3. Note that water slightly dissociates in equilibrium with hydrogen and hydroxide ions: H 2 O  H + + OH - So, the equilibrium constant is: K w = [H + ][OH - ] (product constant of water) Note that water slightly dissociates in equilibrium with hydrogen and hydroxide ions: H 2 O  H + + OH - So, the equilibrium constant is: K w = [H + ][OH - ] (product constant of water) Experimentally, it has been determined in a neutral solution at 25 0 C: [H + ] = [OH - ] = 10 -7 (a very small amount!) Experimentally, it has been determined in a neutral solution at 25 0 C: [H + ] = [OH - ] = 10 -7 (a very small amount!) So, K w = (10 -7 )(10 -7 ) = 10 -14 So, K w = (10 -7 )(10 -7 ) = 10 -14 This shows how far to the left this equilibrium is

4. In the previous slide we saw: K w = (10 -7 )(10 -7 ) = 10 -14 This value applies to 25 0 C In the previous slide we saw: K w = (10 -7 )(10 -7 ) = 10 -14 This value applies to 25 0 C K w changes with temperature variations and so does the pH! Note: this does not mean that water is more acidic at higher temperatures and more alkaline at lower temperatures. Water is still neutral. It simply means that neutral is a different pH value at different temperatures.

5. 1.Given the values in the table, what are the [H + ] and [OH - ] at 0 0 C and 100 0 C? 2.What do the values of Kw tell you about whether the ionisation of water is endothermic or exothermic? Explain. Answers: 1.At 0C : 3.38 x10 -7 ; at 100C: 7.16x10 -7 (square root of Kw) 2.An increase of temperature increases the value of Kw. This suggests the products are favoured. According to LC Principle, this means the forward reaction is endothermic (absorbs heat)

6. Since K w = (10 -7 )(10 -7 ) = 10 -14 We can use this constant value to calculate [H + ] or [OH - ] Since K w = (10 -7 )(10 -7 ) = 10 -14 We can use this constant value to calculate [H + ] or [OH - ] Calculating [H + ] An alkali with a hydroxide ion concentration of 0.01M (10 -2 ) K w = (10 -2 )[H + ] = 10 -14 [H + ] = 10 -12 Calculating [H + ] An alkali with a hydroxide ion concentration of 0.01M (10 -2 ) K w = (10 -2 )[H + ] = 10 -14 [H + ] = 10 -12 Calculating [OH - ] An acid with a hydrogen ion concentration of 0.001M (10 -3 ) K w = (10 -3 )[OH - ] = 10 -14 [OH - ] = 10 -11 Calculating [OH - ] An acid with a hydrogen ion concentration of 0.001M (10 -3 ) K w = (10 -3 )[OH - ] = 10 -14 [OH - ] = 10 -11

7. pH We have seen previously that pH = -log[H 3 0 + ] pH We have seen previously that pH = -log[H 3 0 + ] Calculating pH example 1 An alkali with a hydroxide ion concentration of 0.01M (10 -2 ) K w = (10 -2 )[H + ] = 10 -14 [H + ] = 10 -12 pH = 12 Calculating pH example 1 An alkali with a hydroxide ion concentration of 0.01M (10 -2 ) K w = (10 -2 )[H + ] = 10 -14 [H + ] = 10 -12 pH = 12 Calculating pH example 2 A substance has been found to have a hydroxide ion concentration of 10 -11 [OH - ] = 10 -11 pOH = 11 pH = 14-11 = 3 Calculating pH example 2 A substance has been found to have a hydroxide ion concentration of 10 -11 [OH - ] = 10 -11 pOH = 11 pH = 14-11 = 3 pOH This is similar to pH pOH = -log[OH - ] pOH This is similar to pH pOH = -log[OH - ] The ‘p’ is a mathematical operation that means ‘-log’ pKw Since Kw = [H+][OH-] pKw = pH + pOH = 14 pKw Since Kw = [H+][OH-] pKw = pH + pOH = 14

8. Recall the inverse of the pH calculation: pH = -log[H 3 0 + ] The inverse: [H 3 0 + ] = 10 -pH or [H + ] = 10 -pH On your calculator, this may be ‘2 nd’ LOG or ‘10 x ’ Recall the inverse of the pH calculation: pH = -log[H 3 0 + ] The inverse: [H 3 0 + ] = 10 -pH or [H + ] = 10 -pH On your calculator, this may be ‘2 nd’ LOG or ‘10 x ’ Example The pH of a solution is found to be 4.6 Find[H + ] [H + ] = 10 -pH = 10 -4.6 = 2.5x10 -5 mol dm -3 Example The pH of a solution is found to be 4.6 Find[H + ] [H + ] = 10 -pH = 10 -4.6 = 2.5x10 -5 mol dm -3

