Advanced A level theoretical chemistry - acid-base equilibrium

5.6 Definition of a weak base, theory and examples of K_b, pK_b, K_w weak base calculations - examples explained and worked through

[Author ©  Dr Phil Brown PhD: Doc Brown's exam revision notes suitable for A level chemistry students of advanced pre–university/college advanced level theoretical–physical chemistry courses:  acid–base equilibrium revision notes on weak base theory [updated April 29th 2026 *]

5.0 Basic notes and equations on acids, bases, salts, uses of acid–base titrations

5.1 Equilibria:  Lewis and Bronsted–Lowry acid–base theories

5.2 Self–ionisation of water and pH scale

5.3 Strong acids – examples and pH calculations

5.4 Weak acids – examples & pH, K_a and pK_a calculations

5.5 Strong bases – examples and pH calculations

5.6 Weak bases – examples and pH, K_b and pK_b calculations (sub-index for this page)

5.6 Definition, examples and pH, K_b, pK_b and K_w calculations of weak bases

Sub-index for this page on the chemistry of weak bases

5.6.0 Introduction to weak bases

5.6.1 Definition and examples of WEAK BASES

5.6.2 Expressing the equilibrium of a weak base in terms of the conjugate acid

5.6.3 Examples of organic weak bases

5.6.4 Comparison of weak and strong bases

5.6.5 A variety of weak base calculations e.g. calculating the pH or K_b of a weak base

5.6.0 Introduction

What is a weak base? What is the K_b of a weak base – base dissociation constant?

How do we calculate the pH of a weak base–alkali solution?

What is the pK_b of a base? Why do we need to use K_w in weak base pH calculations?

How do we write equilibrium expressions to show the dissociation–ionisation of a weak acids?

How do we calculate the K_b of a weak base?

All of these terms are defined and explained below with suitable worked out examples.

5.6.1 Definition and examples of WEAK BASES

• A weak base is only weakly or partially ionised in water to give an alkaline solution

• A good example is ammonia solution, which is only about 2% ionised :

  • NH_3(aq) + H₂O_(l) NH₄⁺_(aq) + OH^–_(aq)

    • Ammonia is the base and the ammonium ion is its conjugate acid.

    • Water is the acid and the hydroxide ion is its conjugate base.

    • This equilibrium is sometimes referred to as a base hydrolysis.

  • The low % of ionisation gives a less alkaline solution of lower pH than for strong soluble bases (alkalis), but pH is still > 7.

  • Again, the concentration of water is considered constant in a similar manner to that for weak acid equilibrium, and to solve simple problems, the base ionisation equilibrium expression is written as:

  • Kb =	[NH4+(aq)] [OH–(aq)]
    	–––––––––––––––––––––––––      mol dm–3
    	[NH3(aq)]

  • K_b is the base ionisation/dissociation constant (mol dm^–3) for any base i.e.

  • B: + H₂O_(l) BH⁺_(aq) + OH^–_(aq)

  • Note [H₂O_(l)] is omitted from the K_b expression, i.e. it is incorporated into K_b in a similar manner to that for weak acid equilibrium expression, where an argument is presented to justify this assumption.

  • pK_b = –log(K_b/mol dm^–3)

  • The bigger K_b or the smaller the pK_b value, the stronger the base.

  • note, sometimes the pK_b isn't quoted, but the pK_a for the conjugate acid is!

  • i.e. pK_a for BH⁺_(aq) B:_(aq) + H⁺_(aq) 

  • In which case it is useful to know that pK_a + pk_b = 14 or pK_b = 14 – pK_a

  • -

5.6.2 Expressing the equilibrium of a weak base in terms of the conjugate acid

• The weak base – water interaction can be expressed in terms of the acidity of the conjugate acid e.g.

  • NH₄⁺_(aq) + H₂O_(l) NH_3(aq) + H₃O⁺_(aq)

  • Ka =	[NH3(aq)] [H3O+(aq)]
    	–––––––––––––––––––––     mol dm–3
    	[NH4+(aq)]

  • Note that: K_{a–conj. acid} x K_b–base = K_w and pK_a + pK_b = pK_w

  • Check it out for yourself.

  • This is why magnesium metal ribbon fizzes when added to a solution of the salt ammonium chloride.

  • -

5.6.3 Examples of organic weak bases

• Aliphatic amines

  • e.g. methylamine, ethylamine etc. which are quite soluble in water but only ionise by a few % like ammonia.

  • R–NH_2(aq) + H₂O_(l) R–NH₃⁺_(aq) + OH^–_(aq) (R = alkyl = CH₃, CH₃CH₂ etc.)

