Doc Brown's Advanced A Level Chemistry Revision Notes

Theoretical–Physical Advanced Level Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 5.6

5.6 Definition of a weak base, theory and examples of Kb, pKb, Kw weak base calculations

What is a weak base? What is the Kb of a weak base – base dissociation constant? How do we calculate the pH of a weak base–alkali solution? What is the pKb of a base? Why do we need to use Kw 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 Ka of a weak base? All of these terms are defined and explained below with suitable worked out examples. What is a weak acid? How do we write equilibrium expressions to show the dissociation–ionisation of a weak acids? What is a weak acid's pKa? How do we calculate the pH of a solution of a weak acid? How do we calculate the Ka of a weak acid?

Chemical Equilibrium Notes Parts 5 & 6 Index


5.6 Definition, examples and pH, Kb, pKb and Kw calculations of weak bases

  • 5.6.1 Definition and examples of WEAK BASES

    • A weak base is only weakly or partially ionised in water e.g.

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

      • NH3(aq) + H2O(l) (c) doc b 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)]

        –––––––––––––––––––––––––––––

            [NH3(aq)]

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

      • B: + H2O(l) (c) doc b BH+(aq) + OH–(aq)

      • Note [H2O(l)] is omitted from the Kb expression, i.e. incorporated into Kb in a similar manner to that for weak acid equilibrium expression above.

      • pKb = –log(Kb/mol dm–3)

      • The bigger Kb or the smaller the pKb value, the stronger the base.  more on comparison with SB

      • note, sometimes the pKb isn't quoted, but the pKa for the conjugate acid is!

      • i.e. pKa for BH+(aq) (c) doc b B:(aq) + H+(aq) 

      • In which case it is useful to know that pKa + pkb = 14 or pKb = 14 – pKa

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

      • NH4+(aq) + H2O(l) (c) doc b NH3(aq) + H3O+(aq)

      • Ka =

        [NH3(aq)] [H3O+(aq)]

        –––––––––––––––––––––

            [NH4+(aq)]

      • Note that: Ka–conj. acid x Kb–base = Kw and pKa + pKb = pKw, check it out for yourself.

      • * Apart from the equation, this section is NOT needed by UK AS–A2 syllabus as far as I can tell?

  • 5.6.2 Other examples of weak bases

    • Aliphatic amines

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

      • R–NH2(aq) + H2O(l) (c) doc b R–NH3+(aq) + OH–(aq) (R = alkyl = CH3, CH3CH2 etc.)

  • 5.6.3 Comparison of weak and strong bases

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

      • e.g. ammonia: NH3(aq) + H2O(l) (c) doc b NH4+(aq) + OH–(aq)

      • a few % ionised because Kb = 1.8 x 10–5 mol dm–3 , pKb = 4.8

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

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

      • virtually 100% ionised because Kb is very large, pKb very negative.

  • 5.6.4 Weak base calculations – calculating the pH of a weak base

    • Calculation example 5.6.4a

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

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

      • Kb = [NH4+(aq)] [OH–(aq)]/[NH3(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. [NH3(aq)]initial base = [NH3(aq)]equilibrium since the weak base is only a few % ionised.

      • So we can then write:

      • Kb = [OH–(aq)]2/[NH3(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.

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

      • [H+(aq)] = Kw/[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 = pKw – pH = 14 – 11.4 = 2.6

    • Calculation example 5.6.4b

      • 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 Kb and pKb.

      • [H+(aq)] = 10–pH = 10–9.53.16 x 10–10 mol dm–3

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

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

      • so, using the simplified expression

      • Kb = [OH–(aq)]2/[B(aq)] = (3.16 x 10–5)2 / 0.50 = 2.00 x 10–9 mol dm–3

      • pKb = –log(2.00 x 10–9) = 8.70

    • Calculation example 5.6.4c

      • The pKb value for ethylamine is 3.27

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

        • CH3CH2NH2(aq) + H2O(l) (c) doc b CH3CH2NH3+(aq) + OH–(aq)

        • Kb =

          [CH3CH2NH3+(aq)] [OH–(aq)]

          ––––––––––––––––––––––––––––

               [CH3CH2NH2(aq)]

      • (b) Calculate Kb.

        • Kb = 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

        • Kw = [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 5.6.4d

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

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

      • The pKa 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 pKa = 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) (c) doc b C6H5NH2(aq) + H3O+(aq)

          • conjugate acid + water (c) doc b 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 Kb for the original phenylamine base and use the information to justify the classification of phenylamine as a very weak base.

        • pKa–conj.acid + pKb–orig.base = pKw = 14

        • pKb = 14 – 4.62 = 9.38

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

          • Kb = 10–9.38 = 4.17 x 10–10 mol dm–3

        • so in terms of the equilibrium

        • C6H5NH2(aq) + H2O(l) (c) doc b C6H5NH3+(aq) + OH–(aq)

          • there isn't very much on the RHS, but the solution will be slightly alkaline.

      • Note, that if given a pKa for the conjugate acid of a weak base, its easy to calculate the pKb, then Kb 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, it is essentially a hydrolysed salt situation, which shows that some 'neutral' salts can be quite acid in aqueous media!

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

        • pka = 4.62, so Ka = 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!

  • –


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