Y13 acids & bases lta 2017

for aqa a level chemistry

1. Write down the formulae of the following acids or bases, if you can also write the equation for their dissociation when dissolved in water ; Sulphuric acid Nitric acid Ethanoic acid Potassium hydroxide Calcium hydroxide Ammonia 2) Define the following terms; a) acid b) base c) alkali 3) Draw a mind map of everything you can remember from AS about ‘Acids and Bases’ Starter Questions

2. Learning Objectives: Know the formulae of common acids and bases. State that an acid releases H + ions in aqueous solution. State that common bases are metal oxides, metal hydroxides and ammonia. State that an alkali is a soluble base that releases OH – ions in aqueous solution. Key Words: Acid, base, alkali, pH, neutral, neutralisation Acids and Bases

3. Learning Objectives: Recall and state the formulae of common acids and bases & state that an acid releases H + ions in aqueous solution. State that an alkali is a soluble base that releases OH – ions in aqueous solution. Explain the difference between strong acids and bases and concentrated vs. weak solutions of acids and bases. Describe and use the term conjugate acid-base pairs. Key Words: Acid, base, alkali, pH, neutral, neutralisation, conjugate pair Acids and Bases

4. Acid: A species that is a proton (H + ion) donor Acids The word Acid comes from the Latin ‘ acidus ’ – meaning sour. Acids have been known for hundreds of years but have only recently been understood in detail. The 3 common acids in A-Level chemistry are Sulphuric acid – Hydrochloric acid - Nitric acid – You need to know the name and formula for these H 2 SO 4 HCl HNO 3 In water acids give a pH of less than 7.0

5. Acids Some other common acids are naturally occurring ad weaker than lab acids Ethanoic (acetic) acid – CH 3 COOH – in vinegar Methanoic (formic) acid – HCOOH – in insect bites Citric acid – C 6 H 8 O 7 – in citrus fruits

6. If you look at the formulae of all acids you will notice that they all contain hydrogen; Mono-, di-, and tri- P rotic Acids H 2 SO 4 H Cl H NO 3 When an acid is added to water the acid splits into it’s component ions and releases H + ions . H Cl + aq  H + + Cl - H 2 SO 4 + aq 2 H + + SO4 2- The H + ion (proton) if the active ingredient in acids. All acids are proton donors

7. Depending on their formulae and bonding, different acids can release different numbers of protons . Acids – mono, di and tri protic H Cl is a mono protic acid as each molecule can release 1 proton. HCl ( aq ) H + ( aq ) + Cl - ( aq ) (chloride) H 2 SO 4 is a di protic acid - each molecule can release 2 protons. H 2 SO 4 ( aq ) H + ( aq ) + HSO 4 - ( aq ) ( hydrogensulphate ) HSO 4 - ( aq ) H + ( aq ) + SO 4 2- ( aq ) (sulphate ) H 3 PO 4 is a tri protic acid - each molecule can release 3 protons . H 3 P O 4 ( aq ) H + ( aq ) + H 2 PO 4 - ( aq ) ( dihydrogenphosphate ) H 2 PO 4 - ( aq ) H + ( aq ) + HPO 4 2- ( aq ) ( hydrogenphosphate ) HPO 4 2- ( aq ) H + ( aq ) + PO 4 3- ( aq ) (phosphate)  

8. Base: A species that is a proton (H + ion) acceptor Common bases are metal oxides and hydroxides: Metal Oxides – MgO , CuO Metal Hydroxides – NaOH , Mg(OH) 2 Ammonia – NH 3 is also a base as are all amines e.g. CH 3 NH 2 Bases We can use bases in every day life: For acid indigestion MgO is in milk of magnesia Ca (OH) 2 is used to treat acid soils

9. Alkali: a type of base that dissolves in water to form hydroxide (OH - ) ions. A chemical that gives a solution with a pH greater than 7.0 when dissolved in water. Common alkalis are: Sodium Hydroxide – NaOH Potassium hydroxide – KOH Ammonia – NH 3 Alkalis An alkali is a certain type of base that dissolves in water to give aqueous hydroxide ions – OH - ( aq ) E.g. NaOH + aq  Na + ( aq ) + OH - ( aq ) Alkalis are very corrosive and sometimes more dangerous than acids!

