Sunday, 27 November 2011

Acids, bases and buffers

1)      Acids and Bases

  • The Bronsted-Lowry theory says that acids are proton donors (H+ donors). Bases are proton acceptors.
  • Strong acids and bases are fully dissociated (or ionised)
  • Weak acids and bases are partially dissociated
  • Conjugate acid-base pairs are two species differing by H+
  • For any weak acid, HA:

HA(aq) + H2O(l) ¾     H3O+(aq) +     A-(aq)
Acid          base       conjugate acid    conjugate base
  • In this reaction, A- is the conjugate base of the acid HA because it is formed by loss of H+ from HA
  • In this example, H3O+ is the conjugate acid of the base H2O because it is formed by the gain of H+ by H2O
  • For a weak base, such as NH3

NH3(aq) + H2O(l) ¾   NH4+(aq) +     OH-(aq)
Base         acid       conjugate acid     conjugate base

2)      Quantifying acid and base strength

  • pH is a number that shows the strength of an acid or base
  • pH = -log[H+] and [H+] = 10-pH
  • pH is always given to 2 decimal places
  • [H+] deals with negative powers over a very wide range whereas the pH scale makes the numbers more manageable

a)    pH of a strong acid

Calculate the pH of 0.100 mol dm-3 HCl         
[H+]      = 0.100
pH        = -log[0.100]                                        
                  = 1.00                                                              

b)    pH of a strong base

To calculate the pH of a strong base, we need to take advantage of the fact that water is very slightly dissociated.

H2O(l) ¾ H+(aq) + OH-(aq)

The equilibrium constant for this reaction is:

                        K =      [H+][OH-]

As [H2O] is little changed, we define a constant Kw, which is known as the ionic product of water

                        Kw = [H+] x [OH-]

At 298K (25°C) Kw has the value of 10-14 mol2 dm-6


Calculate the pH of 0.100 mol dm-3 NaOH

[OH-]    = 0.100
[H+]      = Kw/[OH-]
                  = 10-14/0.100
            = 10-13  
pH        = -log[10-13]
            = 13.00

c)    pH of a weak acid

The weak acid HA dissociates as follows.

HA ¾ H+ + A-

The equilibrium constant for the weak acid is:

                        Ka =     [H+][A-]

Ka is a measure of the extent to which a weak acid is ionised. The larger the value of Ka the more the weak acid is ionised

When one mole of HA ionises, one mole of H+ and one mole of A- are produced i.e. [H+] = [A-], so we can write:

                        Ka = [H+]2/[HA]

            Re-arranging gives:

                        [H+] = Ö(Ka x [HA])

Calculate the pH of 0.100 mol dm-3 chloroethanoic acid given that Ka = 1.38 x 10-3 mol dm-3

[H+]      = Ö(1.38 x 10-3 x 0.100)
                  = Ö(1.38 x 10-4)
                  = 0.0117
pH        = -log[0.0117]
                  = 1.93


Percentage dissociation of a weak acid is [H+]/[HA] i.e. hydrogen ion concentration/acid concentration
d)         pH of water

For pure water, [H+] = [OH-], so Kw = [H+]2 and [H+] = ÖKw


At 318K, the value of Kw is 4.02 x 10-14 mol2 dm-6. Calculate the pH of water at this temperature and explain why the water is still neutral.
[H+]      = ÖKw
            = Ö4.02 x 10-14
            = 2.01 x 10-7 mol dm-3
pH        = -log[H+]                                            
                  = 6.70

      Still neutral because [H+] = [OH-]                                                                       

Source:           AQA January 2006 paper

e)         pKa

  • pKa = -log(Ka) and Ka = 10-pKa
  • As with pH, using pKa instead of Ka makes the numbers more manageable

Calculate the pKa of chloroethanoic acid

From the previous example, Ka          = 1.38 x 10-3
                                          pKa       = -log(1.38 x 10-3)
                                                      = 2.86

3)      Acid-base titrations

a)   pH curves

  • pH can be monitored during an acid-base titration and plotted against volume of reagent
  • This produces a pH curve with a shape that depends on whether the acid and base are weak or strong

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