1. 18.2 Buffers

2. Assessment Objectives 18.2.1 Describe the composition of a buffer solution and explain its action. 18.2.2 Solve problems involving the composition and pH of a specified buffer system.

3. Reference Textbook on Moodle (under topic 18 Acids and Bases) Videos on moodle

4. Terms Buffer Equilibrium Acid Base Conjugate acid Conjugate base

5. Assessment Objectives 18.2.1 Describe the composition of a buffer solution and explain its action.

6. Buffers A buffer is a solution characterised by the ability to resist changes in pH when limited amounts of acid or base are added to it. Buffers contain either a weak acid and its conjugate base or a weak base and its conjugate acid. Thus, a buffer solution contains both an acid species and a base species in equilibrium.

7. Before you start it would be helpful to… know that weak acids and bases are only partly ionised in solution be able to calculate pH from hydrogen ion concentration be able to construct an equation for the dissociation constant of a weak acid Buffer solutions

8. Buffer solutions - Brief introduction Acidic Buffer (pH < 7) made from a weak acid + its sodium or potassium salt ethanoic acid sodium ethanoate Alkaline Buffer (pH > 7) made from a weak base + its chloride ammonia ammonium chloride UsesStandardising pH meters Buffering biological systems (eg in blood) Maintaining the pH of shampoos Definition“Solutions which resist changes in pH when small quantities of acid or alkali are added.”

9. Buffer solutions - uses Definition “Solutions which resist changes in pH when small quantities of acid or alkali are added.” Biological Uses In biological systems (saliva, stomach, and blood) it is essential that the pH stays ‘constant’ in order for any processes to work properly. e.g. If the pH of blood varies by 0.5 it can lead to unconsciousness and coma Most enzymes work best at particular pH values. Other Uses Many household and cosmetic products need to control their pH values. Shampoo Buffer solutions counteract the alkalinity of the soap and prevent irritation Baby lotion Buffer solutions maintain a pH of about 6 to prevent bacteria multiplying Others Washing powder, eye drops, fizzy lemonade

10. Acidic buffer solutions - action It is essential to have a weak acid for an equilibrium to be present so that ions can be removed and produced. The dissociation is small and there are few ions. CH 3 COOH(aq) CH 3 COO¯(aq) + H + (aq) relative concs. HIGH LOW LOW NB A strong acid can’t be used as it is fully dissociated and cannot remove H + (aq) HCl(aq) ——> Cl¯(aq) + H + (aq) Adding acid Any H + is removed by reacting with CH 3 COO¯ ions to form CH 3 COOH via the equilibrium. Unfortunately, the concentration of CH 3 COO¯ is small and only a few H + can be “mopped up”. A much larger concentration of CH 3 COO¯ is required. To build up the concentration of CH 3 COO¯ ions, sodium ethanoate is added.

11. Alkaline buffer Very similar but is based on the equilibrium surrounding a weak base; AMMONIA NH 3 (aq) + H 2 O(l) OH¯(aq) + NH 4 + (aq) relative concs. HIGH LOW LOW but one needs ;a large conc. of OH¯(aq) to react with any H + (aq) added a large conc of NH 4 + (aq) to react with any OH¯(aq) added There is enough NH 3 to act as a source of OH¯ but one needs to increase the concentration of ammonium ions by adding an ammonium salt. UseAMMONIA (a weak base) + AMMONIUM CHLORIDE (one of its salts) Alkaline buffer solutions - action

12. Buffer solutions - Ideal Concentration The concentration of a buffer solution is also important If the concentration is too low, there won’t be enough CH 3 COOH and CH 3 COO¯ to cope with the ions added. Summary For an acidic buffer solution one needs... large [CH 3 COOH(aq)]- for dissociating into H + (aq) when alkali is added large [CH 3 COO¯(aq)]- for removing H + (aq) as it is added This situation can’t exist if only acid is present; a mixture of the acid and salt is used. The weak acid provides the equilibrium and the large CH 3 COOH(aq) concentration. The sodium salt provides the large CH 3 COO¯(aq) concentration. One uses a WEAK ACID + its SODIUM OR POTASSIUM SALT

13. Calculating the pH of an acidic buffer solution Calculate the pH of a buffer whose [HA] is 0.1 mol dm -3 and [A¯] of 0.1 mol dm -3. K a = [H + (aq) ] [A (aq ) ] [HA (aq) ] re-arrange [H + (aq)] = [HA (aq) ] x K a [A ¯ (aq) ] from information given[A ¯ ] = 0.1 mol dm -3 [HA] = 0.1 mol dm -3 If the K a of the weak acid HA is 2 x 10 -4 mol dm -3. [H + (aq)]= 0.1 x 2 x 10 -4 = 2 x 10 -4 mol dm -3 0.1 pH= - log 10 [H + (aq)] = 3.699

14. Calculating the pH of an acidic buffer solution Calculate the pH of the solution formed when 500cm 3 of 0.1 mol dm -3 of weak acid HX is mixed with 500cm 3 of a 0.2 mol dm -3 solution of its salt NaX. K a = 4 x 10 -5 mol dm -3.