9. Problem 1 Calculate the pH of solutions with the following H 3 O + concentrations in mol dm -3 : a)10 -8 b)6.8 x 10 -3 c)0.035 Problem 2 Calculate the pH of solutions with the following OH - concentrations in mol dm -3 : a)10 -2 b)7.6 x 10 -3 c)0.055 Problem 3 Calculate the H 3 O + concentrations in solutions with the following pH values: a)0.00b)4.3c)2.35d)13.7 Problem 4 What is the pH of a 3.5M solution of H 2 SO 4 ? (assume complete ionisation) Problem 5 What is the pH of a 0.001M solution of Ba(OH) 2 ? (Hint: use balanced equation)

10. Unlike strong acids, weak acids only partially dissociate: HA + H 2 O  H 3 O + + A - The same is true for weak bases: B + H 2 O  BH + + OH - Dissociation Constants Because of these equilibria, we have known values for the ratio between products and reactants. These are known as dissociation constants: Ka – acid dissociation constant Kb – base dissociation constant Dissociation Constants Because of these equilibria, we have known values for the ratio between products and reactants. These are known as dissociation constants: Ka – acid dissociation constant Kb – base dissociation constant

11. CH 3 COOH (aq) ↔ H + (aq) + CH 3 COO - (aq) K a expression for acetic acid (ethanoic acid) above is: (constant at set temp) K a =[H + ] [CH 3 COO - ] mol dm 3 [CH 3 COOH] K a expression for acetic acid (ethanoic acid) above is: (constant at set temp) K a =[H + ] [CH 3 COO - ] mol dm 3 [CH 3 COOH] K a can be used to find the pK a (like pH) pK a =- log K a K a can be used to find the pK a (like pH) pK a =- log K a The larger the value of K a the stronger the acid (more it dissociates – breaks apart) What about pK a ? The larger the value of K a the stronger the acid (more it dissociates – breaks apart) What about pK a ?

12. CH 3 COOH (aq) ↔ H + (aq) + CH 3 COO - (aq) K a =[H + ] [CH 3 COO - ] mol dm 3 [CH 3 COOH] K a =[H + ] [CH 3 COO - ] mol dm 3 [CH 3 COOH] Using the equilibrium constant expression (above) we are able to calculate [H + ] and thus the pH of the acid CH 3 COOH (aq) ↔ H + (aq) + CH 3 COO - (aq) Initial concx 00 Final concx-y yy Using the equilibrium constant expression (above) we are able to calculate [H + ] and thus the pH of the acid CH 3 COOH (aq) ↔ H + (aq) + CH 3 COO - (aq) Initial concx 00 Final concx-y yy Assumption made based on relative concentrations of ‘x’ vs. ‘y’ to simplify calculations

13. NH 3(aq) + H 2 O (l) ↔ NH 4 + (aq) + OH - (aq) K b =[NH 4 + ] [OH - ] [NH 3 ] K b =[NH 4 + ] [OH - ] [NH 3 ] This works the same as in the acid example NH 3(aq) ↔ NH 4 + (aq) + OH - (aq) Initial concx 0 0 Final concx-y y y This works the same as in the acid example NH 3(aq) ↔ NH 4 + (aq) + OH - (aq) Initial concx 0 0 Final concx-y y y

14. Consider the following generic acid + water reaction: HA + H 2 O  A - + H 3 O + Consider the following generic acid + water reaction: HA + H 2 O  A - + H 3 O + ACID Conjugate BASE Now, if the conjugate base reacts with water in this reaction: A - + H 2 O  HA + OH - Now, if the conjugate base reacts with water in this reaction: A - + H 2 O  HA + OH -

15. From the previous slide, K a x K b = K w If we take the ‘p’ of each of these values (-log) we find, pK a + pK b = pK w = 14

