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Acids and Bases. Objectives Theories of acids and Bases Define acids and bases according to the Brønsted – Lowry and Lewis theories Deduce whether or.

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Presentation on theme: "Acids and Bases. Objectives Theories of acids and Bases Define acids and bases according to the Brønsted – Lowry and Lewis theories Deduce whether or."— Presentation transcript:

1 Acids and Bases

2 Objectives Theories of acids and Bases Define acids and bases according to the Brønsted – Lowry and Lewis theories Deduce whether or not a species could act as a Brønsted – Lowry and / or a Lewis acid or base Deduce the formula of the conjugate acid (or base) of any Brønsted – Lowry base (or acid)

3 The history of acid-base theory There have been many attempts to define acids and bases. The sour taste and the effect on vegetable colourings, such as litmus, characterized acids. The soapy feel and detergent power characterized alkalis. Acids seem to react with alkalis and also with some other compounds to give salts. The term base replaced the term alkali as meaning the opposite of an acid.

4 A base was defined as a substance which would react with an acid to form a salt and water. An acid was defined by Liebig in 1838 as a compound that contains hydrogen that can be displaced by a metal. A big advance was the Ostwald-Arrhenius theory of electrolytic dissociation in They defined acids as substances that produce hydrogen ions in solution. Bases, they said, produce hydroxide ions in solution and neutralize acid by the reaction: H + + OH - H 2 O

5 Before long this theory ran into difficulties. It was noted that, while hydrochloric acid conducts electricity, pure hydrogen chloride does not. What is the significance of conducting electricity? Should hydrogen chloride be classified as an acid or does it only become an acid on contact with water? Also bases such as ammonia neutralize acids by picking up hydrogen ions, rather than by providing hydroxide ions. NH 3 +H + NH 4 +

6 As work proceeded, doubt was cast on the very existence of the hydrogen ion, H +, in solution. The proton H +, is very small ( m diameter) compared with other cations (around m diameter). The electric field around it is so intense that it attracts any molecule with unshared electrons, such as H 2 O The reaction H + + H 2 O H 3 O + was shown to liberate 1300kJ mol -1. As the reaction was so exothermic, it gave evidence that unhydrated protons can not exist in solution. The hydrated proton, H 3 O +, is called the oxonium ion, and is also referred to as the hydrogen ion.

7 The Brønsted-Lowry definition of an acid The Brønsted – Lowry definition of acids and bases are about competition for protons, H + This defines an acid as a substance that can donate a proton (a H + ion ) and a base as a substance that can accept a proton. Acids and bases can only react in pairs – one acid and one base.

8 Example 1 Ammonia acts as a base in water by accepting a proton from a molecule of water, using its lone pair to form the bond. In this case the water is acting as an acid + N H H H: O H H : : N+N+ H H HH : O - H: : +

9 Example 2 Hydrogen chloride is a covalently bonded gas, but when it dissolves in water, the water acts as a base and hydrogen chloride donates a proton to a molecule of water, acting as an acid. HCl+H2OH2OH3O+H3O+ + Cl - A proton has no electrons at all, so to form a covalent bond with another species, that species must have a lone pair of electrons. This is the case with water: O H H : :

10 Conjugate acid-base pairs The Cl - ion left after HCl has donated a proton is itself able to accept a proton (to go back to HCl) and is therefore a base. It is called the conjugate base of HCl. In the same way, the NH 4 + ion is an acid as it can donate a proton to return to being NH 3. It is the conjugate acid of ammonia. So we have two acid-base pairs. HCl+NH 3 Cl - +NH 4 + acidbaseconjugate baseconjugate acid

11 Water as an acid and a base HCl can donate a proton to water, so that water acts as a base. Write the equation to show this. Water may also act as an acid. For example

12 Water as an acid and a base HCl can donate a proton to water, so that water acts as a base. Write the equation to show this. HCl+H2OH2OH3O+H3O+ +Cl - Water may also act as an acid. For example H2OH2O +NH 3 OH - + NH 4 + Here OH - is the conjugate base and H 2 O the acid Water is described as an amphoteric or amphiprotic solvent

13 Quick questions 1. Hydrogen bromide, HBr, is acidic. What is its conjugate base? 2. OH- is a base. What is its conjugate acid? 3. Identify the conjugate acid/base pairs in: HNO 3 + OH- → NO H 2 O 4. Magnesium oxide is described as basic whereas sodium hydroxide is an alkali. Explain the difference? 5. What species are formed when the following bases accept a proton? a) OH - b) NH 3 c) H 2 O d) Cl -

14 The Lewis definition of acids and Bases There are reactions which appear to us, on common-sense grounds, to be acid-base reactions, and which do not come within the scope of the Brønsted-Lowry definition. Such reactions are: CaO+SO 3 CaSO 4 NH 3 +BF 3 NH 3 BF 3

15 To accommodate reactions of this type, G N Lewis (during 1930 to 1940) proposed a fresh definition of acids and bases. He described acid-base reactions as reactions in which an unshared electron pair in the base molecule is accepted by the acid molecule, with the formation of a covalent bond. For example, in this reaction N H H H :+ B F F F N H H H B F F F Ammonia is the base, and boron trifluoride is the acid

16 The Lewis definition of a base includes the Brønsted-Lowry bases because a species with a lone pair of electrons will accept a proton from a Brønsted-Lowry acid. Lewis acids, such as BF 3 and SO 3, are not acids in the Brønsted-Lowry sense, and acids such as HCl, H 2 SO 4 and CH 3 CO 2 H are not acids according to the Lewis definition The Brønsted-Lowry description of acids and bases lends itself readily to a quantitative treatment of the strengths of acids and bases. No such quantitative treatment is possible for Lewis acids and bases.

17 Objectives Strong and weak acids and bases Distinguish between strong and weak acids and bases in terms of the extent of dissociation, reaction with water and electrical conductivity State whether a given acid or base is strong or weak Distinguish between strong and weak acids and bases, and determine the relative strengths of acids and bases using experimental data

18 Strong acids The acidity of a substance depends on the concentration of H + (aq). This is what pH measures. When acids such as hydrochloric and nitric acid dissolve into water they break up completely into ions. This is called complete dissociation. For example:

19 Strong acids The acidity of a substance depends on the concentration of H + (aq). This is what pH measures. When acids such as hydrochloric and nitric acid dissolve into water they break up completely into ions. This is called complete dissociation. For example: HCl (g) H + (aq) + Cl - (aq) HNO 3(aq) H + (aq) + NO 3 - (aq)

20 Acids that completely dissociate into ions in aqueous solutions are called strong acids. The word strong refers only to how much the acid dissociates and not in any way to how concentrated it is. Can you think of a reason why hydrogen chloride dissolved into an organic liquid does not have acidic qualities?

