1. Acids and Bases

2. What are acids? What properties do you associate with acids? What properties do you associate with bases?

3. Bronsted-Lowry Theory An acid is a proton (H + ) donor e.g. in water, hydrochloric acid gives a proton to a water molecule, forming the hydronium ion. HCl (aq) + H 2 O (l)  H 3 O + (aq) + Cl - (aq) hydronium ion A base is a proton acceptor e.g. ammonia accepts an H + ion from hydrogen chloride NH 3 + HCl 

4. What do the bases need in order to be able to accept a proton? A lone pair of electrons to form a covalent bond

5. Conjugate pairs When an acid loses a proton, it does so reversibly. This means that the resulting species can accept a proton and is thus a base. A base formed from an acid is known as the conjugate base. Similarly, the acid formed when a base reversibly gains a proton is a conjugate acid. Note: an acid can only act as an acid in solution if a base is present to accept a proton – the reactions of acids are acid-base reactions. Similarly the reactions of bases are acid-base reactions.

6. HCl + H 2 O  H 3 O + + Cl - acid 1 base 1 base 2 acid 2 H 2 O + NH 3  NH 4 + + OH - acid 1 base 1 base 2 acid 2

7. Acid-Base pairs AcidConjugate base HCl H 2 SO 4 H 2 O CH 3 COO - NH 3 BaseConjugate acid OH - H 2 O NH 3 C 2 H 5 NH 3 + H 2 SO 4

8. Practice! Identify the conjugate acid-base pairs in the following: CH 3 COOH + OH -  CH 3 COO - + H 2 O HSO 4 - + NH 4 +  H 2 SO 4 + NH 3 C 2 H 5 NH 2 + HCl  C 2 H 5 NH 3 + + Cl - H 2 SO 4 + NH 3  HSO 4 - + NH 4 +

9. Note: when an acid loses an H + ion, it does so reversibly to produce its conjugate base. The species resulting from a base gaining an H + ion is called the conjugate acid of that base.

10. Strong or weak? A strong acid or base dissociates completely in solution e.g. HCl + H 2 O  H 3 O + + Cl - for a strong acid the position of equilibrium lies so far to the right that 100% dissociation is assumed i.e. all the molecular HCl, once dissolved in water dissociates (ionises) to yield ions. Strong bases dissolve completely in water and are fully ionised or completely dissociated to form hydroxide ions. e.g.

11. E.g. hydrochloric acid

12. Note: hydrochloric acid is called a monobasic acid since i.e. they give only hydrogen ion per formula unit whereas sulphuric acid is a dibasic acid because i.e. it gives two hydrogen ions per formula unit. Strong acids includeStrong bases include hydrochloric acidoxides or hydroxides of nitric acidmetals, particularly those sulphuric acid (during 1stof group I ionisation only)

13. Weak acids and bases involve partial dissociation in solution – the interaction with water can be shown: CH 3 COOH (aq) + H 2 O (l)  CH 3 COO - (aq) + H 3 O + (aq) NH 3(aq) + H 2 O (l)  NH 4 + (aq) + OH - (aq) In these equilibria, the equilibrium position lies more to the left i.e. in a solution of ethanoic acid, for example, there are more undissociated ethanoic acid molecules than there are ethanoate and hydronium ions. Note: generally, a strong acid gives rise to a weak conjugate base whereas a weak acid gives rise to a strong conjugate base. Similarly for bases.

14. Strong acids and bases are much better conductors of electricity than weak acids or bases – why? Weak acids includeWeak bases include Ethanoic acidammonia Phosphoric (V) acidamines Ethanedioic acid Methanoic acid Carbonic acid What do we use to express how acidic or alkaline a solution is?

