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^-]

   [H₂O]

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

                        K_w = [H⁺] x [OH^-]

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

OO

Calculate the pH of 0.100 mol dm^-3 NaOH

[OH^-]    = 0.100

[H⁺]      = K_w/[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:

                        K_a =     [H⁺][A^-]

  [HA]

K_a is a measure of the extent to which a weak acid is ionised. The larger the value of K_a 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:

                        K_a = [H⁺]²/[HA]

            Re-arranging gives:

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

Calculate the pH of 0.100 mol dm^-3 chloroethanoic acid given that K_a = 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

Source             http://www.chemsheets.co.uk/

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 K_w = [H⁺]² and [H⁺] = ÖK_w

Example

At 318K, the value of K_w 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⁺]      = ÖK_w

            = 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)         pK_a

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

Calculate the pK_a of chloroethanoic acid

From the previous example, K_a          = 1.38 x 10^-3

                                          pK_a       = -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