21 Weak acids Weak acids are not fully dissociated when dissolved in water. Ethanoic acid (the acid in vinegar, also called acetic acid) is a good example. In a 1 mol dm -3 solution of ethanoic acid, only about 4 in every thousand ethanoic acid molecules are dissociated into ions. We say the degree of dissociation is 4/1000; the rest remain dissolved as covalently bonded molecules.

22 In fact an equilibrium is set up: CH 3 CO 2 H(aq)H+(aq)+CH 3 CO 2 -(aq) Before dissociation at equilibrium Acids like this are called weak acids – weak refers only to the degree of dissociation. We can have a very dilute solution of a strong acid, or a very concentrated solution of a weak acid. Strength and concentration are independent.

23 Bases In aqueous solutions, bases dissociate to produce the OH - ion (which can react with a H + ion by forming a bond through the electrons of one of its lone pairs). Bases can also be classified as weak or strong. Strong bases are completely dissociated into ions in aqueous solutions, while weak bases are only partially dissociated.

24 Dissociation of strong and weak bases Sodium hydroxide is a strong base; NaOH (aq) Na + (aq) + OH - (aq) A solution of ammonia in water (NH 3 (aq) ), also called ammonium hydroxide, NH 4 OH, is a weak base: NH 3(aq) + H 2 O (l) NH 4 + (aq) +OH - (aq) Or NH 4 OH (aq) NH 4 + (aq) + OH - (aq)

25 Examples of strong acids and bases Examples of strong acids and bases include: Strong acidsStrong bases

26 Examples of strong acids and bases Examples of strong acids and bases include: Strong acidsStrong bases HCl – hydrochloric acidNaOH – sodium hydroxide HNO 3 – nitric acidKOH – potassium hydroxide H 2 SO 4 – sulphuric acidBa(OH) 2 – barium hydroxide Note: because one mole of HCl produces one mole of H+ ions it is known as monoprotic (monobasic). Sulphuric acid is known as a diprotic (dibasic) acid, as one mole of sulphuric acid produces two moles of hydrogen ions.

27 Examples of weak acids and bases Weak acids and bases are only slightly dissociated (ionized) into their ions in aqueous solution: Weak acidsWeak bases

28 Examples of weak acids and bases Weak acids and bases are only slightly dissociated (ionized) into their ions in aqueous solution: Weak acidsWeak bases CH 3 COOH - Ethanoic acidNH 3 – ammonia H 2 CO 3 - Carbonic acidC 2 H 5 NH 2 - aminoethane

29 Experiments to distinguish between strong and weak acids and bases It is important for us to be able to distinguish between strong and weak acids and bases, ways in which we can do this are: pH measurement Conductivity measurement Using universal indicator Observation from vigour of reaction

30 pH measurement Because a strong acid produces a higher concentration of hydrogen ions than a weak acid, with the same concentration, the pH of a strong acid will be lower than a weak acid. Similarly a strong base will have a higher pH than a weak base, with the same concentration. The most accurate way to determine pH is to use a pH meter 0.1 moldm -3 HCl (aq) pH = moldm -3 CH 3 COOH (aq) pH = 2.9

31 Conductivity measurement Strong acids and strong bases in solution will give much higher readings on a conductivity meter than equimolar (equal concentration) solutions of weak acids or bases, because they contain more ions in solution Using Universal indicator Strong acids produce a red colour with universal indicator weak acids produce orange/yellow colours Strong bases produce a purple colour with U.I. weak bases produce a blue colour

32 Observation from reaction Reacting an acid with a carbonate or a hydrogen carbonate will produce carbon dioxide gas. The reaction will be more vigorous with a stronger acid compared to a weaker acid. HCl + Na 2 CO 3 2NaCl + H 2 O + CO 2 will be more vigorous than 2CH 3 COOH + Na 2 CO 3 2CH 3 COONa + H 2 O + CO 2

33 Acid – base reactions and Le Chatelier’s principle It is not possible to distinguish between strong and weak acids and bases by a simple acid – base titration. 25cm 3 of moldm -3 ethanoic acid will react with exactly the same volume of 0.100moldm -3 alkali as 25cm 3 of moldm -3 hydrochloric acid. Write out each equation, and calculate the volume of alkali needed if the alkali used was sodium hydroxide. Using Le Chatelier’s principle can you explain why a weak acid and a strong acid react equally even though the pH (hydrogen ion concentration) varies?

34 Acid – base reactions and Le Chatelier’s principle CH 3 COOH + NaOH → CH 3 COONa + H 2 O HCl + NaOH → NaCl + H 2 O The volume of alkali in both cases will be 25 cm 3 Although there are fewer hydrogen ions in the ethanoic acid solution, once they react with hydroxide ions then, according to Le Chatelier’s principle, more of the acid dissociates to replace them until eventually all of the acid has been neutralized. 25cm mol dm -3 x cm 3

35 Titrations The only difference observed in titrations involves temperature rather than volume. Strong acids and bases in aqueous solution always release the same amount of heat per mole of hydrogen ions neutralized, because the only reaction taking place is the neutralisation of hydrogen ions by hydroxide ions. Titrations involving weak acids or bases will release a different amount of heat, because heat energy is required to dissociate the molecules to form ions, and heat energy is released when these ions become hydrated, in addition to the heat evolved during neutralisation.

36 Objectives Outline the characteristic properties of acids and bases in aqueous solution With reactive metals With neutralization reactions with bases With carbonates With hydrogen carbonates With indicators

37 Reactions of acids All acids react to form salts and the reactions may be written as balanced or ionic equations. Metal+acidsalt+hydrogen Write the balanced symbol equation for the reaction between magnesium and hydrochloric acid Mg (s) +2HCl (aq) MgCl 2 (aq) +H 2 (aq) Metals above hydrogen in the reactivity series react with acids to produce hydrogen gas and the metal salt

38 Reactions of acids All acids react to form salts and the reactions may be written as balanced or ionic equations. Metal+acidsalt+hydrogen Write the balanced symbol equation for the reaction between magnesium and hydrochloric acid Metals above hydrogen in the reactivity series react with acids to produce hydrogen gas and the metal salt

39 Metal oxide+acidsalt+ water NaOH (aq) + HCl (aq) NaCl (aq) + H 2 O (l) Metal hydroxide +acidsalt+water MgO (s) + 2HCl (aq) MgCl 2 (aq) +H 2 O (l) Acids react with bases such as metal oxides and hydroxides to form a salt and water. For example: Can you write a balanced symbol equation showing an example of each of these reactions?

40 Metal oxide+acidsalt+ water Metal hydroxide +acidsalt+water Acids react with bases such as metal oxides and hydroxides to form a salt and water. For example: Can you write a balanced symbol equation showing an example of each of these reactions?

