point.

Neutral means $[H_{3}O^{+}] = [OH^{-}]$ ; the corresponding pH depends on temperature via Kw.

Using $pH + pOH = 14$ at any temperature.

The sum is pKw and equals 14 only at 25 °C.

Adding the concentrations instead of multiplying.

Kw is a product: if [H₃O⁺] falls by a factor of 10, [OH⁻] rises by a factor of 10.

Exam context

pH of bases via the pOH, neutralization calculations and transfer questions on temperature dependence.

Formula cluster

Acid-base calculation

Kw couples pH and pOH and underpins the pH scale.

pH = −lg[H₃O⁺]

Definition of the pH Value

pKs = −lg(Ks)

pKa Value (Acid Constant)

Kc = [C]ᶜ·[D]ᵈ / ([A]ᵃ·[B]ᵇ)

Law of Mass Action (Equilibrium Constant)

pKb = −lg(Kb)

pKb Value (Base Constant)

Worked example

Sodium hydroxide solution with $[OH^{-}] =\frac{ 10^{-3} mol}{L}$ at 25 °C: $[H_{3}O^{+}] =\frac{ Kw}{[OH^{-}]} =\frac{ 10^{-14}}{10^{-3}} =\frac{ 10^{-11} mol}{L} \to{} pH = 11 (pOH = 3)$ .

Applications

pH calculation of bases and alkaline solutions, neutralization calculations, water chemistry and aquaristics, temperature correction of pH measurements

How do you calculate with Ion Product of Water?

Here is how to work through a typical Ion Product of Water (Kw = [H₃O⁺]·[OH⁻]) task step by step:

1
Task
$[OH^{-}] =\frac{ 10^{-3} mol}{L}$ : what pH results at 25 °C?
Solution path
$[H_{3}O^{+}] =\frac{ 10^{-14}}{10^{-3}} =\frac{ 10^{-11} mol}{L}$ , so $pH = 11 (pOH = 3)$ .
2
Task
Why does neutral water at 100 °C not have pH 7?
Solution path
1. Kw rises with temperature (pKw ≈ 12.3 at 100 °C).
2. Neutral means $[H_{3}O^{+}] = [OH^{-}] = \sqrt{Kw} \approx{}\frac{ 7\times{}10^{-7} mol}{L}$ , so $pH \approx{} 6.1$ ; it is still neutral.

Frequently asked questions about Ion Product of Water

How do you use the ion product of water to calculate the pH of a base?+

First determine the hydroxide concentration, then convert via Kw. Example: sodium hydroxide with $c =\frac{ 0.001 mol}{L}$ is completely dissociated, so $[OH^{-}] =\frac{ 10^{-3} mol}{L}$ . With $Kw = [H_{3}O^{+}]\cdot{}[OH^{-}] =\frac{ 10^{-14} mol^{2}}{L^{2}}$ it follows that $[H_{3}O^{+}] =\frac{ 10^{-14}}{10^{-3}} =\frac{ 10^{-11} mol}{L}$ and thus $pH = 11$ . It is faster via the pOH: $pOH = -\log{}(10^{-3}) = 3$ , and because $pH + pOH = 14$ , $pH = 14 - 3 = 11$ . Both routes are equivalent. For polyvalent bases such as Ba(OH)₂, remember to use twice the hydroxide amount: $c(OH^{-}) = 2\cdot{}c(Ba(OH)_{2})$ .

Why does

pH + pOH = 14

hold?+

The relation follows directly from the ion product. Taking the logarithm of $Kw = [H_{3}O^{+}]\cdot{}[OH^{-}] =\frac{ 10^{-14} mol^{2}}{L^{2}}$ and multiplying by −1 turns the product into a sum: $-\log{}[H_{3}O^{+}] - \log{}[OH^{-}] = 14$ , i.e. $pH + pOH = pKw = 14$ . The 14 is nothing magical, just the numerical value of pKw at 25 °C. Practically this means: if you know one of the two quantities, you automatically know the other. If the hydronium concentration rises by a factor of 10, the hydroxide concentration falls by a factor of 10; the product stays constant. At other temperatures the sum shifts: at 37 °C, $pH + pOH \approx{} 13.6$ .

Is the ion product of water really constant?+

It is constant only at a fixed temperature. Kw is an equilibrium constant of the endothermic autoprotolysis 2 H₂O ⇌ H₃O⁺ + OH⁻ and therefore rises markedly with temperature: pKw is 14.0 at 25 °C, about 13.6 at body temperature (37 °C) and only around 12.3 at 100 °C. Within a solution at a given temperature, however, Kw holds everywhere, whether acidic, neutral or basic; that is exactly what makes it so useful. If you add acid, [H₃O⁺] rises and [OH⁻] falls correspondingly, so the product stays constant. In very concentrated solutions activities deviate from concentrations, and the simple numerical value then holds only approximately.

