Base Dissociation Constants, Kb and pKb

Last updated: 18 August 2026

Diagnose → Learn → Master: Every study-note section includes 15 free MCQs — start with 5 Diagnostic MCQs to identify gaps in your understanding, then complete 10 Mastery MCQs near the end to check what you have learned.

The Brønsted–Lowry definition tells us that a base is a proton acceptor, but it does not tell us how strongly one base accepts protons compared with another. Ammonia, for example, accepts H⁺ from water only to a limited extent, whereas a stronger base has a greater tendency to accept a proton.

We therefore need a quantitative measure of base strength. For weak bases, this is provided by the base dissociation constant, K_b. Just as K_a describes the strength of a weak acid, K_b describes the position of the equilibrium established by a weak base in water.

Once K_b, pK_b and K_w are connected, we can compare bases, relate a base to its conjugate acid and calculate the pH of weak-base solutions using the same equilibrium reasoning used elsewhere in acid–base chemistry.

Diagnostic Quiz

What Do You Already Know About Weak Bases?

Try these five questions before reading the chapter. They check the acid–base ideas that you will need before working with K_b and pK_b.

5 quick questions
Checks important prerequisite ideas
Every answer option is explained

According to the Brønsted–Lowry theory, what does a base do?

Donates a proton

A proton donor is a Brønsted–Lowry acid, not a base.

Accepts a proton

Correct. A Brønsted–Lowry base accepts H⁺ from another species.

Always releases OH⁻ ions by dissociation

Some bases, such as NH₃, do not contain OH⁻. They produce OH⁻ indirectly by accepting H⁺ from water.

Accepts an electron pair from an acid

Electron-pair acceptance describes a Lewis acid. The question asks for the Brønsted–Lowry definition of a base.

Two weak bases have K_b values of 2.0 × 10⁻⁴ and 3.0 × 10⁻⁸ at the same temperature. Which base is stronger?

The base with K_b = 2.0 × 10⁻⁴

Correct. A larger K_b means that proton acceptance occurs to a greater extent, so the base is stronger.

The base with K_b = 3.0 × 10⁻⁸

This is the smaller K_b, so its equilibrium lies less far towards the protonated base and OH⁻. It is the weaker base.

They are equally strong because both are weak bases

Weak bases can have very different strengths. K_b allows those differences to be compared quantitatively.

Their strengths cannot be compared using K_b

K_b is specifically used to compare weak-base strengths when the values refer to the same temperature.

Which statement correctly describes pK_b?

pK_b = log₁₀K_b

The negative sign is essential. The definition is pK_b = −log₁₀K_b.

A larger pK_b always means a stronger base

The relationship is reversed because of the negative logarithm. A stronger base has a larger K_b but a smaller pK_b.

pK_b = −log₁₀K_b

Correct. This logarithmic form makes very small K_b values easier to compare.

pK_b is the pH of the base

pK_b is a measure derived from the equilibrium constant. The pH of a solution also depends on concentration.

Which pair is a conjugate acid–base pair?

NH₃ and OH⁻

These species do not differ by exactly one H⁺ and are therefore not a conjugate acid–base pair.

NH₄⁺ and H₂O

NH₄⁺ loses H⁺ to form NH₃, not H₂O. These two species are not conjugates.

H₂O and NH₃

H₂O and NH₃ can react as acid and base, but they are not conjugates of each other because they do not differ by one proton.

NH₄⁺ and NH₃

Correct. NH₄⁺ and NH₃ differ by exactly one H⁺. NH₄⁺ is the conjugate acid of NH₃.

When K_b is used to calculate the pH of a weak-base solution, which concentration is usually found first?

[H⁺]

K_b directly describes an equilibrium that produces OH⁻, so [OH⁻] is usually found first.

[OH⁻]

Correct. The weak-base equilibrium gives [OH⁻]. This is converted to pOH and then to pH.

[H₂O]

Water is the solvent and its concentration is treated as effectively constant in the K_b expression.

K_w

K_w is an equilibrium constant rather than a concentration.

What does the strength of a base mean?

In Brønsted–Lowry terms, a base is a species that accepts a proton. Consider ammonia dissolved in water:

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)

NH₃ accepts H⁺ from H₂O. The NH₃ molecule becomes NH₄⁺, while the water molecule that loses H⁺ becomes OH⁻.