15. pH The pH of a solution is defined as the negative logarithm (to the base 10) of the hydrogen ion concentration in mol dm -3 pH = -log [H + ]{or pH = -log [H 3 O + ] } e.g. what is the pH with [H + ] = 1.23 x 10 -2 mol dm -3 ? Note: always give pH values to two decimal places

16. Determine the pH of the following [H + ] = 1.0 x 10 -4 M [H + ] = 2.5 x 10 -2 M [H + ] = 7.1 x 10 -11 M [HCl] = 1.5 M So, for strong acids, the initial [acid] gives the [H + ]

17. If pH = -log [H + ], then [H + ] = 10 -pH i.e the antilog of the negative pH e.g. find the hydrogen ion concentration of a solution of pH 5.20 [H+ ] = 10 -5.20 = 6.31 x 10 -6 mol dm -3 Try these: pH = 12.31pH = 1.38pH = 4.00

18. pOH The pOH of a solution is defined as the negative logarithm (to the base 10) of the hydroxide ion concentration in mol dm -3 pOH = -log [OH - ] and [OH - ] = 10 -pOH e.g. 1what is the pOH with [OH - ] = 3.6 x 10 -3 mol dm -3 ? e.g. 2determine the hydroxide ion concentration in a solution of pOH = 11.2

19. What is the pOH of a 0.24 mol dm -3 solution of sodium hydroxide? NaOH + aq → Na + (aq) + OH - (aq) pOH = -log [OH - ] =-log [0.24] = 0.62 What is the pOH of a 0.031 mol dm -3 solution of barium hydroxide? Ba(OH) 2 + aq → Ba 2+ (aq) + 2OH - (aq) pOH = -log [OH - ] =-log [2 x 0.031] =1.2

20. Kw Pure water has a pH of @25 O C therefore [H 3 O + ] = Where does the [H 3 O + ] come from? Remember that water is amphoteric. H 2 O + H 2 O  H 3 O + + OH - In other words, water dissociates to a slight (but measurable) extent. How can the [H 3 O + ] be measured?

22. An equilibrium expression can thus be written: K = [H 3 O + (aq) ] eq [OH - (aq) ] eq [H 2 O (l) ] 2 eq Combining the constants (K and the [H 2 O (l) ]– why?) gives the dissociation constant (ionic product) of water, Kw. Kw = [H 3 O + (aq) ] eq [OH - (aq) ] eq = 1.0 x 10 -14 mol 2 dm -6 at 25 o C why? Is pure water always neutral? Will the pH always be 7.00?

23. Pure water is neutral A neutral solution is defined as one in which So, [H + ] = [OH - ] = √ Kw = √ (1.0 x 10 -14 ) = 1.0 x 10 -7 mol dm -3 @25 o C therefore, the pH of a neutral solution @25 o C = -log[1.0 x 10 -7 ] = 7.00 An acidic solution is defined as one in which therefore, [H + ] > 1.0 x 10 -7 and pH < 7 An alkaline solution is defined as one in which therefore, [H + ] 7

24. Using Kw Calculate the [H 3 O + ], pH and pOH for the following solutions: i)[OH - ] = 8.4 x 10 -9 M Kw = [H 3 O + ] [OH - ] [H 3 O + ] = Kw = 1.0 x 10 -14 = 1.2 x 10 -6 mol dm -3 [OH - ] 8.4 x 10 -9 pH = -log [H 3 O + ] = -log [1.2 x 10 -6 ] = 5.92 pOH = -log [OH - ] = -log [8.4 x 10 -9 ] = 8.08 ii)[OH - ] = 6.1 x 10 -2 M

25. Summary pH = -log [H 3 O + ] [H 3 O + ] = 10 -pH pOH = -log [OH - ] [OH - ] = 10 -pOH Kw= [H 3 O + ] [OH - ] = 1.0 x 10 -14 mol 2 dm -6 @25 o C pKw = -log Kw = 14 @25 o C pH + pOH = 14 @25 o C

26. 1.What is the pH of a solution with [H 3 O + ] = 5.8 x 10 -3 M? 2.What is the pH of a solution with [OH - ] = 1.0 x 10 -4 M? 3.What is the [OH - ] of a solution of pH = 5.66? 4.What is the [H 3 O + ] of a solution of pH = 13.86? 5.What is the pOH of a solution of pH = 8.92? Answers: 1.2.24 2.10 3.4.6 x 10 -9 4.1.4 x 10 -14 5.5.08

27. What about weak acids (and bases)? To calculate the pH of a solution of a weak acid (or base), the concentration of the solution is insufficient information. The extent to which the acid (or base) is dissociated (or ionised) must be known. Remember, the ionisation is a reversible reaction: HA (aq) + H 2 O (l)  H 3 O + (aq) + A - (aq) The equilibrium expression for this reaction can be given by But [H 2 O (l) ] is a constant (as it is in huge excess), therefore is absorbed in the constant – Ka – the acid dissociation constant i.e.Ka = [H + (aq) ] eq [A - (aq) ] eq [HA (aq) ] eq

28. Using Ka The higher the value of Ka, the stronger / weaker the acid. Why? e.g HClKa = 1 x 10 7 HNO 3 Ka = 40 HFKa = 5.6 x 10 -4 CH 3 COOHKa = 1.7 x 10 -5 C 6 H 5 OHKa = 1.3 x 10 -10 Calculate the pH of a 0.01mol dm -3 solution of hydrofluoric acid..