41 If both the acid and the base are strong,– that is, they both completely dissociate in water – for example, the reaction between nitric acid and sodium hydroxide can be written: HNO 3(aq) + NaOH (aq) NaNO 3(aq) + H 2 O (aq) This can also be written as: H + (aq) + NO 3- (aq) + Na + (aq) + OH - (aq) Na + (aq) + NO 3- (aq) + H 2 0 (l) We can remove the spectator ions to leave us with: H + (aq) + OH - (aq) H 2 0 (l)

42 If both the acid and the base are strong,– that is, they both completely dissociate in water – for example, the reaction between nitric acid and sodium hydroxide can be written: HNO 3(aq) + NaOH (aq) NaNO 3(aq) + H 2 O (aq)

43 The nitrate ions and the sodium ions are ‘spectator ions’, and so the only reaction that is actually taking place is the neutralization of hydrogen ions by hydroxide ions to form water molecules. The enthalpy change for the reaction is -57.3kJ mol -1, and it will therefore always be the same when a strong acid is neutralized by a strong base in aqueous solution. A base can be defined as a substance that neutralizes an acid to form a salt and water, so other acid-base reactions include the reactions of acids with most metal oxides.

44 All carbonates and hydrogen carbonates will neutralize acids to produce carbon dioxide and water. For example: CaCO 3(s) + 2HCl (aq) CaCl 2(aq) + H 2 O (l) + CO 2(g) 2NaHCO 3(aq) + H 2 SO 4(aq) NaSO 4(aq) 2H 2 O (l) + 2CO 2(g) The carbonate ion (the conjugate base of the hydrogen carbonate ion, HCO 3 - ) and the hydrogen carbonate ion (the conjugate base of carbonic acid, H 2 CO 3 ) both behave as Brønsted-Lowry bases and accept protons from the acid Metal carbonate+acidsalt+water+carbon dioxide Metal hydrogen carbonate +acidsalt+water+carbon dioxide

45 All carbonates and hydrogen carbonates will neutralize acids to produce carbon dioxide and water. For example: The carbonate ion (the conjugate base of the hydrogen carbonate ion, HCO 3 - ) and the hydrogen carbonate ion (the conjugate base of carbonic acid, H 2 CO 3 ) both behave as Brønsted-Lowry bases and accept protons from the acid Metal carbonate+acidsalt+water+carbon dioxide Metal hydrogen carbonate +acidsalt+water+carbon dioxide

46 With indicators Acid- base indicators can be used to determine whether or not a solution is acidic. Common indicators include: IndicatorColour in acidic solution Colour in alkaline solution Litmus phenolphthalein Methyl orange

47 With indicators Acid- base indicators can be used to determine whether or not a solution is acidic. Common indicators include: IndicatorColour in acidic solution Colour in alkaline solution Litmusredblue phenolphthaleincolourlesspink Methyl orangeredyellow

48 Characteristic reactions of bases The simplest definition of a base is that it is a substance that can neutralize an acid. A base that is soluble in water is known as an alkali. Typical reactions of bases in solution include the following. Neutralization of acids Displacement of ammonia from ammonium salts With indicators

49 Displacement of ammonia from ammonium salts NH 4 Cl (s) + NaOH (aq) NaCl (aq) + NH 3(g) + H 2 O (l) Ammonia is very soluble in water, so it may be formed in solution, or, if small amounts of sodium hydroxide are added to solid ammonium chloride, it can be collected as a gas. In this reaction the chloride ions and sodium ions are spectator ions, so the actual reaction taking place in solution is: NH 4 + (aq) + OH - (aq) NH 3(g) + H 2 O (l)

50 Displacement of ammonia from ammonium salts NH 4 Cl (s) + NaOH (aq) NaCl (aq) + NH 3(g) + H 2 O (l) Ammonia is very soluble in water, so it may be formed in solution, or, if small amounts of sodium hydroxide are added to solid ammonium chloride, it can be collected as a gas. In this reaction the chloride ions and sodium ions are spectator ions, so the actual reaction taking place in solution is:

51 The equilibrium lies very much on the right, so a single arrow has been shown, but technically the reaction should be written NH 4 + (aq) + OH - (aq) NH 3(g) + H 2 O (l) Under the Brønsted-Lowry definition, ammonium ions (NH 4 + ) and water molecules (H 2 O) are acting as acids, and hydroxide ions (OH - ) and ammonia molecules (NH 3 ) are acting as bases. As water is a weak acid, its conjugate bases is strong, so the position of equilibrium lies very much to the right.

52 With indicators If a base is insoluble in water, for example copper (II) oxide (CuO), then there will be no reaction with indicators. Indicators actually measure the concentration of hydrogen ion in aqueous solution, H + (aq), so only soluble bases (alkalis) that produce hydroxide ions in water will change the colour of indicators. Increasing the concentration of hydroxide ions will lower the concentration of hydrogen ions H + (aq) + OH - (aq) H 2 O (l)

53 Objectives The ionisation of water Understand how water ionises and give the ionic product of water at 298K AHL State the expression for the ionic product of water (Kw). Deduce [H + (aq)] and [OH - (aq)] for water at different temperatures given Kw values.

54 The ionisation of water Water is slightly ionised: H 2 O (l) H + (aq) + OH - (aq) Or this may be written: H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) This equilibrium is established in water and all aqueous solutions, write the equilibrium expression for this reaction: [H + (aq) ] eqm [OH - (aq) ] eqm [H 2 O (l) ] eqm Kc=

55 The ionisation of water Water is slightly ionised: H 2 O (l) H + (aq) + OH - (aq) Or this may be written: H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) This equilibrium is established in water and all aqueous solutions, write the equilibrium expression for this reaction:

56 The equilibrium lies very much over to the left hand side and so we say the concentration of H 2 O stays constant. To avoid having two constants (Kc and the concentration of water) in the same expression, a new equilibrium constant is defined. This is called the ionic product for water and is given the symbol Kw Kw = [H + (aq) ] eqm [OH - (aq) ] eqm The value of Kw is usually taken to be 1x mol 2 dm -6 at 298K

57 Each H 2 O that dissociates (splits up) gives rise to one H + and one OH - so, in pure water at 298K [H + (aq) ] eqm = [OH - (aq) ] eqm So, 1x = [H + (aq) ] 2 eqm [H + (aq) ] = 1x10 -7 mol dm -3 = [OH - (aq) ] eqm This means that only 2 in every 1000,000,000 water molecules is split up into hydrogen and hydroxide ions!!!

58 Kw and temperature If Kw is 5.13 x mol 2 dm -6 what is the concentration of [H + ] Kw = [H + ] 2 [H + ] 2 = 5.13 x [H + ] = 5.13 x [H + ] =

59 Objectives The pH scale Distinguish between aqueous solutions that are acidic, neutral or alkaline using the pH scale Identify which of two or more aqueous solutions is more acidic or alkaline using pH scales. State that each change of one pH unit represents a 10-fold change in the hydrogen ion concentration [H + (aq) ] Deduce changes in [H + (aq) ] when the pH of a solution changes by more than one pH unit

60 The pH Scale The acidity of a solution depends on the concentration of H + (aq) and is measured on the pH scale. pH is defined as –log 10 [H+ (aq) ] Remember that square brackets, [ ], mean the concentration in mol dm -3. This expression is more complicated than simply stating the concentration of H + (aq), but it does away with awkward numbers like , which occur because the concentration of H + (aq) solutions is so small. The minus sign makes almost all pH values positive (because the logs of numbers less than 1 are negative).