The equilibrium arrow matters. Ammonia is a weak base, so only a fraction of the NH₃ present accepts protons. At equilibrium, a substantial amount of NH₃ therefore remains unreacted.

Weak does not mean dilute.

The strength of a base describes its tendency to accept H⁺. Concentration describes how much solute is present per unit volume. A weak base can be concentrated, and a strong base can be dilute.

Strong bases and weak bases

A strong base reacts essentially completely in the relevant proton-transfer process. Soluble metal hydroxides such as NaOH also dissociate completely in water to provide OH⁻ ions:

NaOH(s) → Na⁺(aq) + OH⁻(aq)

A weak molecular base such as NH₃ behaves differently. It does not simply split into NH₄⁺ and OH⁻. Instead, NH₃ reacts with water by accepting a proton.

Feature	Strong base	Weak base
Extent of reaction	Essentially complete	Partial
Equilibrium	Products strongly favoured in the relevant proton-transfer process	Significant amounts of both reactants and products remain
Typical example	OH⁻ supplied by NaOH	NH₃
K_b	Normally unnecessary in routine strong-base calculations	Used to quantify weak-base strength

Strength: tells us how extensively the base accepts H⁺.

Concentration: tells us how much base is present in a given volume of solution.

The base dissociation constant, K_b

Consider a general weak base B reacting with water:

B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)

B accepts H⁺ from H₂O and becomes BH⁺. The species B and BH⁺ therefore form a conjugate base–acid pair.

Since the reaction reaches dynamic equilibrium, its position can be described by an equilibrium constant. For this base equilibrium, the constant is called the base dissociation constant, K_b.

K_b = [BH⁺][OH⁻] / [B]

The concentrations used in this expression are equilibrium concentrations, not necessarily the concentrations present before the equilibrium was established.

Why is H₂O not included?

If we began with the ordinary equilibrium expression, water would appear as a reactant. However, water is the solvent and is present in a very large excess. Its concentration changes by a negligible proportion as a small amount of weak base reacts.

The concentration of water is therefore treated as effectively constant and is incorporated into K_b.

K_b = [BH⁺][OH⁻] / [B]

K_b refers to a particular equilibrium at a particular temperature.

Like other equilibrium constants, K_b changes when temperature changes. K_b values should therefore be compared at the same temperature.

Why is it called a dissociation constant?

The traditional name can be slightly misleading for a molecular base such as NH₃. NH₃ does not simply split apart into NH₄⁺ and OH⁻. It accepts H⁺ from water.

The term base dissociation constant is nevertheless the conventional name for K_b, and the equilibrium expression is based on the concentrations of the species present after proton transfer has occurred.

What does the magnitude of K_b tell us?

Look again at the general expression:

K_b = [BH⁺][OH⁻] / [B]

A base that accepts H⁺ more readily produces greater equilibrium concentrations of BH⁺ and OH⁻ and leaves a smaller proportion of B unreacted.

The numerator becomes larger relative to the denominator, so K_b is larger.

Value of K_b	Equilibrium behaviour	Base strength
Larger K_b	Greater formation of BH⁺ and OH⁻	Stronger base
Smaller K_b	More B remains unprotonated	Weaker base

Larger K_b → stronger base

Example: comparing three bases

Suppose three weak bases have the following K_b values at the same temperature:

Base	K_b	Relative strength
Base P	4.0 × 10⁻³	Strongest
Base Q	2.0 × 10⁻⁵	Intermediate
Base R	8.0 × 10⁻⁹	Weakest

We do not need to calculate the pH of three solutions to rank their intrinsic base strengths. The equilibrium constants already provide the comparison.

The use of pK_b

K_b values for weak bases are often very small. Comparing several numbers written in scientific notation can become inconvenient, so chemists commonly use the logarithmic pK_b scale.

pK_b = −log₁₀K_b

The reverse conversion is:

K_b = 10^−pK_b

The direction of the pK_b scale

The negative sign reverses the comparison. A larger K_b gives a smaller pK_b.

Stronger base → larger K_b → smaller pK_b

Base	K_b	pK_b	Relative strength
X	1.0 × 10⁻³	3.00	Strongest
Y	1.0 × 10⁻⁵	5.00	Intermediate
Z	1.0 × 10⁻⁸	8.00	Weakest

K_b: larger value = stronger base.

pK_b: smaller value = stronger base.

