Buffer Solutions A buffer solution is a mixture that minimises pH changes on the addition of small amounts of acid or base. No buffer solution can cope with the addition of large concentrations of acid or alkali. pH changes are minimised as long as some of the buffer solution remains. Buffers can be acidic or basic but we only need to be concerned with acid buffers.
Acidic buffers Are made from a mixture of A weak acid HA Its conjugate base, A- E.g. sodium ethanoate in ethanoic acid. The salt dissociates completely since it is ionic: CH 3 COONa (aq) → Na + (aq) + CH 3 COO - (aq) The acid is partially dissociated: CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq)
Acidic buffers This results in an equilibrium mixture containing large concentrations of the undissociated acid, CH 3 COOH and its conjugate base, CH 3 COO -. The very large concentration of the base shifts the acid equilibrium LEFT, so the concentration of H + (aq) ions is very small. Adding acid An increase in H + (aq) concentration would rapidly lower the pH of water, but in the buffer it simply combines with the ethanoate ions and shifts the acid dissociation back to the LEFT. i.e. CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) ←
Acidic buffers The hydrogen ions are transferred to the ethanoate ions so that ethanoic acid is formed. So a moderate input of H + (aq) ions has a very minimal effect on overall pH. Adding Alkali OH - (aq) ions react with the H + (aq) ions present: H + (aq) + OH - (aq) → H 2 O (l) Some of the acid dissociates, returning the [H + (aq) ] to near its original concentration: CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq)
Calculating the pH of Buffer Solutions Since there is an acid dissociation taking place the pH of a buffer solution must depend on the K a of the acid present AND The equilibrium concentrations of the conjugate acid- base pair. The equation for the acid equilibrium is: HA (aq) H + (aq) + A - (aq) K a = [H + (aq) ][A - (aq) ] [HA (aq) ] [H + (aq) ]= K a x [HA] [A - (aq) ] As before we assume that [HA (aq) ] equil = [HA (aq) ] undiss
Henderson-Hasselbalch This can be derived but the derivation is not needed. It is sometimes easier to use, depending on the form in which data is supplied. pH = pK a + log [A - (aq) ] [HA (aq) ] Where pK a = -logK a This is often written as: pH = pK a +log [salt] [acid]
Examples A buffer solution can be made containing 0.600 mol dm -3 propanoic acid and 0.800 mol dm -3 sodium propanoate. The equilibrium constant, K a, for propanoic acid is 1.3 x 10 -5 mol dm -3. Substituting in: [H + (aq) ] = K a x [HA] gives us [A - (aq) ] [H + (aq) ] = 1.3 x 10 -5 x 0.600 mol dm -3 0.800 = 9.75 x 10 -6 mol dm -3 And pH = -log(9.75 x 10 -6 ) = -(-5.01) = 5.01
Calculate the pH of a solution containing: a) 0.0500 mol dm -3 methanoic acid and 0.100 mol dm -3 sodium methanoate. K a =1.6 x 10 -4 mol dm -3 Ans = 4.10 b) 0.0100 mol dm -3 benzoic acid and 0.0400 mol dm -3 sodium benzoate Ka = 6.3 x 10 -5 mol dm -3. Ans = 4.8 c) 0.100 mol dm -3 ethanoic acid and 0.200 mol dm -3 sodium ethanoate (pK a ethanoic acid = 4.8) Ans = 5.1
Calculate the pH of a solution containing: d) 0.02 mol dm -3 ethanoic acid and 0.05 mol dm -3 sodium ethanoate. K a = 1.7 x 10 -5 mol dm -3 Ans = 5.17 e) 0.200mol dm -3 citric acid and 0.100 mol dm -3 sodium citrate. Ka = 7.24 x 10 -4 mol dm -3 Ans = 2.8 f) Calculate the pH of a buffer made by dissolving 18.5g of propanoic acid, C 2 H 5 COOH and 12.0g of sodium propanoate, C 2 H 5 COONa, in water and then making up the volume to 250cm 3. pK a for propanoic acid = 4.87
Working 1 mole C 2 H 5 COOH weighs 74g. 1 mole C 2 H 5 COONa weighs 96g. [C 2 H 5 COOH] = 18.5 x 4 74 = 1.00mol dm- 3. [C 2 H 5 COONa] = 12.0 x4 96 = 0.500 mol dm -3 K a = 10 -pKa K a = 10 -4.87 = 1.35 x 10 -5 mol dm -3
Now we can do the calculation Ka = [C 2 H 5 COO - ] [H + ] [C 2 H 5 COOH] 1.35 x 10 -5 = 0.5 x [H + ] 1.00 [H + ] = 1.35 x 10 -5 0.500 = 2.70 x 10 -5 mol dm -3 pH = 4.57 OR Use Henderson Hasselbalch to arrive at the same answer.