61 On the pH scale: The smaller the pH, the greater the concentration of H+ (aq) A difference of one pH number means a tenfold difference in acidity, so that pH 2 is ten times as acidic as pH 3 pH measures alkalinity as well as acidity, because as [H + (aq) ] goes up, [OH - (aq) ] goes down. If a solution contains more H + (aq) than OH - (aq), its pH will be less than 7 and we call it acidic. If a solution contains more OH - (aq) than H + (aq), its pH will be greater than 7 and we call it alkaline

62 Finding the pH of strong acids and bases Acids that completely dissociate into ions in aqueous solutions are called strong acids. The word strong refers only to the extent of dissociation and not in any way to the concentration. So it is perfectly possible to have a very dilute solution of a strong acid. The same argument applies to bases. Strong bases are completely dissociated into ions in aqueous solutions.

63 [H+ (aq) ] =1x10 -3 mol dm -3 pH[H + ] The pH scale. We can work out the conc n of hydrogen ions, [H+], in an aqueous solution if we know the pH. It is the antilog of the pH value For example, an acid has pH = 3 pH = –log 10 [H+ (aq) ] 3 = –log 10 [H+ (aq) ] – 3 = log 10 [H+ (aq) ] Take the antilog of both sides Complete the [H+] for pH 0,1,6,7,8,13 and 14

64 [H+ (aq) ] =1x10 -3 mol dm -3 pH[H + ] x x x x x x The pH scale. We can work out the conc n of hydrogen ions, [H+], in an aqueous solution if we know the pH. It is the antilog of the pH value For example, an acid has pH = 3 pH = –log 10 [H+ (aq) ] 3 = –log 10 [H+ (aq) ] – 3 = log 10 [H+ (aq) ] Take the antilog of both sides 1x10 -3 Complete the [H+] for pH 0,1,6,7,8,13 and 14

65 Recall [H + (aq) ] [OH - (aq) ]= 1x mol 2 dm -6 [OH - (aq) ] [1x ] = 1x mol 2 dm -6 [OH - (aq) ] = 1x /1x =1x10 -4 moldm -3 [H+ (aq) ] =1x mol dm -3 [OH - ]pH[H + ] pH = –log 10 [H+ (aq) ] 10 = –log 10 [H+ (aq) ] -10 = log 10 [H+ (aq) ] Take the antilog of both sides Copy and complete [OH+] for pH = 0, 1, 6,7, 8, 13 and 14 With bases we need two steps, suppose the pH of a solution = 10

66 Recall [H + (aq) ] [OH - (aq) ]= 1x mol 2 dm -6 [OH - (aq) ] [1x ] = 1x mol 2 dm -6 [OH - (aq) ] = 1x /1x =1x10 -4 moldm -3 [H+ (aq) ] =1x mol dm -3 [OH - ]pH[H + ] x x x x x x x x x x x x pH = –log 10 [H+ (aq) ] 10 = –log 10 [H+ (aq) ] -10 = log 10 [H+ (aq) ] Take the antilog of both sides 1x10 -3 Copy and complete [OH+] for pH = 0, 1, 6,7, 8, 13 and 14 With bases we need two steps, suppose the pH of a solution = 10 1x10 -4

67 Acid Rain Rain is naturally acidic, because it dissolves and reacts with carbon dioxide as it falls through the air. However, carbon dioxide forms only a weak acid in water, and even if water is completely saturated with carbon dioxide the pH will only be 5.6. Acid rain is therefore defined as precipitation (rain, snow, hail etc) that has a pH lower than 5.6 There are two main reasons why acid rain is fomed: The release of sulphur dioxide into the atmosphere from the combustion of fossil fuels The release of oxides of nitrogen from jet engines and internal combustion engines

68 Sulphur dioxide Sulphur dioxide is released into the atmosphere from the combustion of fossil fuels such as coal, which contain sulphur, or the smelting of sulphide ores. The sulphur dioxide is slowly oxidised in the atmosphere to form sulphur trioxide, which dissolves in the rain to form sulphuric acid: 2SO 2(g) + O 2(g) → 2SO 3(g) SO 3(g) + H 2 O (l) → H 2 SO 4(aq)

69 Oxides of nitrogen Oxides of nitrogen are formed in jet engines and in internal combustion engines where the temperature reached is high enough to combine with the oxygen in the air to form nitrogen monoxide: N 2(g) + O 2(g) → 2NO (g) Oxidation in the atmosphere then occurs to produce nitrogen dioxide, which then can dissolve and react with water to form nitric acid and nitrous acid: 2NO 2(g) + H 2 O (l) → HNO 3(aq) + HNO 2(aq) Or be oxidised directly to nitric acid by oxygen in the presence of water 4NO 2(g) + O 2(g) + 2H 2 O (l) → 4HNO 3(aq)

70 Problems with acid rain Acid rain is a world wide problem, because it often precipitates many miles away from its source as a result of air currents. It has a massive impact the planet, for example; It causes damage to buildings by reacting with carbonates in building materials. It affects vegetation by leaching out important minerals from the soil. It affects aquatic life by altering the pH of rivers and lakes. It directly affects human health by increasing the risk of respiratory diseases.

71 Activity: Read through the article on pages 149 – 150 in Chemistry Companion and answer the questions at the end.

72 Titration curves Activity: Watch the acid-base titration. Calculate the concentration of hydrogen ions [H + ] in the Hydrochloric acid Using the graph calculate the volume of hydrochloric acid needed to neutralise the sodium hydroxide. What volume of an equimolar solution of ethanoic acid is need to neutralise the same volume and concentration of sodium hydroxide?

73 AHL Objectives Calculations involving acids and bases Solve problems involving [H + (aq)], [OH - (aq)], pH and pOH. State the equation for the reaction of any weak acid or base with water, and hence deduce the expressions for Ka and Kb. Solve problems involving solutions of weak acids and bases using the expressions: Ka x Kb = Kw pKa + pKb = pKw pH + pOH = pKw Identify the relative strengths of acids and bases using values of Ka, Kb, pKa and pKb.

74 Calculating pH of a strong monoprotic acid. Strong acids are completely dissociated in water thus the concentration of hydrogen ions for a solution of a strong monoprotic acid will be the same as the concentration of the undissociated acid. Example: What is the pH of a 0.10 moldm -3 solution of hydrochloric acid? pH = -log 10 [H+] =-log 10 (0.1) =1.0

75 Calculating pH of a strong monoprotic acid. Strong acids are completely dissociated in water thus the concentration of hydrogen ions for a solution of a strong monoprotic acid will be the same as the concentration of the undissociated acid. Example: What is the pH of a 0.10 moldm -3 solution of hydrochloric acid?