10. When the protons (H + ) from acids and the hydroxide ions (OH - ) from bases meet in solution. This is called a neutralisation reaction. Water is formed: H + ( aq ) + OH - ( aq ) H 2 O(l) E.g. HCl + NaOH  NaCl + H 2 O Acid + Base (or alkali)  Salt + Water Acid + Carbonate  Salt + Water + carbon dioxide Acid + metal  salt + hydrogen gas (redox reaction – not strictly acid-base) Neutralisation

11. Ammonia – NH 3 is a gas that dissolves in water to form a weak alkaline solution. It reacts with the water; Ammonia as a W eak Base Ammonia is a weak base because only a small proportion of the dissolved NH 3 actually reacts with the water. As indicated by the equilibrium sign   NH 3 ( aq ) + H 2 O(l) NH 4 + ( aq ) + OH - ( aq )  

12. 3) Write the equations for dissociation of the following acids in water; Nitric acid Chromic acid H 2 CrO 4 4 ) Write the full and ionic equations for the following acid-base reactions; a) sulphuric acid and solid magnesium carbonate b) sulphuric acid and aqueous potassium carbonate c) hydrochloric acid and solid calcium oxide d) nitric acid and aqueous sodium hydroxide e) hydrochloric acid and aluminium. Questions – 4 minutes!

13. Learning Objectives: Describe and use the term conjugate acid-base pairs. Key Words: Acid-base pair, alkali, pH, neutral, neutralisation Conjugate Acid-Base pairs

14. Acid: A species that is a proton (H + ion) donor Acids are Proton Donors A molecule of acid must contain a hydrogen that can be released as H + However this does not just happen on its own. Acids only release protons if there is something there to accept them. Formation of the hydronium ion from HCl and H 2 O Hydronium ions, H 3 O + - water molecules accept protons. HCl + H 2 O  H 3 O + ( aq ) + Cl - ( aq )

15. The equation for the process of HCl dissolving in water is often simplified to: HCl + ( aq )  H + ( aq ) + Cl - ( aq ) Using just H + makes it e asier to understand but remember that H + ( aq ) and H 3 O + both represent protons in aqueous solution. Acids are Proton Donors Formation of the hydronium ion from HCl and H 2 O Hydronium ions, H 3 O + - water molecules accept protons. HCl + H 2 O  H 3 O + ( aq ) + Cl - ( aq )

16. An acid-base pair: is a set of 2 species that transform into each other by gain or loss a proton. Below shows the acid-base dissociation of nitrous acid, HNO 2 . It shows that the acid-base pair are linked by H + Acid-Base Pairs Conjugate acid–base pair

17. A cid-base equilibria involve 2 acid-base pairs. The equilibrium below shows the dissociation of nitrous acid in water. HNO 2 ( aq ) + H 2 O(l) H 3 O + ( aq ) + NO 2 - ( aq ) In the forward direction: The acid HNO 2 releases a proton to form its conjugate base NO 2 - The base H 2 O accepts the proton to form its conjugate acid H 3 O + In the reverse direction: The acid H 3 O + releases a proton to form its conjugate base H 2 O - The base NO 2 - accepts the proton to form its conjugate acid HNO 2 HNO 2 and NO 2 - only differ by H + and make up acid-base pair 1 H 3 O + and H 2 O only differ by H + and make up acid-base pair 2   Acid-Base Equilibria

18. Identify the acid-base pairs in the following acid-base equilibria . HIO 3 + H 2 O H 3 O + + IO 3 - CH 3 COOH + H 2 O H 3 O + + CH 3 COO - NH 3 + H 2 O NH 4 + + OH - 2) Write and equation for the formation of the hydronium ion from hydrofluoric acid, HF, and water. Checkpoint sheet   Questions