Worked example: converting K_b to pK_b

A weak base has:

K_b = 1.8 × 10⁻⁵

Therefore:

pK_b = −log₁₀(1.8 × 10⁻⁵)

pK_b = 4.74

Do not rank K_b and pK_b in the same direction.

The strongest base has the largest K_b but the smallest pK_b. This reversal is one of the most common errors in questions involving pK values.

The relationship between K_a, K_b and K_w

An important connection appears when an acid and base form a conjugate pair.

Consider the weak base B:

B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)

Its base dissociation constant is:

K_b = [BH⁺][OH⁻] / [B]

BH⁺ is the conjugate acid of B. Its acid dissociation can be represented as:

BH⁺(aq) + H₂O(l) ⇌ B(aq) + H₃O⁺(aq)

Hence:

K_a = [B][H₃O⁺] / [BH⁺]

Multiplying K_a by K_b gives:

K_aK_b = ([B][H₃O⁺] / [BH⁺])([BH⁺][OH⁻] / [B])

[B] and [BH⁺] cancel:

K_aK_b = [H₃O⁺][OH⁻]

Since:

K_w = [H₃O⁺][OH⁻]

we obtain:

K_aK_b = K_w

K_aK_b = K_w applies only to a conjugate acid–base pair.

The acid and base must differ by exactly one H⁺. You cannot combine the K_a of one acid with the K_b of an unrelated base.

The pK form of the relationship

Taking negative logarithms gives:

pK_a + pK_b = pK_w

At 25 °C:

K_w = 1.00 × 10⁻¹⁴

pK_w = 14.00

Therefore:

pK_a + pK_b = 14.00

What does this tell us about conjugate strength?

Suppose an acid has a very large K_a. Since K_aK_b must equal K_w, the K_b of its conjugate base must be small.

This leads to an important general pattern:

Stronger acid → weaker conjugate base.

Stronger base → weaker conjugate acid.

Worked example: finding K_b from pK_a

The conjugate acid of a weak base has pK_a = 9.25 at 25 °C. Find the pK_b and K_b of the base.

pK_b = 14.00 − 9.25 = 4.75

K_b = 10⁻⁴·⁷⁵

K_b = 1.78 × 10⁻⁵

Answer: pK_b = 4.75 and K_b = 1.78 × 10⁻⁵.

Calculating the pH of a weak base

A weak-base pH calculation must take the equilibrium into account. We cannot assume that every mole of weak base produces one mole of OH⁻ because only a fraction of the base reacts with water.

Consider 0.100 mol dm⁻³ NH₃ with:

K_b = 1.8 × 10⁻⁵

The equilibrium is:

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)

Step 1: represent the equilibrium concentrations

Suppose x mol dm⁻³ of NH₃ reacts. For every mole of NH₃ that reacts, one mole of NH₄⁺ and one mole of OH⁻ are formed.

	NH₃ / mol dm⁻³	NH₄⁺ / mol dm⁻³	OH⁻ / mol dm⁻³
Initial	0.100	0	0
Change	−x	+x	+x
Equilibrium	0.100 − x	x	x

Step 2: substitute into the K_b expression

K_b = [NH₄⁺][OH⁻] / [NH₃]

1.8 × 10⁻⁵ = x² / (0.100 − x)

Step 3: use the small-change approximation when justified

Because K_b is small, we expect only a small fraction of NH₃ to react. We can therefore test the approximation:

0.100 − x ≈ 0.100

The equation becomes:

1.8 × 10⁻⁵ ≈ x² / 0.100

x² = 1.8 × 10⁻⁶

x = 1.34 × 10⁻³ mol dm⁻³

Since x represents the equilibrium concentration of OH⁻:

[OH⁻] = 1.34 × 10⁻³ mol dm⁻³

The value of x is only about 1.34% of 0.100 mol dm⁻³, so neglecting x in 0.100 − x has only a small effect here.

Write C − x before deciding whether it can be approximated as C.

The equilibrium concentration of the base is really C − x. The approximation C − x ≈ C should be made only when x is sufficiently small compared with C. Otherwise, retain C − x and solve the full equation.

