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Acids, Bases & Aqueous EquilibriaAcids and BasesContent level: Core 20 min

Weak Bases, Kb and the Ka-Kb Relationship

What you'll be able to do: Calculate the pH of a weak base solution and convert between Ka and Kb for a conjugate pair.

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Introduction

Weak bases are solved with the same machinery as weak acids, with one extra step at the end: the ICE table gives you hydroxide, not hydronium.

These are recommended, not required. You can start this lesson at any time.

Learning objectives

  • Write the Kb expression for a weak base
  • Calculate the pH of a weak base solution through pOH
  • Convert between Ka and Kb using Ka x Kb = Kw
  • Identify common weak bases and explain why amines are basic

Lesson

Bases take a proton from water

Ammonia has no hydroxide in its formula, yet its solution is basic because the nitrogen lone pair pulls a proton off water, leaving OH⁻ behind. The equilibrium constant for that process is Kb.

Kb = [HB⁺][OH⁻] / [B]

Solving is the same, with one extra step

Build the ICE table with x as the amount of base that reacts. Under the small-x conditions x = sqrt(Kb x C). The trap is stopping there: x is [OH⁻], so take pOH = -log x and then pH = 14.00 - pOH.

A weak base problem that ends with pH below 7 almost always skipped the pOH conversion.

Ka and Kb are locked together

Multiply the Ka expression of HA by the Kb expression of A⁻ and every concentration except the water ions cancels, leaving Kw. So Ka x Kb = 1.0 × 10⁻¹⁴ and pKa + pKb = 14.00 at 25 C.

Ka x Kb = Kw = 1.0 × 10⁻¹⁴ at 25 C

Why this matters

Reference tables usually list only Ka. To find the pH of a sodium acetate solution you look up Ka for acetic acid, divide Kw by it to get Kb for acetate, and proceed. This one conversion links the whole unit together.

Kb = Kw / Ka of the conjugate acid. Never use the Ka of the base itself in the base ICE table.

Families of weak bases

Amines are ammonia with organic groups replacing hydrogens, and they are all weak bases: methylamine has Kb = 4.4 × 10⁻⁴, somewhat stronger than ammonia at 1.8 × 10⁻⁵. Anions of weak acids, such as CN⁻, F⁻ and CH₃COO⁻, are also weak bases, while anions of strong acids are not.

Key ideas

Definition
Kb

The equilibrium constant for a base accepting a proton from water.

Rule
Conjugate relationship

Ka x Kb = Kw for an acid and its conjugate base.

Rule
Two-step conversion

A weak base ICE table gives [OH⁻]; find pOH, then pH.

Rule
pKa + pKb

Equals 14.00 at 25 C.

Rule
Which anions are basic

Only conjugate bases of weak acids; Cl⁻, Br⁻, I⁻, NO₃⁻ and ClO₄⁻ are not.

Equation
Base ionization constant

Kb = [HB⁺][OH⁻] / [B]

  • [B] = equilibrium concentration of the un-protonated base
Equation
Conjugate pair relation

Ka x Kb = Kw

  • Kw = 1.0 × 10⁻¹⁴ at 25 C
Equation
Approximate hydroxide

[OH⁻] = sqrt(Kb x C)

  • C = initial base concentration

Worked examples

Worked example 1

Calculate the pH of 0.25 M NH₃ (Kb = 1.8 × 10⁻⁵.

Try it first: Decide at the start whether x will be hydronium or hydroxide.

    0 of 3 steps revealed.

    Worked example 2

    Acetic acid has Ka = 1.8 × 10⁻⁵. Find Kb for the acetate ion.

    Try it first: Write which species is the conjugate acid of acetate.

      0 of 3 steps revealed.

      Worked example 3

      Find the pH of 0.10 M NaF given Ka(HF) = 6.8 × 10⁻⁴.

      Try it first: Decide which ion in the salt actually reacts with water.

        0 of 3 steps revealed.

        Common mistakes

        Reporting the weak base x value as the pH directly.

        Why it's wrong: x is hydroxide, so -log x is the pOH.

        Check instead: Subtract pOH from 14.00.

        Using Kw / Kb where Kw / Ka is needed, or the reverse.

        Why it's wrong: The relationship connects a pair, so you must divide by the partner constant.

        Check instead: Name the conjugate acid explicitly before dividing.

        Treating Cl⁻ as a weak base.

        Why it's wrong: It is the conjugate base of a strong acid and has no measurable basicity.

        Check instead: Only conjugate bases of weak acids react with water.

        Assuming NH₃ must contain OH⁻.

        Why it's wrong: It generates hydroxide by taking a proton from water.

        Check instead: Write the Bronsted equilibrium.

        Adding Ka and Kb instead of multiplying.

        Why it's wrong: Combining equilibria multiplies constants; only the p-values add.

        Check instead: Ka x Kb = Kw and pKa + pKb = 14.00.

        Practice this skill

        No practice questions are available for this topic yet. You can still practice the whole unit.

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        What you should now know

        A weak base accepts a proton from water: B + H₂O ⇌ HB⁺ + OH⁻, with Kb = [HB⁺][OH⁻]/[B]. The ICE table is identical in form to the weak acid case, giving x = sqrt(Kb x C) under the same small-x conditions, but x is [OH⁻], so you must find pOH first and then subtract from 14.00. For any conjugate pair, Ka x Kb = Kw, which means pKa + pKb = 14.00 at 25 C; this is why a table of Ka values is all you ever need. Common weak bases include ammonia and the amines, which act through the lone pair on nitrogen, and the conjugate bases of weak acids such as acetate and fluoride.

        • Kb describes a base taking a proton from water
        • The weak base ICE table gives [OH⁻], so convert through pOH
        • Ka x Kb = Kw for a conjugate pair, and pKa + pKb = 14.00
        • Kb for an anion comes from the Ka of its parent acid
        • Anions of strong acids are not bases
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