Ka, Kb, Kw, pKa, pKb

Explain what Ka, Kb, Kw, pKa and pKb mean and convert between them.

GCE A-Level H2 Chemistry 9476-2027

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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

E. Conjugate acid–base pair link (same temperature)

For a conjugate pair HA/A⁻:

KₐK_b = K_w and pKₐ + pK_b = pK_w

Key Ideas (What Earns Marks)

Strong acids/bases dissociate completely; Kₐ and K_b are for weak acids/bases.
Kₐ and K_b are equilibrium constants: the expression uses equilibrium concentrations (in mol dm⁻³).
Smaller pKₐ means stronger acid because it means larger Kₐ.
K_w (and pK_w) depends on temperature. At 25°C, K_w = 1.0 × 10⁻¹⁴ and pK_w = 14.00.
For conjugate pairs, use pKₐ + pK_b = pK_w to avoid algebra/log mistakes (must be same temperature).

Neutral means [H+] = [OH-]. When pKw changes with temperature, neutral pH is not always 7.

Data table

Temperature	Neutral pH
25°C	7
50°C	6.63

Because Kₐ measures “how far the acid ionises”, therefore larger Kₐ gives more H⁺ at equilibrium (lower pH), for the same starting concentration.

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

Confirm the temperature (use the given K_w/pK_w, or 25°C if stated).
Use KₐK_b = K_w (or pKₐ + pK_b = pK_w) for the conjugate pair.
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

Problem

A weak acid has pKₐ = 4.80. Calculate Kₐ.

Study the worked solution

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).
Substitute
Method
Evaluate 10^(-4.8).
Reason
A positive pKₐ gives a dissociation constant below one.
Working
Kₐ = 1.58 × 10⁻⁵.

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

Conjugate relationship, temperature value and result

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

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.
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.

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

Acid PAcid Q

View solution step by step

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).
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.

Problem

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

Expression, rearrangement, value and unit

View solution step by step

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⁻].
Rearrange
1 mark
Method
Divide by the hydrogen-ion concentration.
Reason
This isolates the requested hydroxide concentration.
Working
[OH⁻] = K_w/[H⁺].
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.

Self-mark with the mark scheme

Compare your response with each mark point. Select a point only when your response contains that evidence.

Credit the Kw expression, rearrangement and correctly unitised concentration.

Mark points

1 mark: Writes Kw = [H+][OH−].
1 mark: Rearranges for [OH−].
1 mark: Gives 5.0 × 10−10 mol dm−3.

Problem

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

Temperature-specific pOH and hydroxide concentration

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

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.
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.

Next step

Try the neutral-water-at-50°C Mind Stretcher independently.

Common Mistakes

Using K_w = 1.0 × 10⁻¹⁴ at temperatures other than 25°C without being told it’s 25°C.
Mixing up Kₐ and K_b expressions (wrong species in numerator/denominator).
Thinking smaller Kₐ means stronger acid (it is weaker).
Using pKₐ + pK_b = 14.00 without stating 25°C (use pK_w at the stated temperature).

Use the Aqueous Equilibria topic check to practise and check your understanding.

Exam Tips

Always write the equilibrium first, then the K expression.
For conjugate pairs, write: pKₐ + pK_b = pK_w (same temperature) to avoid sign errors.
If the question says “at 25°C”, state pK_w = 14.00 before you use it.

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.

Syllabus and review details

GCE A-Level H2 Chemistry 9476-2027 · 9476-2027
9476 (2027), complete syllabus

Last reviewed: 2026-08-29

Ka, Kb, Kw, pKa, pKb

Definitions (Must Know)

A. Acid dissociation constant, Kₐ

B. Base dissociation constant, K_b

C. Ionic product of water, K_w

D. pKₐ and pK_b

E. Conjugate acid–base pair link (same temperature)

Key Ideas (What Earns Marks)

Detailed Explanations

A. What Kₐ and K_b actually measure

B. Why we use pKₐ and pK_b

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

Worked Examples

(convert pKₐ → Kₐ)

Invert the logarithmic definition

Substitute

(conjugate pair at 25°C)

State the relationship

Calculate

Relate Kₐ, pKₐ, and acid strength

Use the logarithmic direction

Link constant to strength

(K_w calculation)

Write the ionic product

Rearrange

Substitute

Use pK_w away from 25°C

Find pOH

Convert to concentration

Common Mistakes

Exam Tips

Mind Stretchers

Mind stretcher 1Extension