29. When HF ionises, HF (aq) + H 2 O (l)  H 3 O + (aq) + F - (aq) I0.01 xs 0 0mol dm -3 C-x +x +x E 0.01-x x x Ka = [H 3 O + (aq) ] eq [F - (aq) ] eq [HF (aq) ] eq Since one F - ion is formed for each H 3 O + ion [H 3 O + (aq) ] = [F - (aq) ] and [H 3 O + (aq) ] [F - (aq) ] = [H 3 O + (aq) ] 2 As the degree of dissociation is small, we assume that the value of x is so small that we can ignore it, so (0.01-x) ≈ 0.01 5.6 x 10 -4 = x 2 0.01

30. So, (5.6 x 10 -4 ) x 0.01 = x 2 x = 2.37 x 10 -3 = [H 3 O + ] pH = - log [H 3 O + ] = - log 2.37 x 10 -3 = - (-2.63) = 2.63(0.01 mol dm -3 HF solution)

31. Calculate the pH of a 0.123 mol dm -3 solution of ethanoic acid. Ka = 1.74 x 10 -5 mol dm -3

32. Calculate the pH of a 0.423 mol dm -3 solution of methanoic acid. pKa = 3.75

33. Calculation of Ka from pH Calculate the acid dissociation constant for a weak acid, HA, given that a 0.106 mol dm -3 solution has a pH of 3.21. Ka = [H + ][A - ] [HA] [H + ] = 10 -pH = 10 -3.21 = 6.166 x 10 -4 mol dm -3 = [A - ] [HA] = 0.106 – (6.166 x 10 -4 ) = 0.1054 mol dm -3 Ka = (6.166 x 10 -4 ) 2 = 3.61 x 10 -6 mol dm -3 0.1054

34. Acid – base titrations The end point is reached when enough base has been added to react completely with the acid (or vice versa). The pH at the end point is not necessarily 7. For a strong acid – strong base titration, the end point has pH = 7 For strong acid – weak base titration the end point has pH < 7 why? For a weak acid – strong base titration the end point has pH > 7 why? A titration curve is a measurement of the recorded pH in a solution (on the y-axis) and volume of base added (on the x – axis) {or volume of acid added}. The pH remains virtually constant while the base is added and then the pH changed dramatically as equivalence is approached. The pH reaches a plateau in the xs alkaline region (for base added to acid).

35. pH at end point = 7 Common indicators: methyl orange, methyl red, phenolphthalein

36. pH at end point = 5 – 6why? Common indicators: methyl orange, methyl red

37. pH at end point = 8 - 9 why? Common indicator - phenolphthalein

38. Sketching titration curves Decide whether the acid and base are strong or weak Draw and label axes Calculate the volume that has to be added to reach the equivalence point (the point at which the base reacts exactly with the measured amount of acid). Note, indicators are chosen so that the end point (the point at which the indicator changes colour) coincides with the equivalence point. Estimate the pH - at the start - at the equivalence point - at the end Draw a smooth curve – make sure you know the range of the vertical section expected.

39. Explaining titration curves Consider the addition of a 0.10 mol dm -3 NaOH solution from a (burette) to 20.0 cm 3 of a 0.10 mol dm -3 solution of HCl. The pH at various point during the titration can be calculated: 1. before the addition of NaOH 2. after the addition of 10.0 cm 3 NaOH 3. after the addition of 15.0 cm 3 NaOH 4. after the addition of 18.0 cm 3 NaOH 5. after the addition of 19.0 cm 3 NaOH 6. after the addition of 19.5 cm 3 NaOH 7. after the addition of 19.8 cm 3 NaOH