76 Calculating the pH of a strong diprotic acid For a strong diprotic acid the hydrogen ion concentration will be twice the concentration of the acid: Example: what is the pH of 0.10 moldm -3 sulphuric acid? 1mole of H 2 SO 4 produces 2 moles of H + thus pH = -log 10 [H + ] = -log 10 (2 x 0.1) = 0.7

77 Calculating concentration of a strong acid from its pH Simple reverse the process and convert pH into hydrogen ion concentration and use the hydrogen ion concentration to calculate the concentration of the acid. Example: What is the concentration of some hydrochloric acid whose pH is 1.60? antilog the pH to give [H+] = moldm -3 1mole of hydrochloric acid gives 1 mole of hydrogen ions thus concentration of hydrochloric acid = moldm -3

78 Calculating concentration of a strong acid from its pH Simple reverse the process and convert pH into hydrogen ion concentration and use the hydrogen ion concentration to calculate the concentration of the acid. Example: What is the concentration of some hydrochloric acid whose pH is 1.60?

79 Calculating the pH of a strong diprotic acid For a strong diprotic acid the hydrogen ion concentration will be twice the concentration of the acid: Example: what is the pH of 0.10 moldm-3 sulphuric acid?

80 Activity: Complete problems 4 and 5 from page 187 Calculations in A ’Level chemistry

81 Calculating pOH of a strong base For a strong base such as an aqueous solution of sodium hydroxide, the hydroxide ion concentration will be the same as the concentration of the sodium hydroxide. Example; what is the pOH of a 0.10moldm-3 solution of sodium hydroxide? We know; [H + ] x [OH - ] = Kw = 1.0 x mol 2 dm -6 at 298K If [OH-] = 0.10 moldm-3 So pOH = -log10[OH-] = -log10(0.1) = 1

82 Calculating pOH of a strong base For a strong base such as an aqueous solution of sodium hydroxide, the hydroxide ion concentration will be the same as the concentration of the sodium hydroxide. Example; what is the pOH of a 0.10moldm-3 solution of sodium hydroxide? We know; [H + ] x [OH - ] = Kw = 1.0 x mol 2 dm -6 at 298K

83 Using pOH to calculate pH We can calculate pH from pOH using the following relationship: pOH + pH = pKw (Remember p just means –log 10 ) We know that Kw = 1.0 x thus pKw = 14 So.... pH + pOH = 14 Example; calculate the pH for a 0.10 moldm -3 solution of sodium hydroxide pOH = -log 10 (0.1) = 1 Thus pH = pKw – pOH = 14 – 1 = 13

84 Using pOH to calculate pH We can calculate pH from pOH using the following relationship: pOH + pH = pKw (Remember p just means –log 10 ) We know that Kw = 1.0 x thus pKw = 14 So.... pH + pOH = 14 Example; calculate the pH for a 0.10 moldm -3 solution of sodium hydroxide

85 Calculations involving weak acids A weak acid is one which does not ionise to a very large extent in solution. Typical weak acids include most organic acids and some inorganic ones like hydrofluoric acid, HF. One of the commonest weak acids is ethanoic acid, CH 3 COOH. Depending on its concentration, about 1% of the ethanoic acid molecules are ionised in solution. An equilibrium is set up: Write the equation for this reaction; CH 3 COOH + H 2 O CH 3 COO - + H 3 O +

86 Calculations involving weak acids A weak acid is one which does not ionise to a very large extent in solution. Typical weak acids include most organic acids and some inorganic ones like hydrofluoric acid, HF. One of the commonest weak acids is ethanoic acid, CH 3 COOH. Depending on its concentration, about 1% of the ethanoic acid molecules are ionised in solution. An equilibrium is set up: Write the equation for this reaction;

87 The dissociation constant Ka The equilbrium constant for this is called the dissociation constant, and is given the symbol, Ka Write the equilibrium expression for this reaction. In general the equilibrium expression for all weak acids is:

88 The dissociation constant Ka The equilibrium constant for this is called the dissociation constant, and is given the symbol, Ka Write the equilibrium expression for this reaction. In general the equilibrium expression for all weak acids is: Ka= [CH 3 COO - ][H 3 O + ] [CH 3 COOH] Ka= [CH 3 COO - ][H + ] [CH 3 COOH] or Ka= [A - ][H + ] [HA] mol dm -3

89 Using Ka to calculate the pH of a weak acid We can use the Ka expression to find the pH of a solution of a weak acid. For example; calculate the pH of 0.10 moldm -3 ethanoic acid (CH 3 COOH (aq) ) given that the acid dissociation constant, Ka for CH 3 COOH = 1.8 x moldm -3 at 298K CH 3 COOH CH 3 COO - + H + the acid is only very slightly dissociated so it is assumed that the concentration of the acid at equilibrium is the same as the initial concentration Ka= [CH 3 COO - ][H + ] [CH 3 COOH] becomes Ka ≈ x2x x + x At eqmAt start 00 so

90 Using Ka to calculate the pH of a weak acid We can use the Ka expression to find the pH of a solution of a weak acid. For example; calculate the pH of 0.10 moldm -3 ethanoic acid (CH 3 COOH (aq) ) given that the acid dissociation constant, Ka for CH 3 COOH = 1.8 x moldm -3 at 298K CH 3 COOH CH 3 COO - + H +

91 Putting the numbers in... If... Then... Ka ≈ x2x x ≈ x2x x = x2x x x2x2 = 1.8 x x = = Rearrange the equation To find x we need to find the square root x = [H + ] thus pH = -log 10 [H + ] = 2.87

92 Activity: Calculations in A’Level Chemistry page 201 complete problem 14 a – c.

93 pKa The values for Ka, for weak acids are very small – for example, Ka for ethanoic acid is 1.74 x moldm -3. To tidy these numbers up, pKa is often used instead. pKa = -log 10 Ka The relationship between pKa and Ka is exactly the same as that between pH and [H + ]. If Ka = 1.74 x pKa = -log 10 (1.74 x ) = 4.76 Going the other way, you need to “unlog” the pKa value; If pKa =4.76 -log 10 Ka = 4.76 log 10 Ka =-4.76 Ka= 1.74 x 10 -5

94 pKa The values for Ka, for weak acids are very small – for example, Ka for ethanoic acid is 1.74 x moldm -3. To tidy these numbers up, pKa is often used instead. pKa = -log 10 Ka The relationship between pKa and Ka is exactly the same as that between pH and [H + ]. If Ka = 1.74 x pKa = -log 10 (1.74 x ) = Going the other way, you need to “unlog” the pKa value; If pKa =4.76 -log 10 Ka = 4.76 log 10 Ka =-4.76 Ka=Ka=

95 Activity: Calculations in A’Level Chemistry page 201 complete problems 13, 14 (d-e) and 15

96 The pH of weak bases Ammonia is a typical weak base. It reacts with water to produce ammonium ions and hydroxide ions. Write the equation for this reaction NH 3(aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) The ammonia is acting as a base because it is combining with a hydrogen ion from the water. Because most of it remains in solution as unreacted ammonia molecules, ammonia is described as a weak base. The stronger the base, the further the position of equilibrium lies to the right.