19. Learning Objectives: Recall and state the formulae of common acids and bases & state that an acid releases H + ions in aqueous solution. State that an alkali is a soluble base that releases OH – ions in aqueous solution. Explain the difference between strong acids and bases and concentrated vs. weak solutions of acids and bases. Describe and use the term conjugate acid-base pairs. Key Words: Acid, base, alkali, pH, neutral, neutralisation, conjugate pair Acids and Bases

20. Learning Objectives: Define pH as pH = –log[H + ( aq )]. Define [H+] = 10 – pH . Convert between pH values and [H + ( aq )]. Key Words: pH, log, concentration What is pH?

21. Aqueous solutions have concentrations of H + ( aq ) ions in the range of; From 10 mol dm -3 (10 1 mol dm -3 ) To 0.000,000,000,000,001 mol dm -3 (10 -15 mol dm -3 ) The pH Scale This range is huge! It is therefore difficult to work with such numbers. So the Danish chemist Søren Sørensen devised the pH scale as a more convenient way of measuring the concentration of H + ions . This scale is still used by all scientists today! pH 1 2 3 4 5 6 7 8 9 10 11 12 13 14 [H + ( aq )]/ mol dm -3 1 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

22. If you compare the pH and [H + ] below you will see a pattern. pH is the negative power of [H + ( aq )] [H + ( aq )] = 10 -pH The pH scale is logarithmic – the difference between each successive number value is a factor of 10. E.g. the difference between pH 3 and pH 5 is a factor of 10x10 = 100 pH is equal to the negative logarithm to the base 10 of [H + ( aq )] pH = -log [H + ( aq )] The pH Scale pH 1 2 3 4 5 6 7 8 9 10 11 12 13 14 [H + ( aq )]/ mol dm -3 1 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

23. pH 1 2 3 4 5 6 7 8 9 10 11 12 13 14 [H + ( aq )]/ mol dm -3 1 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 What does a pH value mean? The relationship between pH and [H + ( aq )] is sometimes called a ‘see-saw’ As one goes up – the other goes down A low pH means a large [H + ( aq )] A high pH means a small [H + ( aq )] A pH change of 1 means [H + ( aq )] changes by 10 times An acid with a pH 2 contains 1000x more [H + ( aq )] of an acid with a pH 5

24. The table shows some examples of the relationship between pH & [H+( aq )] pH 4.52 5.79 10.63 13.41 [H + ( aq )]/ mol dm -3 10 -4.52 10 -5.79 10 -10.63 10 -13.41 Converting between pH and [H + ( aq )] The [H+( aq )] values are not shown in conventional standard form. To get them into standard form we can use our calculator. Input 10 -pH pH 4.52 5.79 10.63 13.41 [H + ( aq )]/ mol dm -3 3.02 x 10 -5 1.62 x 10 -6 2.34 x 10 -11 3.89 x 10 -14

25. Converting between pH and [H + ( aq )] Use this button on your calculator to convert between pH and [H + ( aq )] pH 4.52 5.79 10.63 13.41 [H + ( aq )]/ mol dm -3 3.02 x 10 -5 1.62 x 10 -6 2.34 x 10 -11 3.89 x 10 -14

26. Hints for pH calculations pH calculations are easy if you learn to use your calculator! Learn how to use your particular brand of calculator and stick to it! Once you have an answer, check if it looks sensible Practical : Now carry out some practice pH readings Ilpac 7.1 & 7.2

27. Questions How many more times more hydrogen ions are in a solution with a pH 2 than one of pH 4? Calculate the pH of solutions with the following hydrogen ion concentrations. To 2 d.p. 3.33 x 10 -3 mol dm -3 4.73 x 10 -4 mol dm -3 2.39 x 10 -12 mol dm -3 2.30 mol dm -3 3. Calculate [H+( aq )] of solutions with the following pH values. To 3 s.f pH 6.53 pH 2.87 pH 9.58 pH -0.12 2.48 3.32 11.6 -0.36 2.95 x10 -7 mol dm -3 1.35 x10 -3 mol dm -3 2.63 x10 -10 mol dm -3 1.32 mol dm -3

