1. CONTENTS Brønsted-Lowry theory of acids and bases Lewis theory of acids and bases Strong acids and bases Weak acids Weak bases Hydrogen ion concentration and pH Ionic product of water K w Relation between pH and pOH Introduction to buffer solutions Check list Acid & Bases

2. Before you start it would be helpful to… know the simple properties of acids, bases and alkalis Acid & Bases

3. BRØNSTED-LOWRY THEORY ACIDproton donorHCl ——> H + (aq) + Cl¯(aq) BASEproton acceptor NH 3 (aq) + H + (aq) ——> NH 4 + (aq) ACIDS AND BASES

4. BRØNSTED-LOWRY THEORY ACIDproton donorHCl ——> H + (aq) + Cl¯(aq) BASEproton acceptor NH 3 (aq) + H + (aq) ——> NH 4 + (aq) Conjugate systems Acids are related to bases ACID PROTON + CONJUGATE BASE Bases are related to acids BASE + PROTON CONJUGATE ACID ACIDS AND BASES

5. BRØNSTED-LOWRY THEORY ACIDproton donorHCl ——> H + (aq) + Cl¯(aq) BASEproton acceptor NH 3 (aq) + H + (aq) ——> NH 4 + (aq) Conjugate systems Acids are related to bases ACID PROTON + CONJUGATE BASE Bases are related to acids BASE + PROTON CONJUGATE ACID For an acid to behave as an acid, it must have a base present to accept a proton... HA + B BH + + A¯ acid base conjugate conjugate acid base example CH 3 COO¯ + H 2 O CH 3 COOH + OH¯ base acid acid base ACIDS AND BASES

6. Conjugate Pairs HCl (aq) + H 2 O (aq) H 3 O + (aq) + Cl - (aq) Acid Base Conjugate pair

7. The negative ion is referred to as the conjugate base of the acid HCl (aq) + H 2 O (aq) H 3 O + (aq) + Cl - (aq) AcidConjugate Acid Conjugate Base Base The positive ion is referred to as the conjugate acid of the base

8. CH 3 COOH (aq) + NH 3 (aq) NH 4 + (aq) + CH 3 COO - (aq) Acid Conjugate Acid Conjugate Base Base

9. H 2 O (aq) + NH 3 (aq) NH 4 + (aq) + OH - (aq) Acid Conjugate Acid Conjugate Base Base

10. HCOOH (aq) + H 2 O (aq) H 3 O + (aq) + HCOO - (aq) Acid Conjugate Acid Conjugate Base Base

11. Water can act as an acid and a base – it can be described as amphoteric HCOOH (aq) + H 2 O (aq) H 3 O + (aq) + HCOO - (aq) H 2 O (aq) + NH 3 (aq) NH 4 + (aq) + OH - (aq) Pure water will always contain some H + & OH - ions it is described as self protonating 2H 2 O (aq) H 3 O + (aq) + OH - (aq)

12. HNO 3 (aq) + HF (aq) H 2 F + (aq) + NO 3 - (aq) Acid Conjugate Acid Conjugate Base Base

13. H 2 SO 4 (aq) + HNO 3 (aq) H 2 NO 3 + (aq) + HSO 4 - (aq) AcidConjugate Acid Conjugate Base Base

14. LEWIS THEORY ACIDlone pair acceptorBF 3 H + AlCl 3 BASElone pair donorNH 3 H 2 O ACIDS AND BASES LONE PAIR DONOR LONE PAIR ACCEPTOR LONE PAIR DONOR LONE PAIR ACCEPTOR

15. STRONG ACIDS completely dissociate (split up) into ions in aqueous solution e.g. HCl ——> H + (aq) + Cl¯(aq) MONOPROTIC1 replaceable H HNO 3 ——> H + (aq) + NO 3 ¯(aq) H 2 SO 4 ——> 2H + (aq) + SO 4 2- (aq) DIPROTIC 2 replaceable H’s STRONG ACIDS AND BASES

16. STRONG ACIDS completely dissociate (split up) into ions in aqueous solution e.g. HCl ——> H + (aq) + Cl¯(aq) MONOPROTIC1 replaceable H HNO 3 ——> H + (aq) + NO 3 ¯(aq) H 2 SO 4 ——> 2H + (aq) + SO 4 2- (aq) DIPROTIC 2 replaceable H’s STRONG BASES completely dissociate into ions in aqueous solution e.g. NaOH ——> Na + (aq) + OH¯(aq) STRONG ACIDS AND BASES