17. Calculate the H 3 O + concentrations, and the pH value for the following weak acids: a)HCO 2 Hconc. = 0.0100 mol dm -3 K a = 2.04 x 10 -4 b)CH 3 CO 2 Hconc. = 0.1 mol dm -3 K a = 1.77 x 10 -5 c)HCNconc. = 1.00 mol dm -3 K a = 3.96 x 10 -10 Calculate K a for the following acids: d)HFconc. = 0.200 mol dm -3 pH = 2.23 e)HClO 2 conc. = 0.0200 mol dm -3 pH = 1.85 Calculate the OH - and H 3 O + concentrations, and the pH value for the following weak bases: f)NH 3 conc. = 0.0100 mol dm -3 K b = 1.80 x 10 -5 g)C 5 H 5 Nconc. = 0.1 mol dm -3 K b = 1.40 x 10 -9 h)CH 3 NH 2 conc. = 1.00 mol dm -3 K b = 4.38 x 10 -4

19.  A buffer is a solution that resists changes in pH when small amounts of acid or base are added to it  There are 2 types: › Acidic › Alkaline

20. An acidic buffer has a pH less than 7 It is often made from equal molar concentrations of a weak acid and it’s conjugate base Example: ethanoic acid and ethanoate ion CH 3 COOH  CH 3 COO - + H + (weak acid) (conj. base) (weak acid) (conj. base)

21. Take a 0.1M solution of ethanoic acid: CH 3 COOH  CH 3 COO - + H + CH 3 COOH  CH 3 COO - + H + At equil:(0.09583M) (0.00417M) (0.00417M) What’s the pH of this solution? Now, say you add HCl to this equilibrium. The acetate ion is the only species available to reduce the added H + and these are very low in concentration, so the pH will drop dramatically. What can you do to reduce this effect? -log [H+] = 2.38 Add acetate ions (sodium acetate) in equimolar amounts

22. CH 3 COOH  CH 3 COO - + H + If we buffer the solution by adding 1M sodium acetate, what will happen? This will have the following effect: 1. Increase the amount of acetate ions, shifting the equilibrium to the left 2. Increase the pH to 4.76 3. When adding H + ions, the extra acetate ions will react to reduce the [H + ] moving the equilibrium to the left – producing more weak acid. 4. The solution resists changes to pH, meaning it is buffered

23. CH 3 COOH  CH 3 COO - + H + Adding base results in more OH- being added to the equilibrium. This extra amount of ions is removed by:  CH 3 COOH + OH -  CH 3 COO - + H 2 O  hydroxide ions are removed by reaction with undissociated ethanoic acid, shifting equilibrium right making a weak base (CH 3 COO - ) Add OH -

24. An alternative way to make an acidic buffer is to add a strong base to the weak acid to produce high proportions of the weak acid and the conjugate base. MOH + HA   H 2 O + MA This drives the equilibrium to the right producing more salt (MA) which dissociates. This leads to: M + + OH - + HA  H 2 O + M + + A -  Adding acid  Adding acid – conjugate base A - reacts with the added H + to minimise the effect  Adding base  Adding base – weak acid HA dissociates further and the H + reacts with the OH - to minimise the effect A 2:1 molar ratio of a weak acid and a strong base of the same concentration is used to make this buffer solution.

25. pHvol. of 0.1M acetic acid (ml) vol. of 0.1M sodium acetate (ml) 3982.317.7 4847.0153.0 5357.0653.0 652.2947.8 Note: you can also get various pH buffers by changing the acid/base pair.

26. An alkaline buffer has a pH greater than 7 It is often made from a equal molar concentrations of a weak base and it’s conjugate acid Example: ammonia and ammonium ion NH 3 + H 2 O   NH 4 + + OH - (weak base) (conj. acid) (weak base) (conj. acid)

27. NH 3 + H 2 O   NH 4 + + OH - NH 3 + H 2 O   NH 4 + + OH - (weak base) (conj. acid) (weak base) (conj. acid) Due to the weak nature of ammonia, this equilibrium will be well to the left. We can create a buffered solution by adding ammonium chloride or a strong acid (like the acid buffer). What will this do to the equilibrium? Addition of ammonium ions (adding ammonium chloride) will shift the equilibrium even further to the left. The pH of this solution would be 9.25 You can also add ~half the moles of a strong acid (e.g. HCl) which will react with ammonia to form more ammonium ions (NH 3 + H +  NH 4 + )

28. NH 3 + H 2 O   NH 4 + + OH - What will happen if you add acid to this solution? Reaction of the acid with the weak base  NH 3 + H +   NH 4 +  removal by reaction with ammonia to produce more ammonium ion – a weak acid)

29. NH 3 + H 2 O   NH 4 + + OH - What will happen to the equilibrium if you add base? Adding base effectively adds OH -. This means: The ammonium ion reacts with the OH - to shift the equilibrium to the left, consuming most of the OH - ions. NH 4 + + OH -   NH 3 + H 2 O