28. Learning Objectives: Explain qualitatively the differences between strong and weak acids. Explain that the acid dissociation constant, K a , shows the extent of acid dissociation. Deduce expressions for K a and p K a for weak acids. Key Words: Strong acid, weak acid, pH, K a , p K a Strong and W eak Acids

29. In aqueous solution acids dissociate and an equilibrium is set up. The equilibrium below shows the dissociation of nitrous acid in water. HNO 2 ( aq ) + H 2 O(l) H 3 O + ( aq ) + NO 2 - ( aq ) The strength of an acid is the extent to which this dissociation occurs – or how far to the right the equilibrium is. Strong acids: dissociate 100% - the equilibrium cam be assumed to be completely over to the right. There are comparatively few strong acids   Acid-Base Eqilibria

30. A weak acid: only partially dissociated in water. Many naturally occurring acids are weak acids. CH 3 COOH ( aq ) + H 2 O(l) H 3 O + ( aq ) + CH 3 COO - ( aq ) When methanoic acid dissociates in water the equilibrium lies well over to the left. There is only a small concentration of the dissociated ion compared to the un-dissociated methanoic acid molecule. We can express the equilibrium constant for the dissociation of an acid – this is called K a – the acid dissociation constant   Weak Acids

31. Learning Objectives: Calculate pH from [H + ( aq )] and [H + ( aq )] from pH for strong and weak acids. Calculate K a for a weak acid. Key Words: Strong acid, weak acid, pH, K a , p K a Calculating pH for Strong and W eak Acids

32. A strong monobasic acid HA has virtually complete dissociation in water.: HA ( aq )  H + ( aq ) + A - ( aq ) Meaning that the [H+( aq )] of a strong acid is equal to [HA( aq )] [H+( aq )] = [HA( aq )] So pH can be calculated using: pH = -log [H+( aq )] Calculating pH of Strong Acids

33. A sample of HCl has a concentration of 1.22 x 10 -3 mol dm -3 W hat is the pH? HCl ( aq )  H + ( aq ) + Cl - ( aq ) – dissociates completely Therefore; [H +( aq )] = [ HCl ( aq )] = 1.22 x 10 -3 mol dm -3 And; pH = -log [H+( aq )] So; -log( 1.22 x 10 -3 mol dm -3 ) = 2.91 Calculating pH of Strong Acids

34. A sample of HNO 3 has a pH of 5.63 W hat is the concentration of t he HNO 3 ? HNO 3 ( aq )  H + ( aq ) + NO 3 - ( aq ) – dissociates completely Therefore; [H +( aq )] = [ HNO 3 ( aq )] And; [H +( aq )] = 10 -pH = 10 -5.63 = 2.34 x 10 -6 mol dm -3 Calculating pH of Strong Acids

35. In aqueous solution, a weak monobasic acid, HA, partially dissociates, setting up the equilibrium; HA ( aq ) H + ( aq ) + A - ( aq ) K a = To work out the pH we need to find [H + ( aq )], this depends on; The concentration of the acid, [HA( aq )] The value of k a Unlike strong acids [H + ( aq )] is much less than [ HA( aq )] is dissociation not full.   Calculating pH of Weak Acids

36. HA ( aq ) H + ( aq ) + A - ( aq ) K a = However we do know that H+ and A- ions are formed in equal quantities. [ H+( aq )] = [A - ( aq )] So; [H+( aq )][ A - ( aq )] = [H+( aq )] 2 As so few HA molecules dissociate [HA( aq )] will have reduced slightly. The equilibrium, concentration of HA( aq ) will be [HA( aq )] – [H + ( aq )] Therefore; K a =   Calculating pH of Weak Acids