17. Weak acids partially dissociate into ions in aqueous solution e.g. ethanoic acid CH 3 COOH(aq) CH 3 COO¯(aq) + H + (aq) When a weak acid dissolves in water an equilibrium is set upHA(aq) + H 2 O(l) A¯(aq) + H 3 O + (aq) The water stabilises the ions To make calculations easier the dissociation can be written... HA(aq) A¯(aq) + H + (aq) WEAK ACIDS

18. Weak acids partially dissociate into ions in aqueous solution e.g. ethanoic acid CH 3 COOH(aq) CH 3 COO¯(aq) + H + (aq) When a weak acid dissolves in water an equilibrium is set upHA(aq) + H 2 O(l) A¯(aq) + H 3 O + (aq) The water stabilises the ions To make calculations easier the dissociation can be written... HA(aq) A¯(aq) + H + (aq) The weaker the acid the less it dissociates the more the equilibrium lies to the left. The relative strengths of acids can be expressed as K a or pK a values The dissociation constant for the weak acid HA is K a = [H + (aq)] [A¯(aq)] mol dm -3 [HA(aq)] WEAK ACIDS

19. Partially react with water to give ions in aqueous solution e.g. ammonia When a weak base dissolves in water an equilibrium is set up NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH¯ (aq) as in the case of acids it is more simply written NH 3 (aq) + H + (aq) NH 4 + (aq) WEAK BASES

20. Partially react with water to give ions in aqueous solution e.g. ammonia When a weak base dissolves in water an equilibrium is set up NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH¯ (aq) as in the case of acids it is more simply written NH 3 (aq) + H + (aq) NH 4 + (aq) The weaker the basethe less it dissociates the more the equilibrium lies to the left The relative strengths of bases can be expressed as K b or pK b values. WEAK BASES

21. Hydrogen ion concentration [H + (aq) ] Introductionhydrogen ion concentration determines the acidity of a solution hydroxide ion concentration determines the alkalinity for strong acids and bases the concentration of ions is very much larger than their weaker counterparts which only partially dissociate.

22. Hydrogen ion concentration [H + (aq) ] pHhydrogen ion concentration can be converted to pH pH = - log 10 [H + (aq)] to convert pH into hydrogen ion concentration [H + (aq)] = antilog (-pH) pOHAn equivalent calculation for bases converts the hydroxide ion concentration to pOH pOH = - log 10 [OH¯(aq)] in both the above, [ ] represents the concentration in mol dm -3 STRONGLY ACIDIC 10 0 10 -1 10 -2 10 -3 10 -4 10 -5 10 -6 10 -7 10 -8 10 -9 10 -10 10 -11 10 -12 10 -13 10 -14 10 -13 10 -12 10 -11 10 -10 10 -9 10 -8 10 -7 10 -6 10 -5 10 -4 10 -3 10 -2 10 -1 10 -0 01234567891011121314pH OH¯ [H + ] WEAKLY ACIDIC NEUTRAL STRONGLY ALKALINE WEAKLY ALKALINE

23. Ionic product of water - K w Despite being covalent, water conducts electricity to a very small extent. This is due to the slight ionisation...H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH¯(aq) or, more simply H 2 O(l) H + (aq) + OH¯(aq)

24. Ionic product of water - K w Despite being covalent, water conducts electricity to a very small extent. This is due to the slight ionisation...H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH¯(aq) or, more simply H 2 O(l) H + (aq) + OH¯(aq) Applying the equilibrium law to the second equation gives K c = [H + (aq)] [OH¯(aq)] [ ] is the equilibrium concentration in mol dm -3 [H 2 O(l)]

25. Ionic product of water - K w Despite being covalent, water conducts electricity to a very small extent. This is due to the slight ionisation...H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH¯(aq) or, more simply H 2 O(l) H + (aq) + OH¯(aq) Applying the equilibrium law to the second equation gives K c = [H + (aq)] [OH¯(aq)] [ ] is the equilibrium concentration in mol dm -3 [H 2 O(l)] As the dissociation is small, the water concentration is very large compared with the dissociated ions and any changes to its value are insignificant; its concentration can be regarded as constant. This “constant” is combined with (K c ) to get a new constant (K w ). K w = [H + (aq)] [OH¯(aq)] mol 2 dm -6 = 1 x 10 -14 mol 2 dm -6 (at 25°C) Because the constant is based on an equilibrium, K w VARIES WITH TEMPERATURE

26. Ionic product of water - K w The value of K w varies with temperature because it is based on an equilibrium. Temperature / °C 0 20 25 3060 K w / 1 x 10 -14 mol 2 dm -6 0.110.681.01.475.6 H + / x 10 -7 mol dm -3 0.330.821.01.272.37 pH7.487.08 76.926.63 What does this tell you about the equation H 2 O(l) H + (aq) + OH¯(aq) ?