Step 4: convert [OH⁻] into pOH

pOH = −log₁₀[OH⁻]

pOH = −log₁₀(1.34 × 10⁻³)

pOH = 2.87

Step 5: convert pOH into pH

At 25 °C:

pH + pOH = 14.00

pH = 14.00 − 2.87 = 11.13

Answer: pH = 11.13.

Weak acid: K_a usually gives [H⁺] directly, followed by pH.

Weak base: K_b usually gives [OH⁻] directly, followed by pOH and then pH.

Calculating K_b from the pH of a weak base

A question can also work in the opposite direction. If the pH and initial concentration of a weak base are known, the equilibrium concentrations can be reconstructed and used to calculate K_b.

Suppose a 0.100 mol dm⁻³ solution of a weak base B has pH 11.80 at 25 °C.

Step 1: convert pH into pOH

pOH = 14.00 − 11.80 = 2.20

Step 2: calculate [OH⁻]

[OH⁻] = 10⁻²·²⁰

[OH⁻] = 6.31 × 10⁻³ mol dm⁻³

Step 3: use the equilibrium stoichiometry

B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)

BH⁺ and OH⁻ are produced in a 1:1 ratio, so:

[BH⁺] = 6.31 × 10⁻³ mol dm⁻³

The amount of B remaining is:

[B] = 0.100 − 0.00631

[B] = 0.0937 mol dm⁻³

Step 4: substitute the equilibrium concentrations

K_b = [BH⁺][OH⁻] / [B]

K_b = (6.31 × 10⁻³)(6.31 × 10⁻³) / 0.0937

K_b = 4.25 × 10⁻⁴

If pK_b is required:

pK_b = −log₁₀(4.25 × 10⁻⁴)

pK_b = 3.37

Answer: K_b = 4.25 × 10⁻⁴ and pK_b = 3.37.

Use equilibrium concentrations in the K_b expression.

The initial concentration of B was 0.100 mol dm⁻³, but some B reacted. Its equilibrium concentration is therefore 0.0937 mol dm⁻³, not 0.100 mol dm⁻³.

Choosing the correct calculation route

K_b questions often look different because different quantities are provided, but the underlying relationships are the same. Identifying the starting information helps you choose the shortest route.

Information given	Quantity required	Useful route
K_b and initial base concentration	pH	K_b → [OH⁻] → pOH → pH
pK_b and initial base concentration	pH	pK_b → K_b → [OH⁻] → pOH → pH
pH and initial base concentration	K_b	pH → pOH → [OH⁻] → equilibrium concentrations → K_b
pK_a of the conjugate acid	pK_b	pK_b = pK_w − pK_a
K_a of the conjugate acid	K_b	K_b = K_w / K_a
Several K_b or pK_b values	Relative base strength	Larger K_b or smaller pK_b means stronger base

A reliable weak-base pH sequence

Write the equilibrium between the weak base and water.
Write the K_b expression.
Determine the equilibrium concentrations.
Solve for [OH⁻].
Calculate pOH.
Convert pOH to pH using pH + pOH = pK_w.

Common mistakes with K_b and pK_b

Common mistake	Why it is wrong	Correct approach
Treating a weak base as the same thing as a dilute base	Strength and concentration describe different properties	Use strength for extent of proton acceptance and concentration for amount per unit volume
Saying that NH₃ simply dissociates into NH₄⁺ and OH⁻	NH₃ must accept H⁺ from H₂O	Write NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
Including [H₂O] in the K_b expression	Water is the solvent and its concentration is effectively constant	Use K_b = [BH⁺][OH⁻] / [B]
Using initial concentrations directly in K_b	K_b is defined using equilibrium concentrations	Determine the concentration changes before substitution
Saying that a smaller K_b means a stronger base	A larger K_b means greater proton acceptance	Larger K_b = stronger base
Saying that a larger pK_b means a stronger base	The negative logarithm reverses the comparison	Smaller pK_b = stronger base
Using K_aK_b = K_w for unrelated species	The relationship applies to a conjugate acid–base pair	Check that the two species differ by exactly one H⁺
Calculating pH directly using −log[OH⁻]	−log[OH⁻] gives pOH	Calculate pOH first and then use pH + pOH = pK_w
Assuming pH + pOH is always 14.00	14.00 is pK_w specifically at 25 °C	Use the general relationship pH + pOH = pK_w
Automatically replacing C − x with C	The amount reacting may not always be negligible	Write C − x first and check whether the approximation is justified

K_b and pK_b checklist

Write the equilibrium

Identify the weak base B.
Write B + H₂O ⇌ BH⁺ + OH⁻.
Recognise B and BH⁺ as a conjugate pair.
Remember that a weak base reacts only partially.