30.  Buffer solutions resist changes in pH when acids and alkalis are added  Buffers generally contain: › Sufficient concentrations of a weak acid and it’s conjugate base OR weak base and it’s conjugate acid  The pH of buffer solutions depend on the concentrations and type of conjugate acid/base pairs that are used. Go here  http://www.pearsonhotlinks.co.uk/978043599440 2.aspx for good animations of this (chapter 8) http://www.pearsonhotlinks.co.uk/978043599440 2.aspx http://www.pearsonhotlinks.co.uk/978043599440 2.aspx

31. Commercially available buffer solutions are used to calibrate pH meters

32. HA (aq) + H 2 O (l)  A - (aq) + H 3 O + Assumptions: 1.As a large quantity of conjugate base (A - ) has been added, the equilibrium shifts far left, so that equilibrium concentration of the acid is approximately equal to the initial concentration of the weak acid : [HA] eq ≈ [HA] i = [acid] 2.The equilibrium concentration of the conjugate base ion (A - ) is approximately equal to the concentration of the salt that was added to the equilibrium. [A - ] eq ≈ [A - ] i = [salt] Assumptions: 1.As a large quantity of conjugate base (A - ) has been added, the equilibrium shifts far left, so that equilibrium concentration of the acid is approximately equal to the initial concentration of the weak acid : [HA] eq ≈ [HA] i = [acid] 2.The equilibrium concentration of the conjugate base ion (A - ) is approximately equal to the concentration of the salt that was added to the equilibrium. [A - ] eq ≈ [A - ] i = [salt]

33. So, considering our 2 assumptions: [HA] eq ≈ [HA] I = [acid] [A - ] eq ≈ [A - ] i = [salt] So, considering our 2 assumptions: [HA] eq ≈ [HA] I = [acid] [A - ] eq ≈ [A - ] i = [salt] Alternatively, you may solve for [H+] first and then solve for pH using pH=-log[H+] Alternatively, you may solve for [H+] first and then solve for pH using pH=-log[H+] When [acid] = [salt] pH= pKa When [acid] = [salt] pH= pKa

34. Similarly for a base equilibrium B (aq) + H 2 O (l)  HB + (aq) + OH - (aq) Similarly for a base equilibrium B (aq) + H 2 O (l)  HB + (aq) + OH - (aq) Alternatively, you may solve for [OH-] first and then solve for pOH using pOH=-log[OH-] Alternatively, you may solve for [OH-] first and then solve for pOH using pOH=-log[OH-] When [base] = [salt] pOH= pKb When [base] = [salt] pOH= pKb

35. An aqueous solution of 0.1M ammonia and 0.1M ammonium chloride has a pH of 9.3. The reaction is: NH 3 + H 2 O  NH 4 + + OH - a)Calculate the K b for ammonia b)If a pH of 9.0 is needed, what should be added? Explain c)Calculate the new concentration of the substance added in b) Notice the additional volume has a negligible effect on [NH3] and has been ignored

36. If you wanted to make a buffer solution of pH=4.46 using ethanoic acid and sodium ethanoate, what concentrations could you use? pKa (ethanoic acid) = 4.76. The reaction is: CH 3 COOH+ H 2 O  CH 3 COO - + H 3 O + Note: it should be expected that the acid concentration will be higher than the salt. At equal concentrations, the pH = pK a = 4.76 At pH 4.46 (a lower pH), we need to add acid to shift equil right, increasing [H 3 O + ], decreasing pH

37. An aqueous solution of 0.025M ethanoic acid and 0.050M sodium ethanoate is prepared. a)Calculate the pH of this solution given K a = 1.74x10 -5 mol dm -3 b)If 1.0cm 3 of 1.0M NaOH is added to 250cm 3 of buffer, what will happen to the pH? Notice the additional volume has again been ignored as it is assumed to have little effect on the overall pH CH 3 COOH + OH -  CH 3 COO - + H 2 O n i (mol) 0.006250.0010.0125- n c (mol) -0.001 +0.001- n e (mol) 0.0052500.0135- [ ] (mol dm -3 ) 0.02100.054-

39. Titration is an analytical technique that is used to determine the end point of a reaction (often Acid + Base) using an indicator or pH meter. Acid-Base Titration (basic steps): 1.Unknown: Accurately measure a volume of base of unknown concentration using a pipette and place into a flask 2.Indicator: Add an appropriate indicator which will change colour at the end point of the reaction 3.Titrate: Carefully add an acid of known concentration until the end point is reached as indicated by the colour change Acid-Base Titration (basic steps): 1.Unknown: Accurately measure a volume of base of unknown concentration using a pipette and place into a flask 2.Indicator: Add an appropriate indicator which will change colour at the end point of the reaction 3.Titrate: Carefully add an acid of known concentration until the end point is reached as indicated by the colour change Pipette Indicator addition Titration End point

