pH of Weak Bases Using Kb

Weak bases are the mirror image of weak acids. Instead of a molecule partly losing a proton, a weak base partly gains one, taking it from water and leaving hydroxide ions behind. The calculation has exactly the same shape as for a weak acid, with two twists: you work out [OH⁻] rather than [H⁺], and you finish by converting pOH to pH.

If you’ve already worked through pH of weak acids, you’ll recognise every step.

The setup

A weak base B reacts with water:

B + H₂O ⇌ BH⁺ + OH⁻

For ammonia, that’s NH₃ + H₂O ⇌ NH₄⁺ + OH⁻.

The base dissociation constant is:

Kb = [BH⁺][OH⁻] ÷ [B]

Water doesn’t appear in the expression because its concentration is effectively constant. The larger Kb, the stronger the base.

Weak base	Formula	Kb at 25 °C
Methylamine	CH₃NH₂	4.4 × 10⁻⁴
Ammonia	NH₃	1.8 × 10⁻⁵
Hydrazine	N₂H₄	1.3 × 10⁻⁶
Pyridine	C₅H₅N	1.7 × 10⁻⁹
Aniline	C₆H₅NH₂	4.3 × 10⁻¹⁰

The ICE table

Starting concentration c, and let x react:

	B	BH⁺	OH⁻
Initial	c	0	0
Change	−x	+x	+x
Equilibrium	c − x	x	x

So Kb = x² ÷ (c − x), where x = [OH⁻].

When x is small compared with c:

[OH⁻] ≈ √(Kb × c)

Then pOH = −log[OH⁻] and pH = 14.00 − pOH (at 25 °C). The same 5% check applies as for weak acids.

Worked example 1: ammonia

Calculate the pH of 0.20 mol/dm³ ammonia. Kb = 1.8 × 10⁻⁵ mol/dm³.

• [OH⁻] ≈ √(1.8 × 10⁻⁵ × 0.20) = √(3.6 × 10⁻⁶) = 1.90 × 10⁻³ mol/dm³
• pOH = −log(1.90 × 10⁻³) = 2.72
• pH = 14.00 − 2.72 = 11.28
• Check: 1.90 × 10⁻³ ÷ 0.20 = 0.95%. Approximation valid.

Household ammonia cleaners are more concentrated than this, typically around 1–3 mol/dm³, giving a pH of about 11.6–11.9.

Worked example 2: pyridine

Calculate the pH of 0.10 mol/dm³ pyridine. Kb = 1.7 × 10⁻⁹.

• [OH⁻] ≈ √(1.7 × 10⁻⁹ × 0.10) = √(1.7 × 10⁻¹⁰) = 1.30 × 10⁻⁵ mol/dm³
• pOH = 4.89
• pH = 9.11

Pyridine is a much weaker base than ammonia. The nitrogen’s lone pair sits in an sp² orbital, held more tightly than the sp³ lone pair in ammonia, so it’s less available to grab a proton.

Worked example 3: when you need the quadratic (methylamine)

Calculate the pH of 0.10 mol/dm³ methylamine. Kb = 4.4 × 10⁻⁴.

Shortcut:

• [OH⁻] ≈ √(4.4 × 10⁻⁵) = 6.63 × 10⁻³
• Check: 6.63 × 10⁻³ ÷ 0.10 = 6.6%. Over 5%, so solve properly.

Full equation:

x² + (4.4 × 10⁻⁴)x − 4.4 × 10⁻⁵ = 0

x = [−4.4 × 10⁻⁴ + √((4.4 × 10⁻⁴)² + 1.76 × 10⁻⁴)] ÷ 2 = 6.42 × 10⁻³ mol/dm³

• pOH = −log(6.42 × 10⁻³) = 2.19
• pH = 11.81

(The shortcut gives 11.82. Close, but not exact.)

Methylamine is a stronger base than ammonia because the methyl group pushes electron density onto the nitrogen, making its lone pair more available.

Worked example 4: finding Kb from pH

A 0.050 mol/dm³ solution of a weak base has pH 10.50. Calculate Kb.

• pOH = 14.00 − 10.50 = 3.50
• [OH⁻] = 10⁻³·⁵⁰ = 3.16 × 10⁻⁴ mol/dm³, and [BH⁺] is the same
• [B] = 0.050 − 0.000316 = 0.0497 mol/dm³
• Kb = (3.16 × 10⁻⁴)² ÷ 0.0497 = 2.0 × 10⁻⁶

Anions are weak bases too

The conjugate base of any weak acid is a weak base. Sodium ethanoate dissolves to give ethanoate ions, which take protons from water:

CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻

You’re rarely given Kb for anions directly. Instead, use the relationship between conjugate pairs:

Ka × Kb = Kw = 1.0 × 10⁻¹⁴ (at 25 °C)

Our article on the Ka × Kb = Kw relationship explains why this works.