27. Ionic product of water - K w The value of K w varies with temperature because it is based on an equilibrium. Temperature / °C 0 20 25 3060 K w / 1 x 10 -14 mol 2 dm -6 0.110.681.01.475.6 H + / x 10 -7 mol dm -3 0.330.821.01.272.37 pH7.487.08 76.926.63 What does this tell you about the equation H 2 O(l) H + (aq) + OH¯(aq) ? K w gets larger as the temperature increases this means the concentration of H + and OH¯ ions gets greater this means the equilibrium has moved to the right if the concentration of H + increases then the pH decreases pH decreases as the temperature increases

28. Ionic product of water - K w The value of K w varies with temperature because it is based on an equilibrium. Temperature / °C 0 20 25 3060 K w / 1 x 10 -14 mol 2 dm -6 0.110.681.01.475.6 H + / x 10 -7 mol dm -3 0.330.821.01.272.37 pH7.487.08 76.926.63 What does this tell you about the equation H 2 O(l) H + (aq) + OH¯(aq) ? K w gets larger as the temperature increases this means the concentration of H + and OH¯ ions gets greater this means the equilibrium has moved to the right if the concentration of H + increases then the pH decreases pH decreases as the temperature increases Because the equation moves to the right as the temperature goes up, it must be an ENDOTHERMIC process

29. Relationship between pH and pOH Because H + and OH¯ ions are produced in equal amounts when water dissociates[H + ] = [OH¯] = 1 x 10 -7 mol dm -3 their concentrations will be the same. K w = [H + ] [OH¯] = 1 x 10 -14 mol 2 dm -6 take logs of both sideslog[H + ] + log[OH¯] = -14 multiply by minus- log[H + ] - log[OH¯] = 14 change to pH and pOHpH + pOH = 14 (at 25°C)

30. Relationship between pH and pOH Because H + and OH¯ ions are produced in equal amounts when water dissociates[H + ] = [OH¯] = 1 x 10 -7 mol dm -3 their concentrations will be the same. K w = [H + ] [OH¯] = 1 x 10 -14 mol 2 dm -6 take logs of both sideslog[H + ] + log[OH¯] = -14 multiply by minus- log[H + ] - log[OH¯] = 14 change to pH and pOHpH + pOH = 14 (at 25°C) N.B.As they are based on the position of equilibrium and that varies with temperature, the above values are only true if the temperature is 25°C (298K) Neutral solutions may be regarded as those where [H + ] = [OH¯]. Therefore a neutral solution is pH 7 only at a temperature of 25°C (298K) K w is constant for any aqueous solution at the stated temperature

31. Buffer solutions - Brief introduction Definition “Solutions which resist changes in pH when small quantities of acid or alkali are added.” 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

32. CONTENTS What is pH? - a reminder Calculating the pH of strong acids and bases Calculating the pH of weak acids Calculating the pH of mixtures - strong acid and strong alkali pH calculations

33. Before you start it would be helpful to… know the differences between strong and weak acid and bases be able to calculate pH from hydrogen ion concentration be able to calculate hydrogen ion concentration from pH know the formula for the ionic product of water and its value at 25°C pH calculations

34. What is pH or pOH? pH = - log 10 [H + (aq) ] pOH = - log 10 [OH - (aq) ] where [H + ] is the concentration of hydrogen ions in mol dm -3 where [OH - ] is the concentration of hydroxide ions in mol dm -3 to convert pH into hydrogen ion concentration [H + (aq) ] = antilog (-pH) to convert pOH into [OH - (aq) ] = antilog (-pOH) OH- concentration

35. IONIC PRODUCT OF WATER K w = [H + (aq)] [OH¯(aq)] mol 2 dm -6 = 1 x 10 -14 mol 2 dm -6 (at 25°C) pKw = 14 = -log(1 x 10 -14 ) pKw = pH + pOH pH + pOH = 14