97 The pH of weak bases Ammonia is a typical weak base. It reacts with water to produce ammonium ions and hydroxide ions. Write the equation for this reaction The ammonia is acting as a base because it is combining with a hydrogen ion from the water. Because most of it remains in solution as unreacted ammonia molecules, ammonia is described as a weak base. The stronger the base, the further the position of equilibrium lies to the right.

98 The dissociation constant Kb The equilbrium constant for this is called the dissociation constant, and is given the symbol, Kb Write the equilibrium expression for this reaction.

99 The dissociation constant Kb The equilbrium constant for this is called the dissociation constant, and is given the symbol, Kb Write the equilibrium expression for this reaction. Kb= [OH - ][NH 4 + ] [NH 3 ]

100 Using Kb to calculate the pH of a weak base We can use the Kb expression to find the pH of a solution of a weak base. For example; calculate the pH of 0.10 moldm -3 ammonia solution (NH 3(aq) ) given that the base dissociation constant, Kb for NH 3 = 1.8 x moldm -3 at 298K NH 3 NH OH - the base is only very slightly dissociated so it is assumed that the concentration of the base at equilibrium is the same as the initial concentration Kb= [NH 4 + ][OH - ] [NH 3 ] becomes Kb ≈ x2x x + x At eqmAt start 00 so

101 Using Kb to calculate the pH of a weak base We can use the Kb expression to find the pH of a solution of a weak base. For example; calculate the pH of 0.10 moldm -3 ammonia solution (NH 3(aq) ) given that the base dissociation constant, Kb for NH 3 = 1.8 x moldm -3 at 298K NH 3 NH OH -

102 Putting the numbers in... If... Then... Kb ≈ x2x x ≈ x2x x = x2x x x2x2 = 1.8 x x = = Rearrange the equation To find x we need to find the square root x = [OH - ] thus pOH = -log 10 [OH - ] = 2.87 pKw = pOH + pH therefore pH = 14 – 2.87 = 11.13

103 pKb The values for Kb, for weak bases are very small – for example, Kb for ammonia is 1.8 x moldm -3. To tidy these numbers up, pKb is often used instead. pKb = -log 10 Kb The relationship between pKb and Kb is exactly the same as that between pH and [H + ]. If Kb = 1.8 x pKb = -log 10 (1.8 x ) = 4.74 Going the other way, you need to “unlog” the pKb value; If pKb =4.74 -log 10 Kb = 4.74 log 10 Kb = Kb= 1.8 x 10 -5

104 pKb The values for Kb, for weak bases are very small – for example, Kb for ammonia is 1.8 x moldm -3. To tidy these numbers up, pKb is often used instead. pKb = -log 10 Kb The relationship between pKb and Kb is exactly the same as that between pH and [H + ]. If Kb = 1.8 x pKb = -log 10 (1.8 x ) = Going the other way, you need to “unlog” the pKb value; If pKb =4.74 -log 10 Kb = 4.74 log 10 Kb = Kb=Kb=

105 Ka values for weak bases The conjugate acid of ammonia is the ammonium ion NH 4 + (aq). It is a weak acid as it dissociates to form ammonia and hydrogen ions. Write the equation for this reaction. NH 4 + (aq) NH 3(aq) + H + (aq) Write the equilibrium expression for this reaction: Ka= [NH 4 + ] [H + ][NH 3 ]

106 Ka values for weak bases The conjugate acid of ammonia is the ammonium ion NH 4 + (aq). It is a weak acid as it dissociates to form ammonia and hydrogen ions. Write the equation for this reaction. Write the equilibrium expression for this reaction:

107 So if... We can cancel out the ammonium ions and ammonia to give us..... So The relationship between Ka and Kb Kb= [NH 4 + ] [OH - ] [NH 3 ] and Ka= [NH 4 + ] [H + ] [NH 3 ] Kb Ka = [OH - ][H + ] xx Where have we seen this before? KbKa=Kw x take it further pKbpKa=pKw +

108 Using this expression to find the pH of a base Calculate the pH of a 1.0 x moldm -3 aqueous solution of ethylamine, C 2 H 5 NH 2, given that pKa value for ethylamine is at 298K Write the equation for the dissociation of ethylamine in water Write the Kb equilibrium expression for this:

109 Using this expression to find the pH of a base Calculate the pH of a 1.0 x moldm -3 aqueous solution of ethylamine, C 2 H 5 NH 2, given that pKa value for ethylamine is at 298K Write the equation for the dissociation of ethylamine in water C 2 H 5 NH 2(aq) + H 2 O (l) C 2 H 5 NH 3 + (aq) + OH - (aq) Write the Kb equilibrium expression for this: [C 2 H 5 NH 3 + (aq) ][OH - (aq) ] [C 2 H 5 NH 2(aq) ] Kb =

110 Putting the numbers in.... So.... pKa + pKb = pKw Thus pKb = pKw – pKa = 14 – pKb = 3.27 Kb therefore = antilog – 3.27 = 5.37 x And remember... [C 2 H 5 NH 3 + (aq) ][OH - (aq) ] [C 2 H 5 NH 2(aq) ] Kb = = [OH - (aq) ] 2 [C 2 H 5 NH 2(aq) ] 5.37 x [OH - (aq) ] x 10-3 ≈ 5.37 x Rearrange the equation [OH - ] 2 = 5.37 x x 1.0 x = 5.37 x Therefore taking the square root [OH - ] = 7.32 x We can use the expression.. pOH + pH = pKw Therefore pOH = -log10[OH - ] =3.14 So pH = pKw – pOH = 14 – 3.14 = 10.86

111 Relative strengths of acids and bases using Ka, Kb, pKa and pKb Remember that Ka and Kb are just equilibrium constants.... The greater the equilibrium constant the greater the extent of dissociation, i.e. the equilibrium lies to the right (products). So the greater the value of Ka the greater the dissociation of the acid therefore = a stronger acid. pKa and pKb are logarithms of Ka and Kb thus the greater the value of Ka and Kb the lower the value of pKa and pKb.

