--- title: "Kw — AP Chemistry Definition & Exam Guide" description: "Kw is the equilibrium constant for water's autoionization, equal to [H3O+][OH−] = 1.0 × 10⁻¹⁴ at 25°C. It links pH, pOH, Ka, and Kb across Unit 8." canonical: "https://fiveable.me/ap-chem/key-terms/kw" type: "key-term" subject: "AP Chemistry" unit: "Unit 8" --- # Kw — AP Chemistry Definition & Exam Guide ## Definition Kw is the ion product of water, the equilibrium constant for water's autoionization: Kw = [H3O+][OH−] = 1.0 × 10⁻¹⁴ at 25°C. Because pKw = pH + pOH = 14 at 25°C, Kw connects every pH, pOH, and Ka/Kb calculation in AP Chem Unit 8. ## What It Is Kw is the [equilibrium constant](/ap-chem/unit-7/reaction-quotient-le-chateliers-principle/study-guide/JFx1InPfZCZ9SugPKDCE "fv-autolink") for the autoionization of water, the reaction where two water molecules swap a proton: H2O + H2O ⇌ [H3O+](/ap-chem/key-terms/h3o "fv-autolink") + OH−. Like any equilibrium constant, it's a product of concentrations, so Kw = [H3O+][OH−]. At 25°C, Kw = 1.0 × 10⁻¹⁴. Taking the negative log of both sides gives you the equation you'll use constantly: pKw = pH + pOH = 14.0 at 25°C. Here's the part that trips people up. Kw isn't a fixed law of nature, it's an equilibrium constant, which means it changes with [temperature](/ap-chem/unit-5/reaction-rates/study-guide/4V94d3BwjoPaOOyQtDKQ "fv-autolink"). Autoionization is endothermic, so heating water raises Kw, which lowers the pH of pure water below 7. The water is still neutral because [H3O+] = [OH−], it just isn't pH 7 anymore. Neutral means equal concentrations, not pH 7. That distinction is a classic AP trap. ## Why It Matters Kw lives in [Unit 8](/ap-chem/unit-8 "fv-autolink") (Acids and Bases) and sits at the foundation of the whole unit. Learning objective [AP Chem](/ap-chem "fv-autolink") 8.1.A asks you to calculate pH and pOH using Kw and the concentrations of species in neutral water, and the essential knowledge spells out the core equations: Kw = [H3O+][OH−] = 1.0 × 10⁻¹⁴ and pKw = 14 = pH + pOH at 25°C. Kw also powers AP Chem 8.3.A, because once you've solved a weak acid problem for [H3O+], Kw is how you get [OH−] and pOH. And the single most-tested Kw relationship is Ka × Kb = Kw for any conjugate acid-base pair. If you know Ka for an acid, Kw hands you Kb for its conjugate base in one step. ## Connections ### Ka and Kb for conjugate pairs (Unit 8) For any conjugate acid-base pair, Ka × Kb = Kw. This is why a [strong acid](/ap-chem/key-terms/strong-acid "fv-autolink") has a uselessly weak conjugate base and a weak acid has a conjugate base that actually does chemistry. The exam loves asking you to convert Ka to Kb (or vice versa) using this one equation. ### pH and pOH (Unit 8) pKw = pH + pOH = 14 at 25°C is just Kw with logs applied. Every time you flip between pH and pOH, you're using Kw. It's the bridge between the hydronium side and the hydroxide side of any [solution](/ap-chem/key-terms/solution "fv-autolink"). ### Equilibrium constants and temperature (Unit 7) Kw is an equilibrium constant like any K from [Unit 7](/ap-chem/unit-7 "fv-autolink"), so it shifts with temperature. Autoionization is endothermic, so warmer water has a larger Kw and pure water's pH drops below 7 while staying perfectly neutral. This is Le Châtelier's principle applied to water itself. ### [Hydronium ion (Unit 8)](/ap-chem/key-terms/hydronium-ion) Kw exists because water makes its own hydronium and hydroxide ions, even with nothing dissolved in it. That's why pure water at 25°C already has [H3O+] = 1.0 × 10⁻⁷ M. The CED prefers H3O+ over H+, though both are accepted on the exam. ## On the AP Exam Multiple-choice questions test Kw three main ways. First, conversions: given Ka of a weak acid, find Kb of its conjugate base using Ka × Kb = Kw (for example, Ka = 2.5 × 10⁻⁵ gives Kb = 4.0 × 10⁻¹⁰). Second, conceptual