36. Calculating pH - strong acids and alkalis Strong acids and alkalis completely dissociate in aqueous solution only need to know the concentration It is easy to calculate the pH or pOH ; you only need to know the concentration. Calculate the pH of 0.02M HCl HCl completely dissociates in aqueous solution HC l H + + C l ¯ One H + is produced for each HCl dissociating so [H + ] = 0.02M = 2 x 10 -2 mol dm -3 WORKED EXAMPLE WORKED EXAMPLE pH = - log [H + ] = - log [0.02] = 1.7 Note that negative pH vales are also possible For example : The pH of 1.5 mol dm–3 HCl is -0.18

37. Calculating pH - strong acids and alkalis Strong acids and alkalis completely dissociate in aqueous solution only need to know the concentration It is easy to calculate the pH or pOH ; you only need to know the concentration. Calculate the pH of 0.1M NaOH NaOH completely dissociates in aqueous solution NaOH Na + + OH¯ One OH¯ is produced for each NaOH dissociating [OH¯] = 0.1M = 1 x 10 -1 mol dm -3 WORKED EXAMPLE WORKED EXAMPLE pH + pOH = 14 pH + 1 = 14 pH = 13 To find pH of a strong base use pH + pOH = 14 pOH = -log [OH¯] = -log [ 1 ] = 1

38. Calculating pH - weak acids A weak acid, HA, dissociates as followsHA (aq) H + (aq) + A¯ (aq) (1) A weak acid is one which only partially dissociates in aqueous solution

39. Calculating pH - weak acids A weak acid, HA, dissociates as followsHA (aq) H + (aq) + A¯ (aq) (1) Applying the Equilibrium Law K a = [H + (aq) ] [A¯ (aq) ] mol dm -3 (2) [HA (aq) ] A weak acid is one which only partially dissociates in aqueous solution

40. Calculating pH - weak acids A weak acid, HA, dissociates as followsHA (aq) H + (aq) + A¯ (aq) (1) Applying the Equilibrium Law K a = [H + (aq) ] [A¯ (aq) ] mol dm -3 (2) [HA (aq) ] The ions are formed in equal amounts, so [H + (aq) ] = [A¯ (aq) ] therefore K a = [H + (aq) ] 2 (3) [HA (aq) ] A weak acid is one which only partially dissociates in aqueous solution

41. Calculating pH - weak acids A weak acid, HA, dissociates as followsHA (aq) H + (aq) + A¯ (aq) (1) Applying the Equilibrium Law K a = [H + (aq) ] [A¯ (aq) ] mol dm -3 (2) [HA (aq) ] The ions are formed in equal amounts, so [H + (aq) ] = [A¯ (aq) ] therefore K a = [H + (aq) ] 2 (3) [HA (aq) ] Rearranging (3) gives[H + (aq) ] 2 = [HA (aq) ] K a therefore[H + (aq) ] = [HA (aq) ] K a A weak acid is one which only partially dissociates in aqueous solution

42. Calculating pH - weak acids A weak acid, HA, dissociates as followsHA (aq) H + (aq) + A¯ (aq) (1) Applying the Equilibrium Law K a = [H + (aq) ] [A¯ (aq) ] mol dm -3 (2) [HA (aq) ] The ions are formed in equal amounts, so [H + (aq) ] = [A¯ (aq) ] therefore K a = [H + (aq) ] 2 (3) [HA (aq) ] Rearranging (3) gives[H + (aq) ] 2 = [HA (aq) ] K a therefore[H + (aq) ] = [HA (aq) ] K a pH = [H + (aq) ] A weak acid is one which only partially dissociates in aqueous solution

43. Calculating pH - weak acids A weak acid, HA, dissociates as followsHA (aq) H + (aq) + A¯ (aq) (1) Applying the Equilibrium Law K a = [H + (aq) ] [A¯ (aq) ] mol dm -3 (2) [HA (aq) ] The ions are formed in equal amounts, so [H + (aq) ] = [A¯ (aq) ] therefore K a = [H + (aq) ] 2 (3) [HA (aq) ] Rearranging (3) gives[H + (aq) ] 2 = [HA (aq) ] K a therefore[H + (aq) ] = [HA (aq) ] K a pH = [H + (aq) ] ASSUMPTION HA is a weak acid so it will not have dissociated very much. You can assume that its equilibrium concentration is approximately that of the original concentration. A weak acid is one which only partially dissociates in aqueous solution

