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.
Kw is the equilibrium constant for the autoionization of water, the reaction where two water molecules swap a proton: H2O + H2O ⇌ H3O+ + 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. 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.
Kw lives in Unit 8 (Acids and Bases) and sits at the foundation of the whole unit. Learning objective AP Chem 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.
Keep studying AP® Chemistry Unit 8
Ka and Kb for conjugate pairs (Unit 8)
For any conjugate acid-base pair, Ka × Kb = Kw. This is why a strong acid 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.
Equilibrium constants and temperature (Unit 7)
Kw is an equilibrium constant like any K from Unit 7, 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)
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.
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 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.
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.
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.
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.
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.
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.
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.
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