Use K_b

Write K_b = [BH⁺][OH⁻] / [B].
Do not include liquid water.
Use equilibrium concentrations.
Larger K_b means stronger base.

Use pK_b

pK_b = −log₁₀K_b.
K_b = 10^−pK_b.
Smaller pK_b means stronger base.
For conjugates, pK_a + pK_b = pK_w.

Calculate weak-base pH

Use K_b to determine [OH⁻].
Check any small-change approximation.
Convert [OH⁻] into pOH.
Use pH + pOH = pK_w.

Mastery Quiz

K_b and pK_b Mastery Quiz

Now apply the chapter to equilibrium expressions, conjugate pairs and weak-base calculations. These questions focus on reasoning rather than simple definition recall.

10 application-focused questions
Includes quantitative and conceptual reasoning
Every answer option is explained

Three weak bases have K_b values of 1.0 × 10⁻⁵, 5.0 × 10⁻⁴ and 2.0 × 10⁻⁷ at the same temperature. Which is the strongest?

The base with K_b = 1.0 × 10⁻⁵

This base is stronger than the one with K_b = 2.0 × 10⁻⁷, but it does not have the largest K_b.

The base with K_b = 5.0 × 10⁻⁴

Correct. The largest K_b corresponds to the greatest extent of proton acceptance and therefore the strongest base.

The base with K_b = 2.0 × 10⁻⁷

This has the smallest K_b, so it is the weakest of the three bases.

All three are equally strong because K_b values are all less than 1

Most weak bases have K_b values below 1, but their relative magnitudes still distinguish their strengths.

Bases X, Y and Z have pK_b values of 5.2, 3.8 and 8.1 respectively. Which order shows decreasing base strength?

Z > X > Y

This ranks the bases in order of increasing strength because Z has the largest pK_b.

X > Y > Z

Y is stronger than X because Y has the smaller pK_b.

Y > X > Z

Correct. Smaller pK_b means stronger base, so 3.8 corresponds to the strongest and 8.1 to the weakest.

X > Z > Y

Y has the smallest pK_b and therefore cannot be the weakest base.

Methylamine reacts with water according to CH₃NH₂(aq) + H₂O(l) ⇌ CH₃NH₃⁺(aq) + OH⁻(aq). Which is the correct K_b expression?

[CH₃NH₃⁺][OH⁻] / [CH₃NH₂]

Correct. The products appear in the numerator and the weak base in the denominator. Liquid water is omitted because its concentration is effectively constant.

[CH₃NH₂] / ([CH₃NH₃⁺][OH⁻])

This is the reciprocal of the required equilibrium expression.

[CH₃NH₃⁺][OH⁻] / ([CH₃NH₂][H₂O])

Water is the solvent and is not written explicitly in the K_b expression used for aqueous weak bases.

[CH₃NH₃⁺] / [CH₃NH₂]

The OH⁻ concentration must also appear because OH⁻ is one of the equilibrium products.

The conjugate acid of a base has pK_a = 10.64 at 25 °C. What is the pK_b of the base?

10.64

The conjugate acid and base do not generally have equal pK values.

24.64

At 25 °C the two values add to 14.00; they are not added to 14.00.

4.64

Subtracting 10 from the pK_a has no basis in the conjugate-pair relationship.

3.36

Correct. At 25 °C, pK_a + pK_b = 14.00, so pK_b = 14.00 − 10.64 = 3.36.

A 0.200 mol dm⁻³ solution of a weak base has K_b = 4.0 × 10⁻⁶. Using the small-change approximation, what is its approximate pH at 25 °C?

3.05

3.05 is approximately the pOH, not the pH. A final conversion from pOH to pH is required.