40. Equivalence Point The equivalence point in an acid-base reaction is the point where neutralisation occurs. The molar ratios have been reached Equivalence Point The equivalence point in an acid-base reaction is the point where neutralisation occurs. The molar ratios have been reached End point The end point is where an indicator solution just changes colour permanently. End point The end point is where an indicator solution just changes colour permanently. Indicators change colours at specific pH ranges and are chosen to be as close to the equivalence point as possible. More later…

42. KeMsoft06 42 Investigating the titration between: 1M HCl and 1M NaOH

43. KeMsoft06 43 HCl 10 ml NaOH Start with 10ml of alkali and slowly add acid measuring the pH

44. KeMsoft06 44 HCl 10 ml NaOH + almost 10 ml HCl

45. KeMsoft06 45 HCl 10 ml NaOH + 10 ml HCl Now we have equivalent amounts of strong acid and strong base – notice the pH changes dramatically

46. KeMsoft06 46 HCl 10 ml NaOH Now as we continue to add acid past pH = 7, the pH drops quickly at first and then more slowly

47. KeMsoft06 47 Equivalence point at pH7 NaOH + HCl = NaCl + H 2 O 1M 1M 1NaOH + 1HCl  NaCl + H 2 O 10ml Solutions mixed in the right proportions according to the equation.

48. 48 Running acid into alkaliRunning alkali into acid This nearly horizontal section of the graph is called the buffer region. Here the pH stays relatively constant due to a buffering between the acid and the salt pH = pK a pOH = pK b The same applies to a base being neutralised by an acid

49. KeMsoft06 49 Investigating the titration between: 1M HCl and 1M NH 3

50. KeMsoft06 50 1M 1M 1NH 3 + 1HCl  NH 4 Cl 25ml pH starts to fall quickly as acid is added pH falls less quickly as buffer soln formed (excess NH 3 and NH 4 Cl present) Soln at equivalence point is slightly acidic because ammonium ion is slightly acidic NH 4 + + H 2 O  NH 3 + H 3 O + Weak base so initial pH value is less than 14 acid alkali

51. KeMsoft06 51 alkali acid Very low pH indicative of a strong acid solution After equivalence point the soln contains NH 3 and NH 4 Cl – a buffer soln and so resists large increase in pH so graph flattens out. < pH 7

52. KeMsoft06 52 Running acid into alkaliRunning alkali into acid

53. KeMsoft06 53 Investigating the titration between: 1M CH 3 COOH and 1M NaOH

54. KeMsoft06 54 1M 1M 1CH 3 COOH + 1NaOH  CH 3 COONa 25ml pH falls less quickly as buffer soln formed (excess CH 3 COOH and CH 3 COO - present) Soln at equivalence point is slightly alkaline because ethanoate ion is slightly alkaline CH 3 COO - + H 2 O  CH 3 COOH + OH - acid alkali Very high pH – indicative of a strong alkali solution

55. KeMsoft06 55 alkali acid pH starts to rise quickly as alkali is added pH rises less quickly as buffer soln formed (excess CH 3 COOH and CH 3 COO - present) Excess of alkali present – graph same as when adding strong alkali to strong acid

56. KeMsoft06 56 Running acid into alkaliRunning alkali into acid

57. KeMsoft06 57 Investigating the titration between: 1M CH 3 COOH and 1M NH 3

58. KeMsoft06 58 acid alkali 1CH 3 COOH + 1NH 3 = CH 3 COO - + 1NH 4 + About as weak as each other -

59. KeMsoft06 59 acid alkali No steep section – small addition of acid causes a large change in pH, so… Very difficult to do a titration between a weak acid and a weak base.