112 Acid dissociation constants of some weak acids CompoundEquilibrium expressionKa / moldm-3pKa Hydrofluoric acid, HF 6.3 x Ethanoic acid, CH3COOH 1.6 x Benzoic acid, C6H5COOH 6.3 x Phenol, C6H5OH1.3 x

113 Acid dissociation constants of some weak acids CompoundEquilibrium expressionKa / moldm -3 pKa Hydrofluoric acid, HF 6.3 x Ethanoic acid, CH 3 COOH 1.6 x Benzoic acid, C 6 H 5 COOH 6.3 x Phenol, C 6 H 5 OH 1.3 x CH 3 CO 2 H + H 2 O H 3 O + + CH 3 COO - HF + H 2 O H 3 O + + F - C 6 H 5 CO 2 H + H 2 O H 3 O + + C 6 H 5 COO - C 6 H 5 OH + H 2 O H 3 O + + C 6 H 5 O -

114 Base dissociation constants of some weak bases CompoundEquilibrium expressionKa / moldm -3 pKa Ammonia, NH x Methylamine, CH 3 NH x Phenylamine, C 6 H 5 NH x Phenylmethyl amine, C 6 H 5 CH 2 NH x 10 -5

115 Base dissociation constants of some weak bases CompoundEquilibrium expressionKa / moldm -3 pKa Ammonia, NH x Methylamine, CH 3 NH x Phenylamine, C 6 H 5 NH x Phenylmethyl amine, C 6 H 5 CH 2 NH x CH 3 NH 2 + H 2 O CH 3 NH OH - NH 3 + H 2 O NH OH - C 6 H 5 NH 2 + H 2 O C 6 H 5 NH OH - C 6 H 5 CH 2 NH 2 + H 2 O C 6 H 5 CH 2 NH H 3 O +

116 Objectives Salt Hydrolysis Deduce whether salts form acidic, alkaline or neutral aqueous solutions.

117 Salt hydrolysis Salt hydrolysis is just the interaction of a salt in water. You would assume that all salts are neutral but this is not true, only some salts are neutral. Salts are ionic, and are therefore already completely dissociated. When they dissolve in water they are strong electrolytes. For example: Na + Cl - Na + (aq) + Cl - (aq) Salts like sodium chloride, which are derived from a strong acid (hydrochloric acid) and a strong base (sodium hydroxide), form neutral solutions when they dissolve in water. H 2 O (l)

118 Alkaline salts Salts that are derived from a weak acid and a strong base produce alkaline solutions when they are dissolved in water. An example is sodium ethanoate, which can be formed from the reaction between ethanoic acid and sodium hydroxide. When the salt dissolves in water the ethanoate ions combine with hydrogen ions from the water to form mainly undissociated ethanoic acid. This leaves an excess of hydroxide ions in the solution, which do not combine with the sodium ions, because sodium hydroxide is a strong base.

119 The hydrolysis of sodium ethanoate CH 3 COONa (s) CH 3 COO - (aq) + Na + (aq) H 2 O (l) H + (aq) +OH - (aq) + CH 3 COOH (aq) H 2 O (l)

120 The hydrolysis of ammonium chloride NH 4 Cl (s) Cl - (aq) + NH 4 + (aq) H 2 O (l) H + (aq) +OH - (aq) + NH 3(aq) H 2 O (l) Similarly salts such as ammonium chloride which are derived from a strong acid and a weak base, will be acidic in solution. +H 2 O (l)

121 The acidity of salts also depends on: The size of the cation The charge of the cation Aluminium chloride reacts vigorously with water to give a strongly acidic solution: AlCl 3(s) + 3H 2 O (l) → Al(OH) 3(s) + 3HCl (aq) The tripositive (+3) charge is spread over a very small ion, which gives the aluminium ion, Al 3+, a very high charge density. This makes it an excellent lewis acid, and it will attract a non-bonding pair of electrons on six water molecules to form the hexahydrated ion. Can you draw this moelcule?

122 The hydrolysis of hexahydrated aluminium (III) ion The non-bonding pair of one of the six water molecules surrounding the ion is strongly attracted to the ion and loses a hydrogen ion in the process. This process will continue until three hydrogen ions have been released and aluminium (III) hydroxide is formed. The equilibrium can be moved further to the right by adding hydroxide ions, or back to the left by adding hydrogen ions, which explains the amphoteric nature of aluminium hydroxide. Al OH 2 H2OH2O H2OH2O 3+ [Al(H 2 O) 5 OH] 2+ [Al(H 2 O) 4 OH] + [Al(H 2 O) 3 OH 3 ] - H +

123 Other highly charged ions Similar reactions occur with other small, highly charged ions such as the iron (III) ion, Fe 3+. Even magnesium chloride, MgCl 2, is slightly acidic in aqueous solution for the same reason. ionchargeIonic radius/nm Aqueous solution Na neutral Mg acidic Al acidic

124 Objectives Buffer solutions Describe the composition of a buffer solution and explain its action. Solve problems involving the composition and pH of a specified buffer system.

125 Buffers A buffer solution is defined as an aqueous solution that resists a change in pH when a small amount of acid, base or water is added to it. There are two types of buffer; Acidic buffers with a pH less than 7 Alkaline buffers with a pH more than 7 Buffers are formed from an acid, a base and the salt made from the acid – base reaction.

126 Acidic buffers An acidic buffer can be made in the laboratory by mixing a weak acid together with the salt of that acid and a strong base (in other words by combining a weak acid with a salt containing its conjugate base) e.g. ethanoic acid and sodium ethanoate. A practical method of doing this is to add excess ethanoic acid to sodium hydroxide solution. Once all the sodium hydroxide has been neutralised to form the salt, the remaining solution contains the salt and the excess acid. NaOH (aq) + CH 3 COOH (aq) → CH 3 COONa (aq) + H 2 O (l) + CH 3 COOH (aq) limiting reactantsaltexcess weak acid

127 How an acidic buffer works The explanation of buffer action lies in the fact that the weak acid is only slightly dissociated in solution, but the salt is fully dissociated into its ions, so the concentration of ethanoate ions is high. CH 3 COONa (aq) Na + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) CH 3 COO - (aq) + H + (aq)

128 How an acidic buffer works If an acid is added, the extra hydrogen ions from the added acid combine with ethanoate ions to form undissociated ethanoic acid, and so the hydrogen ions remain unaltered If an alkali is added, the hydroxide ions from the alkali are removed by their reaction with the undissociated acid to form water, so again the hydrogen ion concentration again stays constant. CH 3 COONa (aq) CH 3 COO - (aq) Na + (aq) + CH 3 COOH (aq) CH 3 COO - (aq) H + (aq) + CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (aq)

129 How an alkaline buffer works An alkaline buffer is made by combining a weak base with the salt of that base with a strong acid. An example is ammonia with ammonium chloride: If hydrogen ions are added, they will combine with hydroxide ions to form water, and more of the ammonia will dissociate to replace them. If more hydroxide ions are added, they will combine with ammonium ions to form undissociated ammonia. In both cases the hydroxide ion concentration and the hydrogen ion concentration will remain constant. NH 4 Cl (aq) +NH 4 + (aq) Cl - (aq) NH 3(aq) + + H 2 O (aq) NH 4 + (aq) OH - (aq)

130 Uses of buffer solutions Buffer solutions have many applications in the laboratory: for example, they are used to standardize a pH meter. They are also extremely important in biological systems, because enzymes function efficiently only within a very narrow pH region. Outside this region their structure may rapidly become distorted or broken down (denatured).