10.95

Correct. [OH⁻] ≈ √(K_bC) = √[(4.0 × 10⁻⁶)(0.200)] = 8.94 × 10⁻⁴ mol dm⁻³. pOH ≈ 3.05, so pH ≈ 14.00 − 3.05 = 10.95.

8.94

8.94 × 10⁻⁴ is related to the hydroxide concentration, not the numerical pH.

13.30

This would correspond to a much larger hydroxide concentration than is produced by this weak base.

A weak-base solution has pH 11.20 at 25 °C. What is [OH⁻]?

6.31 × 10⁻¹² mol dm⁻³

This corresponds to 10⁻¹¹·² and treats the pH directly as though it were pOH.

1.58 × 10⁻¹¹ mol dm⁻³

An alkaline solution should have an OH⁻ concentration much greater than 10⁻⁷ mol dm⁻³ at 25 °C.

1.58 × 10⁻³ mol dm⁻³

Correct. pOH = 14.00 − 11.20 = 2.80, so [OH⁻] = 10⁻²·⁸⁰ = 1.58 × 10⁻³ mol dm⁻³.

6.31 × 10⁻³ mol dm⁻³

6.31 × 10⁻³ corresponds to pOH 2.20 rather than pOH 2.80.

A 0.0500 mol dm⁻³ solution of weak base B has pH 11.00 at 25 °C. For B + H₂O ⇌ BH⁺ + OH⁻, what is K_b?

Approximately 2.04 × 10⁻⁵

Correct. pOH = 3.00, so [OH⁻] = [BH⁺] = 1.00 × 10⁻³ mol dm⁻³. [B] = 0.0500 − 0.00100 = 0.0490 mol dm⁻³. Hence K_b = (1.00 × 10⁻³)² / 0.0490 = 2.04 × 10⁻⁵.

2.00 × 10⁻²

This does not result from substituting the equilibrium concentrations into K_b = [BH⁺][OH⁻]/[B].

1.00 × 10⁻⁶

This is only the product [BH⁺][OH⁻]. It must still be divided by the equilibrium concentration of B.

4.90 × 10⁻²

0.0490 mol dm⁻³ is approximately the equilibrium concentration of B, not the value of K_b.

A weak-base calculation is carried out at a temperature other than 25 °C. Which statement is always valid?

pH + pOH = 14.00 at every temperature

14.00 is the value of pK_w at 25 °C. K_w, and hence pK_w, changes with temperature.

pH + pOH = pK_w

Correct. This is the general relationship. At 25 °C, pK_w happens to be 14.00.

K_w = 1.00 × 10⁻¹⁴ at every temperature

K_w is an equilibrium constant and therefore changes with temperature.

The neutral pH must always be exactly 7.00

Neutrality means [H⁺] = [OH⁻]. The numerical pH of neutrality depends on the value of K_w at that temperature.

Acid HA has a larger K_a than acid HB at the same temperature. What can be concluded about their conjugate bases A⁻ and B⁻?

A⁻ is the stronger base because HA is the stronger acid

A stronger acid has a weaker conjugate base, not a stronger one.

A⁻ and B⁻ must have equal K_b values

Different K_a values imply different conjugate-base K_b values because K_aK_b = K_w.

A⁻ is the weaker base

Correct. HA has the larger K_a, so its conjugate base A⁻ must have the smaller K_b. Stronger acid therefore means weaker conjugate base.

No relationship exists between acid strength and conjugate-base strength

The strengths are directly linked through K_aK_b = K_w.

Solution P contains a weak base with a smaller K_b than the base in solution Q, yet solution P has a higher pH. Which explanation is possible?

A smaller K_b actually means a stronger base

The opposite is true. A smaller K_b corresponds to a weaker base.

Solution P may contain a much higher concentration of its weaker base

Correct. pH depends on both base strength and concentration. A sufficiently concentrated solution of a weaker base can have a higher pH than a dilute solution of a stronger base.

K_b cannot be used to compare base strength

K_b does compare intrinsic weak-base strength. The issue is that solution pH also depends on concentration.

The pH proves that the K_b values must have been reported incorrectly

There is no contradiction. Solutions of different concentrations can have pH values that do not follow the same order as their K_b values.

Frequently asked questions

Questions students often ask about K_b and pK_b

These questions connect the equilibrium expressions to the strength comparisons and calculations that commonly cause confusion in A-Level Chemistry.

