• Buffers are solutions which can resist changes in acidity or alkalinity
  • When a small volume of acid/alkali is added to them their pH remains at a constant

How Buffers work

  • Buffers are design to maintain the concentration of hydrogen ions and hydroxide ions
  • They are based on an equilibrium reaction which will move in the direction to remove either additional hydrogen ions or hydroxide ions

Acidic Buffers

  • Acidic buffers are made from weak acids
  • The dissociation of a weak acid is an equilibrium reaction
  • e.g.

HA(aq) ⇋ H+(aq) + A–(aq)

  • Therefore, [H+(aq)] = [A–(aq)]
  • As it is a weak acid, [H+(aq)] = [A–(aq)] are both very small as they are most undissociated

Acid Buffer: adding alkali

  • If a little alkali is add, the OH– ions react with the HA to produce water molecules and A‑:

HA(aq) + OH–(aq) ⇋ H2O(aq) + A–(aq)

  • This removes the added OH–so the pH trends to remain almost the same

Acid Buffers: adding acid

  • If H+ is added, the equilibrium will shift to the left as the H+ ions combine with the A– ions to produce undissociated HA.
  • As the [A–] is small the supply of A– soon runs out and there is no subsequent A– left to remove the added H+.
  • Therefore, the buffer has stopped
  • However, by the addition of extra A– by adding a soluble salt of HA, which fully ionises
  • This increased amount of A– allows for more H+ to be used up
  • An Acidic Buffer is made from a mixture of a weak acid and a soluble salt of that acid. It will thus maintain a pH of below 7 (acidic)
  • The function of the weak acid allows there to be a source of HA which can remove any OH–

HA(aq) + OH–(aq) à H2O(aq) + A–(aq)

  • The function of the salt allows there to be a source of A– ions which can be removed by any added H+ ions:

H+(aq) + A–(aq) à HA(aq)

  • Buffers do note ensure that there is no change in pH but rather that the addition of acid/alkali will only slightly change or the effects are reduced with the use of a buffer.
  • If it possible to saturate a buffer by there being too much acid/alkali that all of the available HA or A– is used up
  • Another method of producing a buffer is the mixture of a weak acid with its salt with the addition of an alkali such as sodium hydroxide.
  • By neutralising half of the acid the result is a buffer whose pH is equal to the pKa of the acid

At Half Neutralisation: pH = pKa

  • These buffers are useful as they are equally efficient at resisting a change in pH whether it be an acid of an alkali

Basic Buffer

  • Basic buffers are those that resist change but maintain a pH at above 7
  • They are a mixture of weak base and a salt of that base
  • A mixture of aqueous ammonia and ammonium chloride (NH4+ Cl–) act as a basic buffer
    • Aqueous ammonia removes added H+:

NH3(aq) + H+(aq) à NH4+(aq)

  • The ammonium ion, NH4+ removes added OH–:

NH4+(aq)  + OH–(aq) à NH3(aq) + H2O(l­)

Examples of Buffers

  • Buffers are used in blood storage where they are maintained at approximately pH 7.4
  • Blood must be buffered as a change in pH >0.5 can be fatal
  • Blood is buffered through several mechanisms, however most importantly:

H+(aq) + HCO3–(aq) ⇋ CO2(aq) + H2O(l­)­

  • The addition of extra H+ ions moves this equilibrium to the right, therefore the removing the additional H+
  • Adding exta OH– ions removes the H+ by reacting to form water.
  • The equilibrium moves to the left releasing more H+ions
  • Buffers are also found in detergents and shampoos as if they become too acidic or to alkaline they can damage fabrics, skin or hair

Calculations of Buffers

  • Different buffers can be made which will maintain different pH’s. when weak acid dissociates:

HA(aq) ⇋ H+(aq) + A–­(aq)

  • The expression of this being:

[H+(aq)][A–(aq)]

[HA(aq)]

  • Using this expression, the pH of the buffer can be calculated

e.g. A buffer consist of 0.1 mol dm-3 ethanoic acid and 0.1 mol dm-3 sodium ethanoate

(Ka for ethanoic acid = 1.7 x10-5, pKa = 4.77)

What is the pH of the buffer?

  • Calculate the [H+(aq)] of the equation

Ka = [H+(aq)][A–(aq)]

[HA(aq)]

  • Sodium ethanoate is fully dissociated so [A–(aq)] = 0.100 mol dm-3
  • Ethanoic acid is almost undissociateed so [HA(aq)] = 0.100 mol dm-3

1.7 x 10-5  = [H+(aq)] x 0.100

0.100

1.7 x 10-5  = [H+(aq)] and pH -log10 [H+(aq)]

pH = 4.77

  • When equal concentrations of acid and salts are used the pH of the buffer = pKa of acid used (equal the half neutralisation point)
  • Changing the concentration of HA to A– will affect the pH of the buffer
  • By using 0.2 mol dm-3 ethanoic acid and 0.1 mol dm-3 sodium ethanoate, the pH will be 4.50

pH change when an acid is added to a buffer

Example:

  • In the above calculation, the pH was 4.77 after adding 1.00 mol dm-3 of buffer solution of ethanoic acid at concentration 0.1 mol dm-3 and sodium ethanoate at concentration 0.1 mol dm-3, ka= 1.7 x 10 -5.
  • If 10cm3of 1 mol dm-3 HCl were added to the buffer all of the H+ ions will react with the ethanoate ions [A–] to form a molecule of ethanoic acid [HA]
    • Number of moles of ethanoic acid = 0.1
    • Number of moles of sodium ethanoate = 0.1

Moles of HCl = M x V /1000

=0.010

  • After adding the acid:
    • Number of moles of ethanoic acid [HA]= 0.110 mol dm-3(increased by 0.010)
    • Number of moles of sodium ethanoate [A–] = 0.090 mol dm-3 (decreased by 0.010)
    • The concentration of [HA] is now 0.110 mol dm-3
    • The concentration of [A–] is now 0.090 mol dm-3
  • The above calculations do not factor in the volume of HCl added

Ka = [H+(aq)][A–(aq)]

[HA(aq)]

1.7 x 10-5 = [H+(aq)] 0.090

0.110

[H+(aq)] = 1.7 x10-5 x 0.090 = 2.08 x 10 -5

0.110

pH = 4.68

  • The pH change is small from 4.77 à68

 

pH change when a base is added to a buffer

  • if 10cm3 of 0.1 mol dm-3 NaOH is added to the original buffer it will react with the H+ and more HA will ionise, thereby decreasing the concentration of the acid [HA] by 0.010 and increasing the concentration of ethanoate ions by 0.010