131 Blood Blood, is responsible for carrying oxygen around the body, contains several different buffering systems. One of the components of the system is that the oxygen adds on reversibly to the haemoglobin (Hb) in the blood: HHb (aq) + O 2(g) H + (aq) + HbO 2 - (aq) If the pH increases ([H + (aq) ] falls), the equilibrium will move to the right, and the oxygen will tend to be bound to the haemoglobin more tightly. If the pH decreases ([H + (aq) ] increases), the oxygen will tend to be displaced from the haemoglobin and released. Both of these processes are potentially life-threatening.

132 Buffer calculations Calculations involving buffer solutions are very similar to the calculations done previously for weak acids and bases. However, there is one very important difference. When a weak monoprotic acid on its own in aqueous solution dissociates, the concentration of the hydrogen ions will be the same as the concentration of the acid anions. In a buffer solution the concentration of the acid anions will be much greater than the hydrogen concentration, because the salt that produces the acid anions is completely dissociated. In fact, the assumption is made that the concentration of the acid anions is due completely to the concentration of the salt, because the anions from the dissociation of the weak acid make a negligible contribution.

133 Acid buffer calculation Example: calculate the pH of a buffer solution containing 3.00g of sodium ethanoate in 100cm 3 of moldm -3 ethanoic acid at 298K (given that Ka for ethanoic acid = 1.8 x moldm -3 at 298K. First we need to write the equilibrium expression for this buffer: Ka= [CH 3 COO - ][H + ] [CH 3 COOH] We can assume this number is the same as the concentration of the salt We can assume this number is the same as the concentration of the acid This is what we are trying to find out We can rearrange this equation to find [H+]

134 Ka = 1.8 x moldm -3 [CH 3 COOH] = 0.05 moldm -3 We need to find [CH 3 COO - ] We are given a mass so we can find moles We are given the volume it is dissolved in, so we can find the concentration Ka= [CH 3 COO - ] [H + ] [CH 3 COOH] Moles = RMM mass = g RMM = = 82 = mol in 100cm 3 Therefore mol in 1000cm x moldm = 2.46 x pH = - log 10 [H + ] = -log 10 (2.46 x ) = 5.61

135 Alkaline buffer calculation Example: what mass of ammonium chloride must be dissolved in 1.00dm -3 of moldm -3 ammonia solution to produce a buffer solution with a pH equal to 9.00, pKb for ammonia = 4.75?. First we need to write the equilibrium expression for this buffer: Kb= [OH - ][NH 4 + ] [NH 3 ] we can find this using the expression pOH = pKw - pH We can assume this number is the same as the concentration of the alkali This is what we are trying to find out We can rearrange this equation to find [H+] we can convert pKb to Kb

136 Kb = antilog 4.75 = 1.78x moldm -3 [NH 3 ] = 0.10 moldm -3 We need to find [OH - ] We have been given pH so we can find pOH We can convert pOH to OH pOH = pKw – pH pOH = 14 – 9 = 5 so OH = antilog 5 = 1 x Ka= [OH - ] [NH 4 + ] [NH 3 ] RMM of ammonium chloride = = x moldm -3 1 x = moldm -3 Assuming all ammonium ions originate from ammonium chloride Therefore mol of NH 4 Cl will have a mass of x 53.5 = 9.42g

137 Objectives Indicators Describe qualitatively the action of an acid – base indicator. State and explain how the pH range of an acid – base indicator relates to its pKa value. Identify an appropriate indicator for a titration, given the equivalence point of the titration and the pH range of the indicator.

138 Acid – base indicators An indicator is a solution of a weak acid in which the conjugate base has a different colour from that of the undissociated acid. If we represent an aqueous solution of an indicator by HIn (aq), then the equation for the dissociation in water is what would the equilibrium expression be? HIn (aq) H + (aq) In - (aq) + Colour A Colour B (colour in acid solution)(colour in alkali solution) K in = [H + (aq) ][In - (aq) ] [HIn (aq) ]

139 Indicator end-points Assuming the colour changes when the concentration of the acid is approximately equal to the concentration of the conjugate base ([HIn (aq) ] = [In- (aq) ]) then the end point of the indicator will be when [H + (aq) ] ≈ K a : that is, when pH ≈ pK a. It can be seen that the various indicators have different K a values, and so change colour within different pH ranges. Two common indicators are methyl orange and phenolphthalein, look up their pK a values and compare with their pH ranges.

140 Objectives Acid – Base titrations Sketch the general shapes of graphs of pH against volume for titrations involving strong and weak acids and bases, and explain their important features.

141 Acid – base titrations Acids and bases neutralise each other when mixed together in aqueous solution. An acid – base titration is a technique used to measure the volumes of acid and base required for neutralisation. If the concentration of one solution is known, then the concentration of the other may be calculated. There are four possible combinations of strong and weak acids and bases that may be considered for a titration: Strong acid – Strong base Strong acid – Weak base Weak acid – Strong base Weak acid – Weak base

142 Strong acid – strong base A titration curve shows the pH changes that occur when an aqueous acid reacts with an aqueous base. It is a graph of pH against volume. The titration of a strong acid with a strong base causes a large pH change on neutralisation, as shown in the graph. The equivalence point of this titration occurs when the amounts of acid and base are exactly equal We can use suitable indicators to observe the equivalence point of a titration. The end-point of a titration occurs when the indicator changes colour. The indicator must be chosen with care to ensure that the end-point coincides with the equivalence point.

143 Choosing an indicator The titration curve for a strong acid-stong base neutralisation shows a large change in pH around the equivalence point. Any indicator with a range between pH 3 and pH11 may be used. The titration curves for other combinations of strong/weak acid/base pairs offer a more limited change in pH. As a result, it is more difficult, if not impossible to choose a suitable indicator.

144 Common indicators

145 Titration of a weak acid with a strong base In this case, you can see that the large change in pH around the equivalence point occurs approximately between pH7 and pH 11. What would be a suitable indicator for this titration?

146 Titration of a strong acid with a weak base Sketch a graph of the titration curve you would expect in this acid – base reaction. What indicator could you use and what colour change would you see?

147 Titration of a weak acid and a weak base There is hardly any change in pH around the equivalence point in weak acid – weak base titration. No indicator will detect the end-point accurate to one drop. However, this is not a problem, as it is never necessary to perform this type of titration because, if the unknown is weak, the standard solution can always be chosen to be strong.

148 Summary of titrations The end-point of a titration occurs when the indicator changes colour. The equivalence point occurs when there are equal amounts of H 3 O + and OH - ions in the titration. Indicators with ranges between pH 11 and pH 3 are suitable for strong acid – strong base titrations. The indicator range for a strong acid – weak base titration must fall between pH 3 and pH 7. Methyl orange is a useful choice. The indicator range for a weak acid – strong base titration must fall between pH 7 and pH 11. Phenolphthalein is a useful choice. Indicators cannot be used to find the equivalence point of a weak acid – weak base titration.


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