40. 8. after the addition of 20.0 cm 3 NaOH 9. after the addition of 20.2 cm 3 NaOH 10. after the addition of 20.5 cm 3 NaOH 11. after the addition of 21.0 cm 3 NaOH 12. After the addition of 22.0 cm 3 NaOH So, the addition of a small amount of strong base can change the pH dramatically around the equivalence point

41. Calculation of pKa from titration curves In a titration between a strong base and a weak acid, the pH at the point when half the acid has been neutralised can be used to determine the pKa and hence Ka of the acid. HA  H + + A - Ka = [H + ][A - ]and the value of Ka is constant [HA] At the halfway point: [HA] halfway = [A - ] halfway ஃ Ka = [H + ] halfway and pKa = pH halfway

42. Indicators Indicators are weak acids in which the acid molecule, HInd, is a different colour from its conjugate base, Ind - HInd + H 2 O  Ind - + H 3 O + eq1 colour A colour B In acid solution, the equilibrium will shift ______ producing more of the __________ ________, and therefore we see colour ___ If base is added, H 3 O + + OH -  2H 2 O shifts to the ___________, thus the Ωm of eq 1 shifts _________ and we see colour ____

43. For the reaction : HInd + H 2 O  Ind - + H 3 O + K ind = [Ind - ] eq [H 3 O + ] eq [HInd] eq At the end point, [HInd] = [Ind - ] ஃ K ind = [H 3 O + ] endpoint pK ind = pH endpoint The noticeable colour change for an indicator usually occurs over a range of  1 pH unit. E.g. methyl orange (pK ind = 3.7) is red below pH 2.7, gradually changes to orange at pH 3.7 and then to yellow at pH 4.7.

44. Choosing indicators TitrationVertical range IndicatorpK ind Strong acid/strong base 3.5 – 10.5Methyl red Methyl orange Phenolphthalein 5.1 3.7 9.3 Strong acid/weak base 3.5 – 6.5Methyl red Methyl orange 5.1 3.7 Weak acid/strong base 6.5 – 10.5Phenolphthalein9.3 The choice of indicator depends on the pH at the equivalence point in the titration. The point at which the indicator changes colour is the end point. The pH range over which an indicator changes colour must lie completely within the vertical part of the titration curve.

45. Buffers Buffer solutions are defined as solutions that resist a change in pH on adding small amounts of either acid or base. An acid buffer is a solution of a weak acid and its (strong) conjugate base (i.e. weak acid + salt of weak acid). An acid buffer has a pH less than 7 e.g. ethanoic acid and sodium ethanoate (pH range 4-6) contains CH 3 COOH (pKa = 4.76) and the relatively strong conjugate base, the ethanoate ion, CH 3 COO - A basic buffer is a solution of a weak base and its (strong) conjugate acid (i.e. weak base + salt of weak base). A basic buffer has a pH greater than 7. e.g. ammonia (aq) and ammonium chloride (pH range 8-10) contains NH 3 and relatively strong conjugate acid, the ammonium ion, NH 4 + (pKa = 9.25)

46. How does a buffer work? Consider the ethanoic acid – sodium ethanoate buffer system. Which equilibria are present? What happens if a base (OH - ions) is added? What happens if an acid (H 3 O + ions) is added?

47. What about the ammonia – ammonium chloride buffer?

48. pH of buffers HA (aq) + H 2 O (l)  H 3 O + (aq) + A - (aq) Ka = [H 3 O + ][A - ] [HA] Taking the negative logs of both sides: pKa = pH – log{ [A - ] / [HA]} pKa = pH + log { [HA] / [A - ] } pH = pKa – log [acid] or [H 3 O + ] = Ka[weak acid] [base][salt of weak acid]

49. Example 1 Calculate the pH of a buffer solution made by adding 6.15g of sodium ethanoate, CH 3 COONa (molar mass 82 gmol -1 ), to 50.0cm 3 of a 1.00 mol dm -3 solution of ethanoic acid (Ka = 1.75 x 10 -5 mol dm -3

50. Example 2 Calculate the pH of a buffer solution made by adding 50 cm 3 of 1.00mol dm -3 solution of sodium hydroxide to 80 cm 3 of 1.00 mol dm -3 solution of propanoic acid, CH 3 CH 2 COOH, Ka = 1.35 x 10 -5 mol dm -3.

51. Enthalpy of Neutralisation The enthalpy of neutralisation of a weak acid is more / less exothermic than that for a strong acid.