44. Calculating pH - weak acids Calculate the pH of a weak acid HX of concentration 0.1M ( K a = 4x10 -5 mol dm -3 ) HX dissociates as followsHX (aq) H + (aq) + X¯ (aq) WORKED EXAMPLE WORKED EXAMPLE

45. Calculating pH - weak acids Calculate the pH of a weak acid HX of concentration 0.1M ( K a = 4x10 -5 mol dm -3 ) HX dissociates as followsHX (aq) H + (aq) + X¯ (aq) Dissociation constant for a weak acid K a = [H + (aq) ] [X¯ (aq) ] mol dm -3 [HX (aq) ] WORKED EXAMPLE WORKED EXAMPLE

46. Calculating pH - weak acids Calculate the pH of a weak acid HX of concentration 0.1M ( K a = 4x10 -5 mol dm -3 ) HX dissociates as followsHX (aq) H + (aq) + X¯ (aq) Dissociation constant for a weak acid K a = [H + (aq) ] [X¯ (aq) ] mol dm -3 [HX (aq) ] Substitute for X¯ as ions are formed in [H + (aq) ] = [HX (aq) ] K a mol dm -3 equal amounts and then rearrange equation WORKED EXAMPLE WORKED EXAMPLE

47. Calculating pH - weak acids Calculate the pH of a weak acid HX of concentration 0.1M ( K a = 4x10 -5 mol dm -3 ) HX dissociates as followsHX (aq) H + (aq) + X¯ (aq) Dissociation constant for a weak acid K a = [H + (aq) ] [X¯ (aq) ] mol dm -3 [HX (aq) ] Substitute for X¯ as ions are formed in [H + (aq) ] = [HX (aq) ] K a mol dm -3 equal amounts and the rearrange equation ASSUMPTION HA is a weak acid so it will not have dissociated very much. You can assume that its equilibrium concentration is approximately that of the original concentration WORKED EXAMPLE WORKED EXAMPLE

48. Calculating pH - weak acids Calculate the pH of a weak acid HX of concentration 0.1M ( K a = 4x10 -5 mol dm -3 ) HX dissociates as followsHX (aq) H + (aq) + X¯ (aq) Dissociation constant for a weak acid K a = [H + (aq) ] [X¯ (aq) ] mol dm -3 [HX (aq) ] Substitute for X¯ as ions are formed in [H + (aq) ] = [HX (aq) ] K a mol dm -3 equal amounts and the rearrange equation ASSUMPTION HA is a weak acid so it will not have dissociated very much. You can assume that its equilibrium concentration is approximately that of the original concentration [H + (aq) ] = 0.1 x 4 x 10 -5 mol dm -3 = 4.00 x 10 -6 mol dm -3 = 2.00 x 10 -3 mol dm -3 ANSWER pH = - log [H + (aq) ] = 2.699 WORKED EXAMPLE WORKED EXAMPLE

49. CALCULATING THE pH OF MIXTURES The method used to calculate the pH of a mixture of an acid and an alkali depends on... whether the acids and alkalis are STRONG or WEAK which substance is present in excess STRONG ACID and STRONG BASE - EITHER IN EXCESS

50. pH of mixtures Strong acids and strong alkalis (either in excess) 1. Calculate the initial number of moles of H + and OH¯ ions in the solutions 2. As H + and OH¯ ions react in a 1:1 ratio; calculate unreacted moles species in excess 3. Calculate the volume of solution by adding the two original volumes 4. Convert volume to dm 3 (divide cm 3 by 1000) 5. Divide moles by volume to find concentration of excess the ion in mol dm -3 6. Convert concentration to pH If the excess isH + pH = - log[H + ] If the excess is OH¯ pOH = - log[OH¯] then pH + pOH = 14 or use K w = [H + ] [OH¯] = 1 x 10 -14 at 25°C therefore [H + ] = K w / [OH¯] then pH = - log[H + ]

51. Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l pH of mixtures Strong acids and alkalis (either in excess) WORKED EXAMPLE WORKED EXAMPLE

52. Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 1. Calculate the number of moles of H + and OH¯ ions present pH of mixtures Strong acids and alkalis (either in excess) 25cm 3 of 0.1M NaOH 20cm 3 of 0.1M HC l 2.5 x 10 -3 moles 2.0 x 10 -3 moles moles of OH ¯ = 0.1 x 25/1000 = 2.5 x 10 -3 moles of H + = 20 x 20/1000 = 2.0 x 10 -3 WORKED EXAMPLE WORKED EXAMPLE

53. Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 1. Calculate the number of moles of H + and OH¯ ions present 2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species pH of mixtures Strong acids and alkalis (either in excess) The reaction taking place is… HC l + NaOH NaC l + H 2 O or in its ionic form H + + OH¯ H 2 O (1:1 molar ratio) 25cm 3 of 0.1M NaOH 20cm 3 of 0.1M HC l 2.5 x 10 -3 moles 2.0 x 10 -3 moles WORKED EXAMPLE WORKED EXAMPLE

54. Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 1. Calculate the number of moles of H + and OH¯ ions present 2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species pH of mixtures Strong acids and alkalis (either in excess) The reaction taking place is… HC l + NaOH NaC l + H 2 O or in its ionic form H + + OH¯ H 2 O (1:1 molar ratio) 2.0 x 10 -3 moles of H + will react with the same number of moles of OH¯ this leaves 2.5 x 10 -3 - 2.0 x 10 -3 = 5.0 x 10 -4 moles of OH¯ in excess 5.0 x 10 -4 moles of OH¯ UNREACTED 25cm 3 of 0.1M NaOH 20cm 3 of 0.1M HC l 2.5 x 10 -3 moles 2.0 x 10 -3 moles WORKED EXAMPLE WORKED EXAMPLE

55. Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 1. Calculate the number of moles of H + and OH¯ ions present 2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species 3. Calculate the volume of the solution by adding the two individual volumes pH of mixtures Strong acids and alkalis (either in excess) the volume of the solution is 25 + 20 = 45cm 3 WORKED EXAMPLE WORKED EXAMPLE

56. Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 1. Calculate the number of moles of H + and OH¯ ions present 2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species 3. Calculate the volume of the solution by adding the two individual volumes 4. Convert volume to dm 3 (divide cm 3 by 1000) pH of mixtures Strong acids and alkalis (either in excess) the volume of the solution is 25 + 20 = 45cm 3 there are 1000 cm 3 in 1 dm 3 volume = 45/1000 = 0.045dm 3 WORKED EXAMPLE WORKED EXAMPLE

57. Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 1. Calculate the number of moles of H + and OH¯ ions present 2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species 3. Calculate the volume of the solution by adding the two individual volumes 4. Convert volume to dm 3 (divide cm 3 by 1000) 5. Divide moles by volume to find concentration of excess ion in mol dm -3 pH of mixtures Strong acids and alkalis (either in excess) WORKED EXAMPLE WORKED EXAMPLE [OH¯]= 5.0 x 10 -4 / 0.045 = 1.11 x 10 -2 mol dm -3

58. Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 1. Calculate the number of moles of H + and OH¯ ions present 2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species 3. Calculate the volume of the solution by adding the two individual volumes 4. Convert volume to dm 3 (divide cm 3 by 1000) 5. Divide moles by volume to find concentration of excess ion in mol dm -3 6. As the excess is OH¯ usepOH = - log[OH¯] then pH + pOH = 14 or K w = [H + ][OH¯] so [H + ] = K w / [OH¯] pH of mixtures Strong acids and alkalis (either in excess) K w = 1 x 10 -14 mol 2 dm -6 (at 25°C) WORKED EXAMPLE WORKED EXAMPLE [OH¯]= 5.0 x 10 -4 / 0.045 = 1.11 x 10 -2 mol dm -3 [H + ] = K w / [OH¯] = 9.00 x 10 -13 mol dm -3 pH = - log[H + ] = 12.05

59. CONTENTS What is a buffer solution? Uses of buffer solutions Acidic buffer solutions Alkaline buffer solutions Buffer solutions - ideal concentration Calculating the pH of a buffer solution Salt hydrolysis Check list Buffer solutions

60. 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

61. Buffers A buffer is a solution whose pH is resistant to change on the addition of relatively small quantities of an acid or base. Buffers have the ability to absorb added H + or OH - ions.