15. Calculating the pH of an acidic buffer solution Calculate the pH of the solution formed when 500cm 3 of 0.1 mol dm -3 of weak acid HX is mixed with 500cm 3 of a 0.2 mol dm -3 solution of its salt NaX. K a = 4 x 10 -5 mol dm -3. K a = [H + (aq)] [X¯(aq)] [HX(aq)]

16. Calculating the pH of an acidic buffer solution Calculate the pH of the solution formed when 500cm 3 of 0.1 mol dm -3 of weak acid HX is mixed with 500cm 3 of a 0.2 mol dm -3 solution of its salt NaX. K a = 4 x 10 -5 mol dm -3. K a = [H + (aq)] [X¯(aq)] [HX(aq)] re-arrange[H + (aq)] = [HX(aq)] K a [X¯(aq)]

17. Calculating the pH of an acidic buffer solution Calculate the pH of the solution formed when 500cm 3 of 0.1 mol dm -3 of weak acid HX is mixed with 500cm 3 of a 0.2 mol dm -3 solution of its salt NaX. K a = 4 x 10 -5 mol dm -3. K a = [H + (aq)] [X¯(aq)] [HX(aq)] re-arrange[H + (aq)] = [HX(aq)] K a [X¯(aq)] The solutions have been mixed; the volume is now 1 dm 3 therefore [HX] = 0.05 mol dm -3 and [X¯] = 0.10 mol dm -3

18. Calculating the pH of an acidic buffer solution Calculate the pH of the solution formed when 500cm 3 of 0.1 mol dm -3 of weak acid HX is mixed with 500cm 3 of a 0.2 mol dm -3 solution of its salt NaX. K a = 4 x 10 -5 mol dm -3. K a = [H + (aq)] [X¯(aq)] [HX(aq)] re-arrange[H + (aq)] = [HX(aq)] K a [X¯(aq)] The solutions have been mixed; the volume is now 1 dm 3 therefore [HX] = 0.05 mol dm -3 and [X¯] = 0.10 mol dm -3 Substituting[H + (aq)] = 0.05 x 4 x 10 -5 = 2 x 10 -5 mol dm -3 0.1 pH= - log 10 [H + (aq)] = 4.699

19. SALT HYDROLYSIS Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate.

20. SALT HYDROLYSIS Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate. Salts of strong acids and strong bases - SODIUM CHLORIDE NaCl dissociates completely in waterNa + Cl¯ ——> Na + + Cl¯ Water only ionises to a very small extent H 2 O OH¯ + H + Na + and OH¯ are ions of a strong base so remain apart H + and Cl¯ are ions of a strong acid so remain apart all the OH¯ and H + ions remain in solution therefore[H + ] = [OH¯] and the solution will be NEUTRAL

21. SALT HYDROLYSIS Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate. Salts of strong acids and weak bases - AMMONIUM CHLORIDE NH 4 Cl dissociates completely in waterNH 4 + Cl¯ ——> NH 4 + + Cl¯ Water only ionises to a very small extent H 2 O OH¯ + H + NH 3 + H 2 O Na + and OH¯ are ions of a strong base so tend to be associated H + and Cl¯ are ions of a strong acid so remain apart all the H + ions remain in solution therefore[H + ] > [OH¯] and the solution will be ACIDIC

22. SALT HYDROLYSIS Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate. Salts of weak acids and strong bases - SODIUM ETHANOATE CH 3 COONa dissociates completely in water CH 3 COO ¯ Na + ——> Na + + CH 3 COO¯ Water only ionises to a very small extent H 2 O OH¯ + H + CH 3 COOH Na + and OH¯ are ions of a strong base so remain apart H + and CH 3 OO¯ are ions of a weak acid so tend to be associated all the OH¯ ions remain in solution therefore [OH¯] > [H + ] and the solution will be ALKALINE

23. SALT HYDROLYSIS Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate. Salts of weak acids and weak bases - AMMONIUM ETHANOATE CH 3 COONH 4 dissociates completely in water CH 3 COO ¯ NH 4 + ——> NH 4 + + CH 3 COO¯ Water only ionises to a very small extent H 2 O OH¯ + H + NH 3 + H 2 O CH 3 COOH Na + and OH¯ are ions of a weak base so tend to be associated H + and CH 3 OO¯ are ions of a weak acid so tend to be associated the solution might be alkaline or acidic i.e. APPROXIMATELY NEUTRAL

24. REVISION CHECK What should you be able to do? Recall the definition of a buffer solution Recall the difference between an acidic and an alkaline buffer solution Recall the uses of buffer solutions Understand the action of buffer solutions Calculate the pH of an acidic buffer solution Recall and understand the reactions due to salt hydrolysis CAN YOU DO ALL OF THESE? YES NO