37. Making an Approximation: We now make an approximation to simplify th e pH calculation. Only a very small proportion of the HA dissociates in weak acids. So we can assume that the equilibrium concentration of HA will be very nearly the same as the concentration od un-dissociated HA. So the expression approximates to; K a = And therefore; [ H+( aq )] 2 = K a [HA( aq )] and; [H+( aq )] = and pH = -log[H +( aq )]   Calculating pH of Weak Acids

38. A sample of nitrous acid, HNO 2 is 0.055 mol dm -3 K a = 4.70 x 10 -4 mol dm -3 at 25 ºC. Calculate pH First find [H+( aq )]. HNO 2 ( aq ) H + ( aq ) + NO 2 - ( aq ) K a = And therefore; [H+( aq )] 2 = K a [HNO 2 ( aq )] and; [H+( aq )] = = = 5.08x10 -13 mol dm -3 and pH = -log[H +( aq )] = -log(5.08x10 -13 ) = 2.29   Worked example of Calculating pH of a Weak Acid

39. Find the pH of the following solutions of weak acids: 0.65 mol dm -3 CH 3 COOH ( Ka = 1.7x10 -5 mol dm -3 ) 4.4 x 10 -2 mol dm -3 HClO ( K a = 3.7 x 10 -8 mol dm -3 ) 2. Find the K a and pK a values for the following weak acids. 0.13 mol dm -3 solution with a pH of 3.52 7.8 x 10 -2 mol dm -3 solution with pH of 5.19 Questions

40. Learning Objectives: State and use the expression for the ionic product of water, K w Understand the importance of K w in controlling the concentration of H + ( aq ) and OH - ( aq ) in aqueous solutions. Key Words: Ionic product of water, K w , equilibrium, pH The Ionisation of Water.

41. Pure water ionises into H + ( aq ) and OH - ( aq ) ions but the extent of the ionisation is very very small – only 1 H 2 O molecule out of every 500 000 000 dissociates. H 2 O ( l ) H + ( aq ) + OH - ( aq ) The position of equilibrium is well to the left.   The Ionisation of Water. K c = so K c x [H 2 O(l)] = [H + ( aq )][OH - ( aq )]   [ H 2 O(l)] = = 55.6 mol dm -3  

42. K c and [H 2 O(l )] are both constants – combined together they give a new constant, the ionic product of water, K w . K w = K c x [H 2 O(l)] = [H + ( aq )][OH - ( aq )] constant, K w ionic product The I onic Product of Water, K w At 25 the pH of water is 7 , so [H + ( aq )] = 10 -7 mol dm -3 as [H + ( aq )] = [OH - ( aq )], so [OH - ( aq )] = 10 -7 mol dm -3 And 10 -7 mol dm -3 x 10 -7 mol dm -3 = 10 -14 mol 2 dm -6 At 25 , K w = 1.00 x 10 -14 mol 2 dm -6  

43. K w controls the balance between [H + ( aq )] and [OH - ( aq )] in all aqueous solutions. At 25 the pH of 7 is the neutral point where the [H + ( aq )] = [OH - ( aq )] = 10 -7 mol dm -3   The Significance of K w All aqueous solutions contain H + and OH - ions. In water and neutral solutions [H + ( aq )] = [OH - ( aq )] In acidic solutions [H + ( aq )] > [OH - ( aq )] In alkaline solutions [H + ( aq )] < [OH - ( aq )] The relative concentrations [H + ( aq )] and [OH - ( aq )] are determined by K w At 25 K w = [H + ( aq )][OH - ( aq )] must always equal 1 x 10 -14 mol 2 dm -6  

44. It is easy to find [H + ( aq )] & [OH - ( aq )] if you know the pH. Because [H + ( aq )][ OH - ( aq )] = 1 x 10 -14 mol 2 dm -6 Finding the values of [H + ] & [OH - ] Therefore, [H + ( aq )] = 1 x 10 -14 mol 2 dm -6 / [OH - ( aq )] And, [OH - ( aq )] = 1 x 10 -14 mol 2 dm -6 / [H + ( aq )]