  • These are quite soluble in water.

  • -

• Aromatic amines

  • e.g. the very slightly soluble phenylamine gives the phenylammonium ion.

  • C₆H₅–NH_2(aq) + H₂O_(l) C₆H₅–NH₃⁺_(aq) + OH^–_(aq)

  • -

5.6.4 Comparison of weak and strong bases

• Weak bases are only partially ionised to give the hydroxide ion and corresponding cation and the K_b is small.

  • e.g. ammonia: NH_3(aq) + H₂O_(l) NH₄⁺_(aq) + OH^–_(aq)

  • a few % ionised because K_b = 1.8 x 10^–5 mol dm^–3 , pK_b = 4.8

• Strong bases are virtually ionised completely to form the hydroxide ion and corresponding cation and the K_b is large.

  • e.g. sodium hydroxide: NaOH_(s) + aq ==> Na⁺_(aq) + OH^–_(aq)

  • virtually 100% ionised because K_b is very large, pK_b very negative.

5.6.5 A variety of weak base calculations e.g. calculating the pH or K_b of a weak base

• Calculation example 5.6.5a

  • Calculate the expected hydroxide and hydrogen ion concentrations and the pH of a 0.40 mol dm^–3 solution of ammonia,

    • The K_b value for ammonia at 298K is 1.78 x 10^–5 mol dm^–3.

  • K_b = [NH₄⁺_(aq)] [OH^–_(aq)]_/[NH_3(aq)]

  • As in the case of weak acids, for simple calculations we assume

    1. [NH₄⁺_(aq)] = [OH^–_(aq)], ignoring any OH^– from water

       • (OH^– contribution from water < 1 x 10^–7 mol dm^–3)

    2. [NH_3(aq)]_{initial base} = [NH_3(aq)]_equilibrium since the weak base is only a few % ionised.

  • So we can then write:

  • K_b = [OH^–_(aq)]²_/[NH_3(aq)] = 1.78 x 10^–5 = [OH^–_(aq)]² / 0.40

  • [OH^–_(aq)] = √(0.40 x 1.78 x 10^–5) = 2.67 x 10^–3 mol dm^–3

  • In base calculations you need to use the ionic product of water expression to calculate the H⁺ ion concentration.

  • K_w = [H⁺_(aq)] [OH^–_(aq)] = 1 x 10^–14 mol² dm^–6, so

  • [H⁺_(aq)] = K_w/[OH^–_(aq)] = 1 x 10^–14_/2.67 x 10^–3 = 3.74 x 10^–12 mol dm^–3

  • pH = –log(3.74 x 10^–12) = 11.4

  • Note: pOH = pK_w – pH = 14 – 11.4 = 2.6

  • -

• Calculation example 5.6.5b

  • A 0.50 mol dm^–3 aqueous solution of a very weak base B, has a pH of 9.5.

  • Calculate the hydrogen and hydroxide ion concentrations in the solution and the value of the base dissociation constant K_b and pK_b.

  • [H⁺_(aq)] = 10^–pH = 10^–9.5 =  3.16 x 10^–10 mol dm^–3

  • K_w = [H⁺_(aq)] [OH^–_(aq)] = 1 x 10^–14 mol² dm^–6, so

  • [OH^–_(aq)] = K_w/[H⁺_(aq)] = 1 x 10^–14 / 3.16 x 10^–10 = 3.16 x 10^–5 mol dm^–3

  • so, using the simplified expression

  • K_b = [OH^–_(aq)]²_/[B_(aq)] = (3.16 x 10^–5)² / 0.50 = 2.00 x 10^–9 mol dm^–3

  • pK_b = –log(2.00 x 10^–9) = 8.70

  • -

• Calculation example 5.6.5c

  • The pK_b value for ethylamine is 3.27

  • (a) Give the ionisation equation for ethylamine in water and corresponding equilibrium expression.

    • CH₃CH₂NH_2(aq) + H₂O_(l) CH₃CH₂NH₃⁺_(aq) + OH^–_(aq)

    • Kb =	[CH3CH2NH3+(aq)] [OH–(aq)]
      	––––––––––––––––––––––––––––     mol dm–3
      	[CH3CH2NH2(aq)]

  • (b) Calculate K_b.

    • K_b = 10^–pKb = 10^–3.27 = 5.37 x 10^–4 mol dm^–3

  • (c) Calculate the pH of a 0.25 mol dm^–3 aqueous solution of ethylamine.