Why is boiling water not acidic although its pH is below 7?+

Because acidic is not defined by the number 7 but by comparing hydronium and hydroxide concentrations. Neutral means $[H_{3}O^{+}] = [OH^{-}]$ . At 100 °C the autoprotolysis is stronger $(pKw \approx{} 12.3)$ , so both concentrations rise to $\sqrt{Kw} \approx{}\frac{ 7\times{}10^{-7} mol}{L}$ ; the pH of neutral water thereby drops to about 6.1. But since hydronium and hydroxide are still exactly equally abundant, the water is neutral, not acidic. The rule to remember: pH 7 is the neutral point only at 25 °C. This transfer question is a popular exam trap because it exposes blind memorization of the 7.

Where do the ions in chemically pure water come from?+

From the water itself: in the autoprotolysis one water molecule transfers a proton to another, 2 H₂O ⇌ H₃O⁺ + OH⁻. This equilibrium lies extremely far to the left; at 25 °C only $\frac{10^{-7} mol}{L}$ each of hydronium and hydroxide ions form, meaning only a single ion pair per roughly 555 million water molecules. Nevertheless this suffices for a measurable electrical conductivity, which Kohlrausch and Heydweiller demonstrated as early as 1894 on elaborately purified water. This residual conductivity is the experimental proof of autoprotolysis and the reason why completely non-conducting water does not exist. All pH calculations in aqueous solutions build on this equilibrium.

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How do you rearrange Ion Product of Water for Hydronium concentration from [OH⁻]?

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax Chemistry 2e, §14.2 pH and pOH

More formulas: Acids and bases

pH = pK_a + \log \frac{[A^-]}{[HA]}

Henderson-Hasselbalch Equation

D=[H^+]^2+K_{a1}[H^+]+K_{a1}K_{a2};\;\alpha_0=\frac{[H^+]^2}{D},\;\alpha_1=\frac{K_{a1}[H^+]}{D},\;\alpha_2=\frac{K_{a1}K_{a2}}{D}

Diprotic Acid Species Fractions

\beta=2.303\,C_T\frac{K_a[H^+]}{(K_a+[H^+])^2}

Buffer Capacity of an Acid–Base Pair

V_{eq}=\frac{c_{HA}V_{HA}}{c_B},\quad pH=pK_a+\log_{10}\frac{n_{OH^-}}{n_{HA,0}-n_{OH^-}}\ (0<V_B<V_{eq}),\quad pH_{eq}\approx7+\frac12\log_{10}\!\left(10^{pK_a}\frac{c_{A^-}}{c^\circ}\right)

Titration Curve of a Weak Acid

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Ion Product of Water

The ion product of water links the hydronium and hydroxide concentrations of every aqueous solution: at 25 °C,

Kw =\frac{ 10^{-14} mol^{2}}{L^{2}}

, from which

pH + pOH = 14

follows.

AdvancedExam-relevantUpper secondaryUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

Kw

Ion product of water (temperature-dependent)

mol²/L²

[H₃O⁺]

Hydronium-ion concentration

mol/L

[OH⁻]

Hydroxide-ion concentration

mol/L

Derivation & background

Exam blueprint

Validity range

Applies to all aqueous solutions; the numerical value $\frac{10^{-14} mol^{2}}{L^{2}}$ holds only at 25 °C, because Kw is strongly temperature-dependent.

Derivation steps

The law of mass action of the autoprotolysis is combined with the practically constant water concentration.

12 H₂O ⇌ H₃O⁺ + OH⁻ is an equilibrium with a very small constant.
2The constant water concentration is absorbed: $Kw = [H_{3}O^{+}]\cdot{}[OH^{-}]$ .

How do you calculate with Ion Product of Water?

Here is how to work through a typical Ion Product of Water (Kw = [H₃O⁺]·[OH⁻]) task step by step:

Task

$[OH^{-}] =\frac{ 10^{-3} mol}{L}$ : what pH results at 25 °C?

Solution path

$[H_{3}O^{+}] =\frac{ 10^{-14}}{10^{-3}} =\frac{ 10^{-11} mol}{L}$ , so $pH = 11 (pOH = 3)$ .

Task

Why does neutral water at 100 °C not have pH 7?

Solution path

Kw rises with temperature (pKw ≈ 12.3 at 100 °C).
Neutral means $[H_{3}O^{+}] = [OH^{-}] = \sqrt{Kw} \approx{}\frac{ 7\times{}10^{-7} mol}{L}$ , so $pH \approx{} 6.1$ ; it is still neutral.

Frequently asked questions about Ion Product of Water

How do you use the ion product of water to calculate the pH of a base?+

Why does

pH + pOH = 14

hold?+

Is the ion product of water really constant?+

Why is boiling water not acidic although its pH is below 7?+

Where do the ions in chemically pure water come from?+

Retain Ion Product of Water for exams

Create a curated FSRS exam set for Kw = [H₃O⁺]·[OH⁻]: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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