62. 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

63. Acidic Buffers Contain a weak acid and its salt formed with a strong base Ethanoic acid & Sodium ethanoate CH 3 COOH & CH 3 COO - Na +

64. Buffer Action The important species present in the buffer solution are the undissociated weak acid CH 3 COOH and its conjugate base CH 3 COO -

65. Undissociated ethanoic acid can remove any added OH - ions. CH 3 COOH + OH - → CH 3 COO - + H 2 O The conjugate base can remove any added H + ions CH 3 COO - + H + → CH 3 COOH

66. For a buffer to work effectively it must contain a large reservoir of the weak acid and its conjugate base. CH 3 COOH + H 2 O CH 3 COO - + H 3 O + CH 3 COOH CH 3 COO - + H + Write an expression for K a state units

67. K a = [CH 3 COO - ] [H + ] [CH 3 COOH] CH 3 COOH CH 3 COO - + H + [H + ] = K a x [CH 3 COOH] [CH 3 COO - ] If both [CH 3 COOH] and [CH 3 COO - ] are large………..

68. [H + ] = K a x [CH 3 COOH] [CH 3 COO - ] If both [CH 3 COOH] and [CH 3 COO - ] are large small changes in their concentrations will not affect the overall ratio significantly so [H + ] remains almost constant So very small change in pH

69. Basic Buffers Contain a weak base and its salt formed with a strong acid Ammonia & Ammonium chloride NH 3 & NH 4 + Cl -

70. Buffer Action The important species present in the buffer solution are the undissociated weak base NH 3 and its conjugate acid NH 4 +

71. Undissociated ammonia can remove any added H + ions. NH 3 + H + → NH 4 + The conjugate acid can remove any added OH - ions NH 4 + + OH - → NH 3 + H 2 O

72. Calculating the pH of a Buffer When performing acidic buffer calculations it is assumed that The acid is completely undissociated The A - ions are formed solely from the salt

73. K a = [H + ] [A - ] [HA] HA H + + A - [H + ] = K a x [HA] [A - ] From salt Weak acid completely undissociated

74. Calculate the pH of a buffer solution produced by adding 3.28g of sodium ethanoate to 1dm 3 of 0.01M ethanoic acid. The K a of ethanoic acid is 1.84 x 10 -5 at 300K moles = mass/Mr 1. Find the number of moles of sodium ethanoate Moles = 3.28/82 = 0.04 2. Find the [CH 3 COOH] and [CH 3 COO - ] [CH 3 COOH] = 0.01M[CH 3 COO - ] = 0.04M

75. 3. Find [H + ] [H + ] = K a x [CH 3 COOH] [CH 3 COO - ] [H + ] = 1.84 x 10 -5 x 0.01 0.04 = 4.6 x 10 -6 3. Find pH pH = -log[H + ] pH = -log 4.6 x 10 -6 = 5.34

76. Calculate the pH of this buffer if 10cm 3 of 0.1M HCl are now added 1. Find the number of moles of acid added moles = conc x volume moles of acid = 0.1 x 10/1000= 1 x 10 -3 CH 3 COO - + H + → CH 3 COOH 1 x 10 -3 + 1 x 10 -3 → 1 x 10 -3

78. 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

79. Calculating the pH of an acidic buffer solution K a * concentration of acid concentration of salt [H + ] = K a * concentration of acid concentration of salt Log [H + ] = Log = Log K a + Log concentration of acid concentration of salt => (both sides put log) =>(multiply all by -1) = - Log K a - Log concentration of acid concentration of salt - Log [H + ] pKa - Log concentration of acid concentration of salt pH =

80. Calculating the pH of an acidic buffer solution Calculate the pH of a buffer whose concentration of acid is 0.40 mol dm -3 Methanoic acid and is 1 mol dm -3 sodium ethanoate Ka of ethaoic acid = 1.6*10 -4 K a * concentration of acid concentration of salt [H + ] = 1.6*10 -4 * 0.40 1 = = 6.4*10 -5 pH = -log (6.4*10 -5 ) = 4.19

81. 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.

82. 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

83. 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

84. 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

85. 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

86. 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

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88. WELL DONE! Try some past paper questions

89. REVISION CHECK What should you be able to do? Calculate pH from hydrogen ion concentration Calculate hydrogen ion concentration from pH Write equations to show the ionisation in strong and weak acids Calculate the pH of strong acids and bases knowing their molar concentration Calculate the pH of weak acids knowing their K a and molar concentration Calculate the pH of mixtures of strong acids and strong bases CAN YOU DO ALL OF THESE? YES NO

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91. WELL DONE! Try some past paper questions