Acids and Bases Buffers

Buffers Buffer solutions are defined as solutions that resist a change in pH on adding small amounts of either acid or base. An acid buffer is a solution of a weak acid and its (strong) conjugate base (i.e. weak acid + salt of weak acid). An acid buffer has a pH less than 7 e.g. ethanoic acid and sodium ethanoate (pH range 4-6) contains CH3COOH (pKa = 4.76) and the relatively strong conjugate base, the ethanoate ion, CH3COO- A basic buffer is a solution of a weak base and its (strong) conjugate acid (i.e. weak base + salt of weak base). A basic buffer has a pH greater than 7. e.g. ammonia(aq) and ammonium chloride (pH range 8-10) contains NH3 and relatively strong conjugate acid, the ammonium ion, NH4+ (pKa = 9.25)

How does a buffer work? Consider the ethanoic acid – sodium ethanoate buffer system. Which equilibria are present? What happens if a base (OH - ions) is added? What happens if an acid (H3O+ ions) is added? CH3COOH + H2O  CH3COO- + H3O+ eqm1 also CH3COONa → CH3COO- + Na+ (suppresses acid ionisation – therefore if HA and MA present in similar amounts, [CH3COOH] = [weak acid] So large reservoirs both acid and conj base [CH3COO-] = [salt of weak acid] H3O+ + OH-  2H2O eqm2 eq2 shifts right, eq1 shifts right to replace H3O+ ions i.e. some of large reservoir of CH3COOH ionises i.e. CH3COOH + OH- → H2O + CH3COO- -e.q. 1 shifts left i.e some of large reservoir of CH3COO- ions used to ‘mop up’ H3O+ in both cases, pH remains about the same Note: for a buffer to work well when an acid or base is added, there must be a large reservoir of both base and acid forms.

What about the ammonia – ammonium chloride buffer?

pH of buffers HA(aq) + H2O(l)  H3O+(aq) + A-(aq) Ka = [H3O+][A-] [HA] Taking the negative logs of both sides: pKa = pH – log{ [A-] / [HA]} pKa = pH + log { [HA] / [A-] } pH = pKa – log [acid] or [H3O+] = Ka[weak acid] [base] [salt of weak acid] Assuming [HA] is undissociated acid and [A-] is only from added salt.

Example 1 Calculate the pH of a buffer solution made by adding 6.15g of sodium ethanoate, CH3COONa (molar mass 82 gmol-1), to 50.0cm3 of a 1.00 mol dm-3 solution of ethanoic acid (Ka = 1.75 x 10-5 mol dm-3

Example 2 Calculate the pH of a buffer solution made by adding 50 cm3 of 1.00mol dm-3 solution of sodium hydroxide to 80 cm3 of 1.00 mol dm-3 solution of propanoic acid, CH3CH2COOH, Ka = 1.35 x 10-5 mol dm-3.