A buffer solution resists changes in pH when small quantities of an acid or an alkali are added to it.
Acidic buffer solutions can be made from a mixture of a weak acid and one of its salts, for example a mixture of ethanoic acid and sodium ethanoate in solution.
Alkaline buffer solutions can also be made from a mixture of a weak base and one of its salts, for example a mixture of ammonia solution and ammonium chloride solution.
bHow buffer solutions work/b
Let's consider an acidic buffer solution made up of a mixture of ethanoic acid and sodium ethanoate.
Ethanoic acid is a weak acid, and the position of equilibrium will be well to the left:
(CH3COOH(aq) ⇌ CH_3COO^-{(aq)} + H+{(aq)} )
Sodium ethanoate completely ionises in solution:
(CH3COO-Na+{(s)} ⟶ CH3COO-{(aq)} + Na+(aq) )
This means that there are plenty of ethananoic acid molecules (from the acid) and plenty of ethanoate ions (from the salt) in the mixture.
When an acid is added, the extra Hsup+/sup ions combine with ethanoate ions to form ethanoic acid. Since most of the extra Hsup+/sup ions are removed, the pH does not change much.
(CH3COO-{(aq)} + H+{(aq)} ⟶ CH3COOH(aq) )
When an alkali is added, the extra OHsup-/sup ions can combine with ethanoic acid. Since most of the extra OHsup-/sup ions are removed, the pH does not change much.
(CH3COOH(aq) + OH^-{(aq)} ⟶ CH3COO-{(aq)} + H2O(l) )
Alternatively, the extra OHsup-/sup ions can combine with Hsup+/sup ions (a small amount is present from the dissociation of the weak acid). Any Hsup+/sup ions that are removed in this way are immediately replaced because the equilibrium of the weak acid shifts to the right slightly. Therefore, the pH does not change much.
(H+{(aq)} + OH-{(aq)} ⟶ H2O(l) )
(CH3COOH(aq) ⟶ CH_3COO^-{(aq)} + H+{(aq)} )
bCalculate the pH of a buffer solution/b
The equilibrium constant Ksuba/sub is:
(Ka = \dfrac{[H^+]A-}{HA} )
Asup-/sup comes from the concentration of the salt (the amount coming from the weak acid is negligible as the position of equlibrium lies so far to the left).
HA comes from the concentration of the weak acid (as the position of equilibrium lies so far to the left, the tiny amount of HA that have dissociated is negligible).
Rearranging the equation gives:
(H+ = Ka \dfrac{HA}{A-} )
Taking negative logarithms of both sides:
(-\log{H+} = -\log{Ka} -\log{\dfrac{HA}{A-}} )
(pH = pKa + \log{\dfrac{A-}{HA}})
You can either learn this equation, or learn the steps to derive it in the exam.
For example: Calculate the pH of a buffer solution comprising 0.10 mol dmsup-3/sup of ethanoic acid (Ksuba/sub = 1.7 x 10sup-5/sup mol dmsup-3/sup) and 0.20 mol dmsup-3/sup of sodium ethanoate.
(Ka = \dfrac{[H^+]CH3COO-}{CH3COOH} )
(1.7 × 10-5 = \dfrac{[H^+]0.2}{0.1} )
(H+ = 1.7 × 10-5 \dfrac{0.1}{0.2} = 8.5 × 10-6 )
(pH = - \log{H+} = 5.1 )
bCalculate the concentrations of solutions required to prepare a buffer solution of a given pH/b
For example: Calculate the concentrations of ethanoic acid (Ksuba/sub = 1.7 x 10sup-5/sup mol dmsup-3/sup) and sodium ethanoate required to produce a buffer of pH 4.7.
(H+ = 10-pH = 10^{-4.7} = 2.0 × 10^{-5} )
Substitute this into the Ksuba/sub equation:
(Ka = \dfrac{[H^+]CH3COO-}{CH3COOH} )
(1.7 × 10-5 = \dfrac{[2.0 × 10^{-5}]CH3COO-}{CH3COOH} )
(\dfrac{1.7 × 10-5}{2.0 × 10^{-5}} = \dfrac{CH3COO-}{CH3COOH} = 0.85 )
To achieve a buffer of pH 4.7, the ratio of the concentrations of ethanoate ions to ethanoic acid must be 0.85.
bBlood as a buffer/b
A carbonic acid–hydrogencarbonate buffer system exists in blood, maintaining a pH between 7.35 and 7.45.
( H2CO{33 (aq)} ⇌ H^+{(aq)} + HCO–{3 (aq)} )