45. Questions Define the term ionic product of water . The following solutions have the shown [H + ( aq )]. Calculate [OH - ( aq )]. 10 -6 mol dm -3 10 -2 mol dm -3 10 -11 mol dm -3 3. The following solutions have the shown [OH - ( aq )]. Calculate [H + ( aq )]. 10 -13 mol dm -3 10 -4 mol dm -3 10 -10 mol dm -3

46. Learning Objectives: Calculate pH from [H + ( aq )] for strong bases using K w . Calculate [H + ( aq )] from pH for strong bases using K w . Key Words: Ionic product of water, K w , base, pH, alkali pH values of bases.

47. Bases: a substance that can accept protons, reacts with an acid to form a salt and water. Alkali: a soluble base that releases OH - ions. Base Strength: The strength of a base is a measure of it’s dissociation in water to release OH - ions. NaOH is a strong base which dissociates completely in water. NH 3 is a weak base as it only partially dissociates water in an equilibrium that is well to the left. NH 3 ( aq ) + H 2 O(l) NH 4 + ( aq ) + OH - ( aq )   pH values of bases.

48. To calculate pH you need to know [H + ( aq )]. This depends on; The concentration of base The ionic product of water, K w = 1.00 x 10 -14 mol 2 dm -6 NaOH is a strong monobasic alkali. Meaning is the same as the concentration of the NaOH . NaOH ( aq ) Na + ( aq ) + OH - ( aq ) So [ NaOH ( aq )] = [OH - ( aq )]   Calculating the pH value of Strong B ases. We can find [H+( aq )] from K w and [OH - ( aq )] K w = [H + ( aq )][OH - ( aq )] So [H + ( aq )] = K w / [OH - ( aq )] And pH = -log [H + ( aq )]

49. KOH( aq ) has a concentration of 0.050 mol dm -3 . What is the pH? KOH is a strong base; [ K OH( aq )] = [OH - ( aq )] = 0.050 mol dm -3 Examples Method 1 : Find [H + ( aq )] from [OH - ( aq )] K w = [H + ( aq )][OH - ( aq )] = 1.00 x 10 -14 mol 2 dm -6 So, [H + ( aq )] = K w / [OH - ( aq )] = 1.00 x 10 -14 / 0.050 = 2.0 x 10 -13 mol dm -3 Now find pH = -log [H + ( aq )] = -log (2.0 x 10 -13 ) = 12.70

50. KOH( aq ) has a concentration of 0.050 mol dm -3 . What is the pH? KOH is a strong base; [ K OH( aq )] = [OH - ( aq )] = 0.050 mol dm -3 Examples Method 2: pOH is defined as; pOH = -log[OH - ( aq )] Now find pOH = -log [OH - ( aq )] = -0.050 ) = 1.30 To get pH subtract the pOH value from 14 pH= 14 – 1.30 = 12.70

51. All aqueous solutions contain both H + ( aq ) and OH - ( a q ). We know that K w is important in controlling the balance between [H + ] and [OH - ] in all aqueous solutions. This relationship is critical: At 25 K w = [H + (aq)][OH - (aq)] = 1.00 x 10 -14 mol 2 dm -6   Back to K w So using K w : We can find [OH - ( aq )] if we know [H + ( aq )] We can find [H + ( aq )] if we know [OH - ( aq )]

52. Questions Find the [H + ( aq )] and pH of the following alkalis at 25 2 x 10 -3 mol dm -3 of OH - ( aq ) 5.7 x 10 -1 mol dm -3 of OH - ( aq ) 3. Find the pH of the following solutions at 25 0.0050 mol dm -3 KOH( aq ) 3.56 x 10 -2 mol dm -3 NaOH ( aq ) 3. Find the concentration in mol dm -3 of OH - ( aq ) at 25 pH = 12.43 pH = 13.82  

53. Learning Objectives: Describe what is meant by a buffer solution. State that a buffer solution can be made from the weak acid and a salt of a weak acid. Explain the role of the conjugate acid-base pair in an acid buffer solution. Key Words: Buffer solution, acid , base, pH Buffer Solutions