statements about the Ka-Kb relationship for conjugate pairs. Third, the temperature twist: given Kw = 4.0 × 10⁻¹⁴ at 45°C, find the pH of pure water (set [H3O+] = [OH−], so [H3O+] = √Kw, giving pH ≈ 6.7, still neutral). On FRQs, Kw shows up inside weak acid/base equilibrium problems, like the released questions on nitrous acid and the glycolate ion, where you may need to convert a given Ka into the Kb you actually need, or move between pH and pOH. The skill being tested is fluency, not memorizing a definition. ## Kw vs Ka (acid dissociation constant) Kw is the equilibrium constant for water reacting with itself; Ka is the equilibrium constant for a specific acid donating a proton to water. Kw is the same for every aqueous solution at a given temperature (1.0 × 10⁻¹⁴ at 25°C), while Ka is unique to each acid and measures its strength. They meet in one equation: for a conjugate pair, Ka × Kb = Kw. ## Key Takeaways - Kw is the equilibrium constant for water's autoionization: Kw = [H3O+][OH−] = 1.0 × 10⁻¹⁴ at 25°C. - Taking negative logs gives pKw = pH + pOH = 14.0 at 25°C, which is how you flip between pH and pOH. - For any conjugate acid-base pair, Ka × Kb = Kw, so a stronger acid always has a weaker conjugate base. - Kw increases with temperature because autoionization is endothermic, so pure water above 25°C has a pH below 7 but is still neutral. - Neutral means [H3O+] = [OH−], not pH = 7; that equality only happens to give pH 7 at exactly 25°C. - In pure water at 25°C, [H3O+] = [OH−] = 1.0 × 10⁻⁷ M, found by taking the square root of Kw. ## FAQs ### What is Kw in AP Chemistry? Kw is the ion product of water, the equilibrium constant for water autoionizing into hydronium and hydroxide ions. It equals [H3O+][OH−] = 1.0 × 10⁻¹⁴ at 25°C, and in log form, pH + pOH = 14. ### Is Kw always 1.0 × 10⁻¹⁴? No, only at 25°C. Kw is an equilibrium constant, so it changes with temperature. The exam has used Kw = 4.0 × 10⁻¹⁴ at 45°C, which makes pure water's pH about 6.7 even though the water is still neutral. ### Is pure water always pH 7? No. Pure water is always neutral ([H3O+] = [OH−]), but its pH only equals 7 at 25°C. At higher temperatures Kw is larger, so neutral water has a pH below 7. ### What's the difference between Kw and Ka? Kw describes water reacting with itself and is the same for every aqueous solution at a given temperature. Ka describes a specific acid donating a proton and varies from acid to acid. They're linked by Ka × Kb = Kw for any conjugate pair. ### How do I find Kb if I'm given Ka? Use Kb = Kw / Ka. For example, if a weak acid has Ka = 2.5 × 10⁻⁵, its conjugate base has Kb = (1.0 × 10⁻¹⁴) / (2.5 × 10⁻⁵) = 4.0 × 10⁻¹⁰ at 25°C. This conversion shows up constantly on the exam. ## Related Study Guides - [8.1 Introduction to Acids and Bases](/ap-chem/unit-8/intro-acids-bases/study-guide/rG2ZBgD9evdBohz4a7Mj) ## Structured Data ```json {"@context":"https://schema.org","@graph":[{"@type":"LearningResource","@id":"https://fiveable.me/ap-chem/key-terms/kw#resource","name":"Kw — AP Chemistry Definition & Exam Guide","url":"https://fiveable.me/ap-chem/key-terms/kw","learningResourceType":"Concept explainer","educationalLevel":"AP® / High School","about":{"@id":"https://fiveable.me/ap-chem/key-terms/kw#term"},"audience":{"@type":"EducationalAudience","educationalRole":"student"},"dateModified":"2026-06-11T05:27:14.212Z","isPartOf":{"@type":"Collection","name":"AP Chemistry Key Terms","url":"https://fiveable.me/ap-chem/key-terms"},"publisher":{"@type":"Organization","name":"Fiveable","url":"https://fiveable.me"}},{"@type":"DefinedTerm","@id":"https://fiveable.me/ap-chem/key-terms/kw#term","name":"Kw","description":"Kw is the ion product of water, the equilibrium constant for water's autoionization: Kw = [H3O+][OH−] = 1.0 × 10⁻¹⁴ at 25°C. 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