Mini Chemistry
Part of Mini EducationSwitch subject

Ka, Kb, Kw, pKa, pKb

Learn and apply Ka, Kb, Kw, pKa, pKb in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
On this page

Ka, Kb, Kw, pKa and pKb: Orientation

This lesson turns “acid/base strength” into numbers: what Kₐ, K_b, and K_w mean, how pK helps you compare them quickly, and how conjugate pairs link via KₐK_b = K_w.

If the acid-base foundations feel rusty, revisit Acids and Bases (Theories) and keep the Aqueous Equilibria hub open for linked methods.

Definitions (Must Know)

A. Acid dissociation constant, Kₐ

For a weak acid HA(aq) in water:

HA(aq) ⇌ H + (aq) + A-(aq)
Kₐ = [H⁺][A⁻]/[HA]

B. Base dissociation constant, K_b

For a weak base B(aq) in water:

B(aq) + H₂O(l) ⇌ BH + (aq) + OH-(aq)
K_b = [BH⁺][OH⁻]/[B]

C. Ionic product of water, K_w

For water:

H₂O(l) ⇌ H + (aq) + OH-(aq)
K_w = [H⁺][OH⁻]

D. pKₐ and pK_b

pKₐ = - log ₁₀Kₐ
pK_b = - log ₁₀K_b

For a conjugate pair HA/A⁻:

KₐK_b = K_w and pKₐ + pK_b = pK_w

Detailed Explanations

A. What Kₐ and K_b actually measure

Kₐ and K_b measure the extent of dissociation at equilibrium.

  • If Kₐ is large, the equilibrium HA ⇌ H⁺ + A⁻ lies further to the right (more ions at equilibrium).
  • If K_b is large, the equilibrium B + H₂O ⇌ BH⁺ + OH⁻ lies further to the right.

This is why Kₐ and K_b are used to compare weak acids/bases: they describe equilibrium position, not “speed”.

B. Why we use pKₐ and pK_b

Because pKₐ = - log ₁₀Kₐ, therefore:

  • a tenfold increase in Kₐ decreases pKₐ by 1
  • smaller pKₐ means stronger acid (larger Kₐ)

C. A repeatable workflow: linking Kₐ, K_b, and K_w

  1. Confirm the temperature (use the given K_w/pK_w, or 25°C if stated).
  2. Use KₐK_b = K_w (or pKₐ + pK_b = pK_w) for the conjugate pair.
  3. Convert between K and pK only at the end (to avoid rounding errors).

Mini example (25°C):

  • If pKₐ = 4.80, then pK_b = 14.00-4.80 = 9.20.

Worked Examples

Modelled example 1

(convert pKₐ → Kₐ)

Core

Problem

A weak acid has pKₐ = 4.80. Calculate Kₐ.
Study the worked solution

  1. Invert the logarithmic definition

    Method

    Rearrange pKₐ = - log ₁₀Kₐ.

    Reason

    The inverse of a base-ten logarithm is a power of ten, while the definition supplies the negative sign.

    Working

    Kₐ = 10^(-pKₐ).

  2. Substitute

    Method

    Evaluate 10^(-4.80).

    Reason

    A positive pKₐ gives a dissociation constant below one.

    Working

    Kₐ = 1.58 × 10⁻⁵.

Guided practice 2

(conjugate pair at 25°C)

About 6 min

Problem

At 25°C, a weak acid has pKₐ = 4.80. Find pK_b for its conjugate base, using pK_w = 14.00.

Try this before viewing the solution

Hints

Hint 1: same conjugate pair
Use pKₐ + pK_b = pK_w at the same temperature.
Hint 2: isolate pKb
Subtract 4.80 from the stated 14.00.
View solution step by step

  1. State the relationship

    Method

    Connect conjugate acid and base constants through water.

    Reason

    The pK relationship applies to a conjugate pair at one temperature.

    Working

    pKₐ + pK_b = pK_w = 14.00 at 25°C.

  2. Calculate

    Method

    Rearrange for the conjugate base.

    Reason

    The supplied pK_w fixes the temperature-dependent total.

    Working

    pK_b = 14.00-4.80 = 9.20.

Common misconception 3

Relate Kₐ, pKₐ, and acid strength

Find and correct the mistake

Learner comparison

Acid P has pKₐ = 3.2 and acid Q has pKₐ = 5.1. A learner says Q is stronger because its pKₐ is larger. Diagnose the claim.

Choose the stronger acid

The stronger acid is

View solution step by step

  1. Use the logarithmic direction

    Method

    Recognise the negative sign in pKₐ = - log ₁₀Kₐ.

    Reason

    A smaller pKₐ corresponds to a larger Kₐ.

    Working

    3.2 < 5.1, so Kₐ(P) > Kₐ(Q).

  2. Link constant to strength

    Method

    Select acid P.

    Reason

    A larger Kₐ indicates a greater equilibrium extent of acid dissociation.

    Working

    P is the stronger acid.

Examiner practice 4

(K_w calculation)

3 marks

Problem

At 25°C, calculate [OH⁻] when [H⁺] = 2.0 × 10⁻⁵ mol dm⁻³ and K_w = 1.0 × 10⁻¹⁴ mol²dm⁻⁶. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Write the ionic product

    1 mark

    Method

    Relate hydrogen and hydroxide concentrations.

    Reason

    K_w is defined by their equilibrium product.

    Working

    K_w = [H⁺][OH⁻].

  2. Rearrange

    1 mark

    Method

    Divide by the hydrogen-ion concentration.

    Reason

    This isolates the requested hydroxide concentration.

    Working

    [OH⁻] = K_w/[H⁺].

  3. Substitute

    1 mark

    Method

    Divide the supplied powers of ten.

    Reason

    The concentration unit follows after cancelling one concentration factor from K_w.

    Working

    [OH⁻] = (1.0 × 10⁻¹⁴)/(2.0 × 10⁻⁵) = 5.0 × 10⁻¹⁰ mol dm⁻³.

Challenge 5

Use pK_w away from 25°C

Minimal support

Problem

At 50°C, pK_w = 13.26. A solution has pH = 4.00. Calculate pOH and [OH⁻] at this temperature.

Try this before viewing the solution

Hints

Hint 1: temperature-specific sum
Use pH + pOH = pK_w = 13.26, not 14.00.
Hint 2: invert pOH
After finding pOH, use [OH⁻] = 10^(-pOH).
View solution step by step

  1. Find pOH

    Method

    Subtract the stated pH from the supplied pK_w.

    Reason

    The logarithmic water relationship uses the value at the same temperature.

    Working

    pOH = 13.26-4.00 = 9.26.

  2. Convert to concentration

    Method

    Apply the inverse logarithm.

    Reason

    pOH = - log ₁₀[OH⁻].

    Working

    [OH⁻] = 10^(-9.26) = 5.50 × 10⁻¹⁰ mol dm⁻³.

Mind Stretchers

Mind stretcher 1Extension

At 50°C, pK_w = 13.26 (given). Explain why “neutral water has pH 7” is not always true, and find the pH of neutral water at 50°C.

Show Answer

Mark scheme:

  • Neutral means [H⁺] = [OH⁻], not “pH = 7”.
  • At 50°C, pH + pOH = pK_w = 13.26.
  • If [H⁺] = [OH⁻], then pH = pOH = 13.26/2 = 6.63.