60. KeMsoft06 60

62.  The pH of a salt depends upon the relative strength of the ions that make up the salt.  Very few salts are neutral  Salts completely dissociate into their ions when sufficiently dilute NaCl (s)  Na + (aq) + Cl - (aq)  It is possible for these ions to interact with water to produce H + or OH - ions which results in acidic or alkaline solutions. These are known as hydrolysis reactions Remember: the stronger the acid/base, the weaker it’s conjugate

63.  Neutral anions are formed from strong acids  Neutral cations are formed from strong bases  NaCl is a neutral salt because the ions that are formed derive from a strong acid and base › HCl + NaOH  NaCl + H 2 O  Na + and Cl - are weak conjugates, so there is no tendency for these ions to undergo hydrolysis reactions. pH = 7

64. Weak acids form conjugate bases that can react with water to form hydroxide ions › H 2 CO 3 + 2NaOH  Na 2 CO 3 + H 2 O › Na 2 CO 3  2Na + + CO 3 2- › CO 3 2- + H 2 O  HCO 3- + OH - In the above reactions:  NaOH is a strong base so the weak conjugate, Na + will not react with water to form hydrogen ions  Carbonic acid is weak, so the carbonate ion will react with water to a small extent to form OH - ions Basic anions are formed by weak acids pH > 7

65. HCl + NH 3  NH 4 + + Cl-  A donatable hydrogen must be available on the cation. NH 4 + + H 2 O  NH 3 + H 3 O + NH 4 + + H 2 O  NH 3 + H 3 O + The ammonium ion comes from the weak base, ammonia, so a hydrolysis reaction can occur to a small extent producing hydronium ions. pH < 7

66.  When combining weak acids and bases, the pH of the salt will depend on their relative strengths. › If K a > K b – acidic › If K b > K a – basic › If K a = K b - neutral Example: CH 3 COOH + NH 3  NH 4 + + CH 3 COO -  Both of the salts formed in this reaction will react with water, but the effects cancel resulting in a nearly neutral solution Ammonium Acetate

67. Fe 3+ O O O O O O HH HH H H H H HH H H  + H + Fe 3+ O O O O O O HH HH H H H H HH H An O-H bond may be broken releasing a proton 2+ Small ions with multiple charges can hydrolyse water. The high charge density pulls electrons towards the metal ion causing a proton to be released, decreasing the pH Other metal ions that are able to hydrolyse water include Be 2+ and Al 3+

68. Type of reaction Example Reaction Salt produced Ion that hydrolyses water Nature of final solution Strong acid Strong Base HCl + NaOHNaClNeither ionNeutral pH = 7 Weak acid Strong Base CH 3 COOH + NaOH NaCH 3 COOAnionBasic pH>7 Strong Acid Weak Base HCl + NH 3 NH 4 ClCationAcidic pH<7 Weak Acid Weak Base CH 3 COOH + NH 3 NH 4 CH 3 COOAnion & Cation Depends on K a /K b Metal complex [Al(H 2 0) 6 ] 3+  [Al(H 2 0) 5 OH] 2+ + H + NAMetal complex with multiple charge Acidic pH<7

70. Indicators can change colours depending upon the pH of the solution they are in. Indicators are themselves weak acids or weak bases and as they lose or gain H +, they form substances that have distinct colours. Indicators can change colours depending upon the pH of the solution they are in. Indicators are themselves weak acids or weak bases and as they lose or gain H +, they form substances that have distinct colours. Let’s look at some examples… In general, an indicator molecule (In) Hin (aq)  H + (aq) + In - (aq) In general, an indicator molecule (In) Hin (aq)  H + (aq) + In - (aq)

71. HLit red “HLit” is litmus – a weak acid – is red in solution Lit-blue When hydroxide (base) is added to this weak base, “Lit-” is formed and is blue When hydrogen ions (acid) is added, we shift back to the red HLit When hydrogen ions (acid) is added, we shift back to the red “HLit”. So, bluered Acid turns blue litmus red redblue Alkali turns red litmus blue Base Acid Add base (hydroxide) Add acid (hydrogen ion)

72. Methyl orange is yellow below pH 3.7 Methyl orange is red above pH 3.7 Notice the transfer of a proton between the different coloured compounds

73. Here is a common acid-base indicator that changes colour at pH 9.3 What colour is the acid? What colour is the base?

74. Here is a common acid-base indicator that changes colour at pH 9.3 colourless What colour is the acid? colourless pink What colour is the base? pink

75. Recall the generic equation for an indicator Hin (aq)  H + (aq) + In - (aq) This means K a = [H + ] or pK a = pH This means K a = [H + ] or pK a = pH So the pKa of the indicator, tells us where the colour change will occur

76. KeMsoft06 76 Choosing an Indicator 3.79.3

77. KeMsoft06 77 Strong Acid - Strong Base

78. KeMsoft06 78 Strong Acid - Weak Base

79. KeMsoft06 79 Weak Acid - Strong Base

80. KeMsoft06 80 Weak Acid – Weak Base