Why does a larger K_b mean a stronger base?

For B + H₂O ⇌ BH⁺ + OH⁻, a larger K_b means that the equilibrium contains more products relative to unreacted B. The base has therefore accepted H⁺ to a greater extent and is stronger.

Why does a smaller pK_b mean a stronger base?

pK_b = −log₁₀K_b. The negative sign reverses the direction of comparison. A large K_b therefore corresponds to a small pK_b.

Does NH₃ really dissociate in water?

Not in the same sense that an ionic substance such as NaOH separates into ions. NH₃ is a molecular base and reacts with water by accepting H⁺: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻.

The name base dissociation constant is conventional, but the chemical process should be understood as proton transfer.

Why is water absent from the K_b expression?

Water is the solvent and is present in a very large excess. Its concentration changes negligibly during the weak-base reaction and is therefore incorporated into the value of K_b.

Can K_aK_b = K_w be used for any acid and base?

No. K_a and K_b must belong to a conjugate acid–base pair. The two species must differ by exactly one H⁺.

For example, NH₄⁺ and NH₃ form such a pair, so the K_a of NH₄⁺ and the K_b of NH₃ are related through K_w.

Why do weak-base calculations use pOH before pH?

The K_b equilibrium directly produces OH⁻, so the concentration obtained from K_b is usually [OH⁻]. Taking −log₁₀[OH⁻] gives pOH, which is then converted to pH using pH + pOH = pK_w.

Is pH = 14.00 − pOH always correct?

It is correct at 25 °C because pK_w = 14.00 at that temperature. The general relationship is:

pH + pOH = pK_w.

If the temperature changes, K_w and pK_w change as well.

When can I use [OH⁻] ≈ √(K_bC)?

This shortcut comes from assuming that only a small fraction of a weak base reacts, so C − x ≈ C. Under that approximation, K_b ≈ x²/C and therefore x ≈ √(K_bC).

You should understand where the expression comes from rather than use it automatically. If x is not small compared with C, retain C − x and solve the full equilibrium equation.

Does the stronger base always give the solution with the higher pH?

Not necessarily. K_b describes the intrinsic strength of the base, while solution pH depends on both strength and concentration.

A concentrated solution of a weaker base can therefore have a higher pH than a very dilute solution of a stronger base.

K_b turns the qualitative idea of a weak base into a measurable equilibrium property. A larger K_b means a stronger base, while the logarithmic pK_b scale runs in the opposite direction. By linking K_b to K_a through K_w and treating weak-base pH calculations as equilibrium problems, the same framework can be used to compare bases, analyse conjugate pairs and calculate the composition and pH of their solutions.

What does the strength of a base mean?

Strong bases and weak bases

The base dissociation constant, Kb

Why is H₂O not included?

Why is it called a dissociation constant?

What does the magnitude of Kb tell us?

Example: comparing three bases

The use of pKb

The direction of the pKb scale

Worked example: converting Kb to pKb

The relationship between Ka, Kb and Kw

The pK form of the relationship

What does this tell us about conjugate strength?

Worked example: finding Kb from pKa

Calculating the pH of a weak base

Step 1: represent the equilibrium concentrations

Step 2: substitute into the Kb expression

Step 3: use the small-change approximation when justified

Step 4: convert [OH⁻] into pOH

Step 5: convert pOH into pH

Calculating Kb from the pH of a weak base

Step 1: convert pH into pOH

Step 2: calculate [OH⁻]

Step 3: use the equilibrium stoichiometry

Step 4: substitute the equilibrium concentrations

Choosing the correct calculation route

A reliable weak-base pH sequence

Common mistakes with Kb and pKb

Kb and pKb checklist

Write the equilibrium

Use Kb

Use pKb

Calculate weak-base pH

Questions students often ask about Kb and pKb

The base dissociation constant, K_b

What does the magnitude of K_b tell us?

The use of pK_b

The direction of the pK_b scale

Worked example: converting K_b to pK_b

The relationship between K_a, K_b and K_w

Worked example: finding K_b from pK_a

Step 2: substitute into the K_b expression

Calculating K_b from the pH of a weak base

Common mistakes with K_b and pK_b

K_b and pK_b checklist

Use K_b

Use pK_b

Questions students often ask about K_b and pK_b