Worked example 5: sodium ethanoate

Calculate the pH of 0.10 mol/dm³ sodium ethanoate. Ka of ethanoic acid = 1.8 × 10⁻⁵.

• Kb of ethanoate = 1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰
• [OH⁻] ≈ √(5.6 × 10⁻¹⁰ × 0.10) = √(5.6 × 10⁻¹¹) = 7.5 × 10⁻⁶ mol/dm³
• pOH = 5.13
• pH = 8.87

This is why the end point of a titration between ethanoic acid and sodium hydroxide is above pH 7. At the equivalence point the flask contains sodium ethanoate, and ethanoate is a weak base. That in turn is why phenolphthalein, which changes colour between pH 8.2 and 10, is the right indicator for that titration.

Worked example 6: sodium carbonate

Estimate the pH of 0.10 mol/dm³ sodium carbonate. For HCO₃⁻, Ka = 4.7 × 10⁻¹¹.

Carbonate is the conjugate base of HCO₃⁻:

• Kb = 1.0 × 10⁻¹⁴ ÷ 4.7 × 10⁻¹¹ = 2.1 × 10⁻⁴
• [OH⁻] ≈ √(2.1 × 10⁻⁴ × 0.10) = 4.6 × 10⁻³ mol/dm³ (4.6%, just under the 5% limit)
• pOH = 2.34
• pH ≈ 11.66

Washing soda solutions really are this alkaline, which is why they cut through grease and why you should wear gloves when using them.

Comparing weak and strong bases

0.10 mol/dm³ solution	[OH⁻] (mol/dm³)	pH
NaOH (strong)	0.10	13.00
Methylamine	6.4 × 10⁻³	11.81
Ammonia	1.3 × 10⁻³	11.13
Sodium ethanoate	7.5 × 10⁻⁶	8.87
Pyridine	1.3 × 10⁻⁵	9.11

Concentration alone tells you almost nothing about pH for weak bases. You always need Kb.

Why weak bases matter outside the exam hall

Weak bases are everywhere in biology and medicine, and their weakness is usually the point. Most drug molecules that contain nitrogen are weak bases: antihistamines, local anaesthetics like lidocaine, and many antidepressants. Whether such a molecule is protonated (charged) or neutral depends on how the pH of its surroundings compares with the pKa of its conjugate acid. The neutral form slips through fatty cell membranes; the charged form dissolves well in watery blood plasma. Drug designers deliberately tune the strength of the basic nitrogen so the molecule can do both.

Your own body relies on weak bases too. The amino acid histidine has a side chain with a conjugate acid pKa close to 6, which lets it pick up or release a proton at physiological pH. That makes histidine a favourite “proton shuttle” in the active sites of enzymes. Ammonia made during protein breakdown is converted in the liver to urea, a much weaker base, so it can be carried safely in the blood and excreted.

In the lab, weak bases such as ammonia are preferred when you want a gently alkaline solution, for example to precipitate metal hydroxides selectively without dissolving amphoteric ones like aluminium hydroxide.

Practice questions

1. Calculate the pH of 0.050 mol/dm³ ammonia (Kb = 1.8 × 10⁻⁵).
2. Calculate the pH of 0.20 mol/dm³ sodium methanoate (Ka of methanoic acid = 1.8 × 10⁻⁴).
3. A 0.10 mol/dm³ solution of a base has pH 11.00. Find Kb.

Answers:

1. [OH⁻] = √(9.0 × 10⁻⁷) = 9.5 × 10⁻⁴; pOH = 3.02; pH = 10.98.
2. Kb = 1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁴ = 5.6 × 10⁻¹¹; [OH⁻] = √(1.1 × 10⁻¹¹) = 3.3 × 10⁻⁶; pOH = 5.48; pH = 8.52.
3. pOH = 3.00, so [OH⁻] = 1.0 × 10⁻³; Kb = (1.0 × 10⁻³)² ÷ (0.10 − 0.001) = 1.0 × 10⁻⁵.

Common mistakes

• Stopping at pOH. The question almost always wants pH.
• Using Ka instead of Kb. For anions, convert with Kb = Kw ÷ Ka.
• Using 14.00 at non-standard temperatures. pKw changes with temperature.
• Treating the ammonium ion as a base. NH₄⁺ is the conjugate acid of ammonia; its solutions are slightly acidic.

Key takeaways

• Weak bases take protons from water: B + H₂O ⇌ BH⁺ + OH⁻.
• Use an ICE table: Kb = x² ÷ (c − x), and usually [OH⁻] ≈ √(Kb × c).
• Convert: pOH = −log[OH⁻], then pH = 14.00 − pOH at 25 °C.
• Anions of weak acids are weak bases; find their Kb from Kb = Kw ÷ Ka.
• Check your working with the pH calculator.