Kb (the base dissociation constant) is the equilibrium constant for the reaction of a weak base with water, B + H₂O ⇌ HB⁺ + OH⁻. A larger Kb means more ionization and a stronger base; for a conjugate acid-base pair, Ka × Kb = Kw.
Kb is the equilibrium constant for a weak base grabbing a proton from water. Write the reaction as B + H₂O ⇌ HB⁺ + OH⁻, and Kb = [HB⁺][OH⁻]/[B]. Water doesn't appear in the expression because it's the solvent. Since weak bases only partially ionize, Kb values are small (much less than 1), and most of the base stays in its molecular form. That's the whole point of the constant. It tells you exactly how weak the base is.
Kb is the mirror image of Ka. Where Ka measures how readily a weak acid donates a proton to water (producing H₃O⁺), Kb measures how readily a weak base accepts one (producing OH⁻). You'll often see it reported as pKb, where pKb = −log(Kb), so a smaller pKb means a stronger base. Given an initial base concentration and a Kb value, you can set up an ICE table, solve for [OH⁻], find pOH, and convert to pH. That's the core calculation Topic 8.3 expects you to handle.
Kb lives in Topic 8.3 (Weak Acid and Base Equilibria) in Unit 8 and directly supports learning objective 8.3.A, which asks you to explain the relationship among pH, pOH, and the concentrations of all species in a solution of a monoprotic weak acid or weak base. The essential knowledge for weak acids (only a small fraction ionizes, so [H₃O⁺] is much less than the initial concentration) applies symmetrically to weak bases with Kb and [OH⁻]. Kb is also the bridge to the rest of Unit 8. The relationship Ka × Kb = Kw lets you hop between an acid and its conjugate base, which is exactly the move buffer and salt-hydrolysis problems demand later in the unit.
Keep studying AP® Chemistry Unit 8
Kₐ (acid dissociation constant) (Unit 8)
Ka and Kb are the same idea pointed in opposite directions. Ka measures proton donation by a weak acid; Kb measures proton acceptance by a weak base. For any conjugate pair, Ka × Kb = Kw, so a strong-ish weak acid (big Ka) has a pathetically weak conjugate base (tiny Kb).
Kw (autoionization of water) (Unit 8)
Kw is the glue. Multiply the Ka reaction of an acid by the Kb reaction of its conjugate base and they add up to the autoionization of water, which is why Ka × Kb = Kw (1.0 × 10⁻¹⁴ at 25°C). Tables usually list only Ka values, so this equation is how you actually get Kb on the exam.
Conjugate Base (Unit 8)
Every weak acid's conjugate base has its own Kb. This is why a solution of sodium acetate is basic. The acetate ion (conjugate base of acetic acid) reacts with water via its Kb reaction to produce OH⁻.
Equilibrium constants and ICE tables (Unit 7)
Kb is just a regular equilibrium constant wearing an acid-base costume. Everything from Unit 7 applies, including writing the K expression from a balanced equation, leaving out water, and using ICE tables with the small-x approximation to solve for [OH⁻].
Multiple-choice questions ask you to write the correct Kb expression for a given base, rank bases by strength using Kb or pKb values, or convert between Ka and Kb using Kw. The classic FRQ task gives you a weak base (often ammonia or an amine) with its Kb, then asks you to calculate the pH of the solution. The workflow is consistent. Write the base + water reaction, set up an ICE table, solve Kb = x²/[B]₀ for x = [OH⁻], compute pOH, then subtract from 14 to get pH. Forgetting that last conversion (reporting pOH as pH) is one of the most common point losses, so build the final step into your habit. You may also be asked to justify why most of the base remains un-ionized, which comes straight from the partial-ionization idea in 8.3.A.
Ka describes a weak acid donating a proton to water and producing H₃O⁺; Kb describes a weak base accepting a proton from water and producing OH⁻. They are not interchangeable, but they're linked. For a conjugate acid-base pair, Ka × Kb = Kw. If a problem gives you Ka for an acid but asks about a solution of its conjugate base, you must convert with Kb = Kw/Ka before doing any ICE-table math.
Kb is the equilibrium constant for a weak base reacting with water: B + H₂O ⇌ HB⁺ + OH⁻, with Kb = [HB⁺][OH⁻]/[B].
A larger Kb (or smaller pKb) means a stronger base that ionizes to a greater extent in water.
For any conjugate acid-base pair, Ka × Kb = Kw, so a stronger acid always has a weaker conjugate base.
Weak base pH problems solve for [OH⁻] first using an ICE table, then convert through pOH to get pH (pH = 14 − pOH at 25°C).
Because weak bases only partially ionize, [OH⁻] is much smaller than the initial base concentration, and most of the base stays in molecular form.
Water is the solvent, so it never appears in the Kb expression.
Kb is the base dissociation constant, the equilibrium constant for a weak base reacting with water to form its conjugate acid and OH⁻. It quantifies base strength in Topic 8.3 (Unit 8), with larger Kb meaning a stronger base.
Ka measures a weak acid donating a proton to water (producing H₃O⁺), while Kb measures a weak base accepting a proton from water (producing OH⁻). They're connected through Ka × Kb = Kw for any conjugate pair, so you can always convert between them.
No, it's the opposite. A small Kb means very little of the base ionizes, so the base is weak. Strong bases like NaOH don't even get a Kb because they dissociate essentially completely.
Use Kb = Kw/Ka, where Kw = 1.0 × 10⁻¹⁴ at 25°C. For example, if an acid has Ka = 1.0 × 10⁻⁵, its conjugate base has Kb = 1.0 × 10⁻⁹. This works only for conjugate acid-base pairs.
Set up an ICE table for B + H₂O ⇌ HB⁺ + OH⁻ and solve Kb = x²/[B]₀ for x, which equals [OH⁻]. Then calculate pOH = −log[OH⁻] and finish with pH = 14 − pOH. Skipping that last conversion is the most common mistake.
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