54. Buffer Solution: a mixture that minimises pH changes on addition of small amount of acid or base. A buffer cannot prevent the pH from changing completely, but pH changes are minimised . At least for as long as some of the buffer solution remains. A buffer solution is a mixture of: A weak acid , HA Its conjugate base , A - It can be made from a weak acid and a salt of that weak acid. e.g. ethanoic acid, CH 3 COOH and sodium ethanoate CH 3 COONa Buffer Solutions

55. In the ethanoic acid, CH 3 COOH and sodium ethanoate CH 3 COONa buffer system; the weak acid dissociates partially; CH 3 COOH ( aq ) H + ( aq ) CH 3 COO - ( aq ) The salt dissociates completely, generating the conjugate base; CH3COO - Na + ( aq )  CH3COO - ( aq ) + Na + ( aq ) The mixture formed contains a lot of the un-dissociated weak acid CH 3 COOH and its conjugate base CH 3 COO - . The high concentration of the conjugate base pushes the equilibrium to the left so [H + ] is very small.   Buffer Solutions

56. The resulting buffer solution contains large ‘reservoirs’ of the weak acid and the conjugate base; Buffer Solutions

57. The weak acid and the conjugate base are both responsible for controlling pH. The buffer minimises the pH changes by using the equilibrium; HA( aq ) H + ( aq ) + A - ( aq ) The weak acid, HA, removes added alkali. The conjugate base, A - , removes added acid.   How do Buffers Act? When an acid (H + ) is added to a buffer solution; [H + ( aq )] increases The conjugate base, A - , reacts with the H + ions The equilibrium shifts to the left, removing most of the added H + ions HA( aq ) H + ( aq ) + A - ( aq )  

58. The weak acid and the conjugate base are both responsible for controlling pH. The buffer minimises the pH changes by using the equilibrium; HA( aq ) H + ( aq ) + A - ( aq ) The weak acid, HA, removed added alkali. The conjugate base, A - , removes added acid.   How do Buffers Act? When an alkali (OH - ) is added to a buffer solution; [OH - ( aq )] increases The small concentration of H + ions react with the OH - ions H + + OH -  H 2 O HA dissociates shifting the equilibrium to the right to restore most of the H + ions that have reacted HA( aq ) H + ( aq ) + A - ( aq )  

59. How do Buffers Act? Shifting the buffer equilibrium

60. Questions What is meant by the term buffer solution . 2. A buffer is made from propanoic acid and sodium propanoate . Explain using equations the role of the conjugate acid-base pair in this buffer solution.

61. Learning Objectives: Calculate the pH of a buffer solution from the K a value of a weak acid and the equilibrium concentrations of the conjugate acid–base pair. Explain the role of the carbonic acid– hydrogencarbonate ion system as a buffer in the control of blood pH. Key Words: Buffer solution, acid , base, pH pH values of Buffer Solutions

62. The pH of a buffer solution depends on; The acid dissociation constant, K a , of the buffer system. The concentration ratio of the weak acid to its conjugate base. There are two methods for doing this. Method 1: K a = So, [H + ( aq )] = K a x Only a very small proportion of HA dissociates so we assume; [HA( aq )] equilibrium = [HA( aq )] undissociated and once [H + ( aq )] is known we find pH from -log[H + ( aq )]   Calculations involving Buffer Solutions

63. The pH of a buffer solution depends on; The acid dissociation constant, K a , of the buffer system. The concentration ratio of the weak acid to its conjugate base. There are two methods for doing this. Method 2: the Henderson- Hasselbalch method pH= p K a + log Where p K a = - log K a   Calculations involving Buffer Solutions

64. Calculate the pH at 25 of a buffer containing 0.050 mol dm -3 CH 3 COOH( aq ) and 0.10 mol dm -3 CH 3 COO - Na + ( aq ). For CH 3 COOH( aq ) K a = 1.7 x 10 -5 mol dm -3   Worked example Using Method 1: [H + ( aq )] = K a x = 1.7 x 10 -5 x = 8.5 x 10 -6 mol dm -3 pH = -log( 8.5 x 10 -6 ) = 5.07  