Try this: A buffer solution was made by mixing 50.0cm 3 of 0.300 mol dm -3 ethanoic acid with 100 cm 3 0f 0.600 mol dm -3 sodium ethanoate. Calculate the pH if K a for ethanoic acid is 1.74 x 10 -5 mol dm -3. Number of moles acid = 50 x 0.300 1000 = 0.0150 ( in a total volume of 150 cm 3 ) [CH 3 COOH] = 0.0150 x 1000 150 = 0.100 mol dm -3
Number of moles CH 3 COONa = 100 x 0.600 1000 = 0.0600 ( in a total volume of 150 cm 3 ) [ CH 3 COONa] = 0.0600 x 1000 150 = 0.400 mol dm -3 Plug in as before: K a = [CH 3 COO - ] [H + ] [CH 3 COOH] 1.74 x 10 -5 = 0.400 x [H + ] 0.100 [H + ] =1.74 x 10 -5 x 0.100 0.400 = 4.35 x 10 -6 pH=5.36
The role of a natural buffer in the control of blood pH. To remain healthy human blood has to be maintained at a constant pH 7.4 (7.35-7.45). If the blood becomes too acidic and the pH drops ( as in the medical condition acidosis), we have to breathe rapidly to expel more carbon dioxide. The main mechanism for maintaining pH is the buffering action of several acid/base pairs e.g carbonic acid/hydrogencarbonate and H 2 PO 4 - (aq) and HPO 4 2- (aq), together with the buffering action of plasma proteins and haemoglobin.
The role of a natural buffer in the control of blood pH. The carbonic acid/hydrogencarbonate ion buffer is the most important buffer system in the blood plasma. Carbonic acid is a weak acid. Hydrogencarbonate ions HCO 3 - are the conjugate base. H 2 CO 3(aq) H + (aq) + HCO 3 - (aq) Any increases in [H + ] ions in the blood are removed by the conjugate base. The equilibrium shifts left removing most of the H + ions.
Effect of adding an alkali Any increase in the [OH - ] concentration is removed by the weak acid, H 2 CO 3. The [H+] ions are removed by the OH - ions: H + (aq) + OH - (aq) → H 2 O (l) The carbonic acid dissociates, shifting the equilibrium RIGHT to restore most of the the H + (aq). The shifts in equilibrium are repeated from the book in the following slide. Breathing rate controls CO 2 in the blood.
pH or Titration Curves In an acid-base titration, a solution of an alkali of known concentration is added from a burette to a measured volume of an acid solution until an indicator shows that the acid has been neutralised. We can then work out the concentration of the acid from the volume of the alkali used. We can follow the neutralisation reaction by measuring with a pH meter (if we had one which still worked!) The result is a titration curve.
pH or Titration Curves The shapes of the curves obtained are typical of each possible combination of acid and base. As base is added to acid in a conical flask the pH changes, but not in a regular manner. There are effectively 3 sections to a titration curve. An initial slight rise in pH A sharp rise in pH A final slight rise in pH You MUST be able to sketch and explain the typical shapes obtained.
Equivalence Point The equivalence point in a titration corresponds to the point where the exact number of moles of OH - ions have been added as there are moles of H + ions present. This corresponds to the mixing of the stoichiometric ratios from the balanced formula equation for the reaction. pH change is NOT directly proportional to the amount of alkali added.
Strong Acid-Strong Base E.g. HCl (aq) + NaOH (aq) → NaCl (aq) + H 2 O (aq) Both the acid and the base are fully dissociated. At the beginning, as alkali is added the PROPORTION of the hydrogen ions removed is small and causes a very small increase in pH. As the alkali is added the number of H+ ions removed increases as a total AND as a proportion so that there is a larger change in pH. This change is greatest near the equivalence point, where the curve is steepest. For this curve the equivalence point is at pH 7 because the conjugate acid and base are weak and do not affect the H + (aq) + OH - (aq) →H 2 O (l) equilibrium.
Finding the Equivalence Point This is found from the mid point of the vertical section of the pH curve. This is 7 for a strong/strong titration. This is about 8.9 for a weak acid with a strong base. For weak base/strong acid this point is below 7 For weak/weak the pH changes steadily but there is no vertical section and no sharp, obvious equivalence point.
Strong/Weak For strong acid/weak base the curve looks like the strong/strong curve at the beginning but the equivalence point is below 7 due to the fact that the pH of the base is relatively low. For weak acid/strong base the weak acid is little dissociated so the initial pH is higher than in the others and also as the alkali is added the increasing anion concentration leads to a buffering action so that pH change is slow. After the equivalence point the curve follows the pattern for strong/strong.
Choice of Indicators Indicators are themselves weak acids. Their dissociated and undissociated forms have different colours. Their dissociation can be represented as: HIn H + + In - They are useful as indicators if they change colour over a range of 1-2 pH units with a recognised end point somewhere in the middle. This is the point where the colour is between the 2 extremes of its colour and where the concentration of the 2 forms are equal.
Choice of Indicators For Titration The colour of at least one form needs to be intense so that 1 or 2 drops can be used to give a clearly visible change without affecting [H + ]. End point and equivalence point do not need to be the same but the result will be most precise when the 2 coincide or nearly do so. Suitable indicators change colour within the vertical section of a titration curve which often corresponds to the addition of a single drop of the base. For weak/weak reactions there is no suitable indicator because the colour change would be gradual. http://www.chemit.co.uk/upload/java/rsc_indi cator/applet.htmhttp://www.chemit.co.uk/upload/java/rsc_indi cator/applet.htm Answer q.p. 155 and q6 p.162