    • substituting in the Kb expression:

    • 5.37 x 10–4 =	[OH–(aq)]2
      	–––––––––––––––
      	0.25

    • therefore: [OH^–_(aq)] = √(5.37 x 10^–4 x 0.25) = 0.0116

    • K_w = [H⁺_(aq)] [OH^–_(aq)] = 1 x 10^–14 mol² dm^–6, so rearranging

    • [H⁺_(aq)] = 1 x 10^–14/0.0116 = 8.62 x 10^–13 mol dm^–3

    • pH = –lg(8.62 x 10^–13) = 12.1

    • -

• Calculation example 5.6.5d

  • 5.6.4d is an example of approaching weak base pH calculations from the point of view of the K_a of the conjugate acid of the weak base.

    • Remember 'pK_x' data can often be presented in two ways i.e. from the point of view of an acid/base OR its conjugate base/acid species.

  • The pK_a of the conjugate acid of the aromatic weak base phenylamine is 4.62

    • (note this is comparable to the 'acidity' of the weak acid ethanoic acid, whose pK_a = 4.76).

  • (a) Give an ionic equation to show what happens when phenylammonium chloride is dissolved in water and explain why the solution is acidic.

    • C₆H₅NH₃⁺_(aq) + H₂O_(l) C₆H₅NH_2(aq) + H₃O⁺_(aq)

      • conjugate acid + water weak base + oxonium ion

    • In aqueous media the solution becomes acidic because hydrogen ion/oxonium ions are formed, so lowering the pH by proton donation from the conjugate acid to the water molecules, which in this case act as the base.

  • (b) Calculate the value of K_b for the original phenylamine base and use the information to justify the classification of phenylamine as a very weak base.

    • pK_{a–conj.acid} + pK_{b–orig.base} = pK_w = 14

    • pK_b = 14 – 4.62 = 9.38

    • A relatively high pK_b value means a very weak base and a stronger conjugate acid (relatively low pK_a), but the 'weakness' of the base is best appreciated by students if the value of K_b is worked out.

      • K_b = 10^–9.38 = 4.17 x 10^–10 mol dm^–3

    • so in terms of the equilibrium

    • C₆H₅NH_2(aq) + H₂O_(l) C₆H₅NH₃⁺_(aq) + OH^–_(aq)

      • there isn't very much on the right-hand side, but the solution will be slightly alkaline.

  • Note, that if given a pK_a for the conjugate acid of a weak base, its easy to calculate the pK_b, then K_b and then perform pH and concentration calculations as exemplified by 5.6.4a–c,

    • but, equally, you can readily calculate the pH of a salt solution of the salt of a weak base and strong acid using a weak acid calculation (section 5.4) – in the example below (5.5.6e), it is essentially a hydrolysed salt situation, which shows that some 'neutral' salts can be quite acid in aqueous media!

  • Calculation example 5.6.5e

    • e.g. What is the pH of a 0.100 mol dm^–3 solution of phenylammonium chloride?

    • pk_a = 4.62, so K_a = 10^–4.62 = 2.40 x 10^–5 mol dm^–3

    • In general for a weak acid

    • Ka =	[H+(aq)] [A–(aq)]
      	–––––––––––––––––
      	[HA(aq)]

    • so, making the assumptions described in section 5.4,

    • 2.40 x 10–5 =	[H+(aq)]2
      	––––––––––––––––
      	0.100

    • [H⁺_(aq)]² = 2.40 x 10^–5 x 0.100

    • [H⁺_(aq)]² = (2.40 x 10^–5 x 0.100) = 2.40 x 10^–6

    • [H⁺_(aq)] = √( 2.40 x 10^–6) = 1.55 x 10^–3 mol dm^–3

    • pH = –lg(1.55 x 10^–3) = 2.81

    • so, (i) very definitely an acid solution!, and,

    • (ii) if you did a theoretical pH calculation on a 0.1 molar phenylamine solution (like 5.6.4c), you would get a pH value above 7, but not that high!

    • -

WHAT NEXT?

Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principle–rules * Part 2. K_c and K_p equilibrium expressions and calculations * Part 3. Equilibria and industrial processes * Part 4 Partition between two phases, solubility product K_sp, common ion effect, ion–exchange systems * Part 5. pH, weak–strong acid–base theory and calculations * Part 6. Salt hydrolysis, acid–base titrations–indicators, pH curves and buffers * Part 7. Redox equilibria, half–cell electrode potentials, electrolysis and electrochemical series * Part 8. Phase equilibria–vapour pressure, boiling point and intermolecular forces watch out for sub–indexes to multiple sections or pages