65. Calculate the pH at 25 of a buffer containing 0.050 mol dm -3 CH 3 COOH( aq ) and 0.10 mol dm -3 CH 3 COO - Na + ( aq ). For CH 3 COOH( aq ) K a = 1.7 x 10 -5 mol dm -3   Worked example Using Method 2: pH= p K a + log = - log K a + log = 4.77 + 0.30 = 5.07  

66. Questions Two different buffer solutions, A and B, were made using CH 3 COOH and CH 3 COONa. Ethanoic acid had a K a of 1.7 x 10 -5 mol dm -3 at 25 degrees. Calculate the pH of each buffer solution if; A has the same concentrations of the acid and the salt B has the concentrations 0.75 mol dm -1 of acid and 0.25 mol dm -3 of the salt 2. What ratio of [ CH 3 COOH ]:[CH 3 COONa] Would be needed to make a solution that buffers at a pH of 5.47

67. Human blood plasma needs to have a pH of between 7.35 and 7.45 . If it falls below a condition called acidosis is experienced. If it goes above it is called alkalosis The carbonic acid- hydrogencarbonate buffer Blood pH is controlled by a mix of buffers. This one is the most important. Carbonic acid, H 2 CO 3 is the weak acid Hydrogencarbonate , HCO 3 - is the conjugate base H 2 CO 3 ( aq ) HCO 3 - (aq) + H + ( aq )   Any increase in H + is removed by the base and the equilibrium shifts to the left. Any increase in OH - is removed by the acid ions reacting with the OH - to form water H + + OH -  H 2 O More H 2 CO 3 dissociates and equilibrium moves to the right to restore H +

68. The carbonic acid- hydrogencarbonate buffer H 2 CO 3 ( aq ) HCO 3 - (aq) + H + ( aq ) K a for this equilibrium is 4.3 x 10 -7 mol dm -3 Most materials released into the blood are acidic. The hydrogencarbonate ions remove these by being converted to H 2 CO 3 . The carbonic acid is converted to dissolved CO 2 through the action of an enzyme and it is excreted ay the lungs as CO 2 gas. The CO2 content of the blood can be changed by changing breathing rate. Faster and heavier breathing removes more CO 2 Calmer, slower breathing removes less CO 2  

69. The carbonic acid- hydrogencarbonate buffer H 2 CO 3 ( aq ) HCO 3 - (aq) + H + ( aq ) The blood contains about 10 times more HCO 3 - ( aq ) than H 2 CO 3 ( aq ). The pH of healthy blood is 7.4 Therefore [H + ( aq )] = 10 -7.4 = 3.98 x 10 -8 mol dm -3 [H + ( aq )] = K a x = = 4.3 x 10 -7 / 3.98 x 10 -8 = 10.8 / 1  

70. Questions 1. Calculate the pH of a buffer solution containing 0.15 mol dm -3 methanoic acid HCOOH( aq ), and 0.065 mol dm -3 sodium methanoate , HCOONa . For HCOOH K a = 1.6 x 10 -4 mol dm -3 2. If extra acid is produced in the blood how do buffers prevent the pH from falling.

71. Learning Objectives: Interpret and s ketch acid-base titration pH curves for strong and weak acids and bases. Explain the choice of suitable indicators for acid-base reactions. Key Words: , acid , base, pH, Equivalence point, end point, Neutralisation – Titration Curves

72. Neutralisation – Titration Curves In a titration, a solution of known concentration is titrated with a solution of unknown concentration . An acid and a base can be titrated to find the concentration of one of them. During an acid-base titration, an indicator is often used to show when the solution has been neutralized. This point is called the equivalence point . If a pH meter is used, a pH curve can be plotted. This is a graph showing how the pH of a solution changes as base (or acid) is added.

73. pH Curves