p H and IONIC EQUILIBRIA A guide for

p. H and IONIC EQUILIBRIA A guide for A level students KNOCKHARDY PUBLISHING 2015 SPECIFICATIONS

p. H and Ionic Equilibria INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A 2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards. Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available. Accompanying notes on this, and the full range of AS and A 2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at. . . www. knockhardy. org. uk/sci. htm Navigation is achieved by. . . either or clicking on the grey arrows at the foot of each page using the left and right arrow keys on the keyboard

p. H and Ionic Equilibria CONTENTS • Brønsted-Lowry theory of acids and bases • Lewis theory of acids and bases • Strong acids and bases • Weak acids • Weak bases • Hydrogen ion concentration and p. H • Ionic product of water Kw • Relationship between p. H and p. OH • Calculating the p. H of strong acids • Calculating the p. H of weak acids • Calculating the p. H of mixtures • A brief introduction to buffer solutions • Check list

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

ACIDS AND BASES BRØNSTED-LOWRY THEORY ACID proton donor HCl ——> H+(aq) + Cl¯(aq) BASE proton acceptor NH 3(aq) + H+(aq) ——> NH 4+(aq)

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

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

ACIDS AND BASES LEWIS THEORY ACID lone pair acceptor BF 3 H+ BASE lone pair donor NH 3 H 2 O LONE PAIR DONOR LONE PAIR ACCEPTOR Al. Cl 3

STRONG ACIDS AND BASES STRONG ACIDS e. g. completely dissociate (split up) into ions in aqueous solution HCl ——> H+(aq) + Cl¯(aq) MONOPROTIC 1 replaceable H DIPROTIC 2 replaceable H’s HNO 3 ——> H+(aq) + NO 3¯(aq) H 2 SO 4 ——> 2 H+(aq) + SO 42 -(aq)

STRONG ACIDS AND BASES STRONG ACIDS e. g. completely dissociate (split up) into ions in aqueous solution HCl ——> H+(aq) + Cl¯(aq) MONOPROTIC 1 replaceable H DIPROTIC 2 replaceable H’s HNO 3 ——> H+(aq) + NO 3¯(aq) H 2 SO 4 ——> 2 H+(aq) + SO 42 -(aq) STRONG BASES e. g. completely dissociate into ions in aqueous solution Na. OH(s) ——> Na+(aq) + OH¯(aq)

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

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

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

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

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

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

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

Hydrogen ion concentration [H+(aq)] p. H p. OH hydrogen ion concentration can be converted to p. H = - log 10 [H+(aq)] to convert p. H into hydrogen ion concentration [H+(aq)] = antilog (-p. H) An equivalent calculation for bases converts the hydroxide ion concentration to p. OH = - log 10 [OH¯(aq)] in both the above, [ ] represents the concentration in mol dm -3

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

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

Ionic product of water - Kw Despite being covalent, water conducts electricity to a very small extent. This is due to the slight ionisation. . . H 2 O(l) + H 2 O(l) or, more simply Applying the equilibrium law to the second equation gives [ ] is the equilibrium concentration in mol dm-3 H 3 O+(aq) + OH¯(aq) H 2 O(l) Kc H+(aq) + OH¯(aq) = [H+(aq)] [OH¯(aq)] [H 2 O(l)] As the dissociation is small, the water concentration is very large compared with the dissociated ions and any changes to its value are insignificant; its concentration can be regarded as constant.

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

Ionic product of water - Kw The value of Kw varies with temperature because it is based on an equilibrium. Temperature / °C 0 20 25 30 60 Kw / 1 x 10 -14 mol 2 dm-6 0. 11 0. 68 1. 0 1. 47 5. 6 H+ / x 10 -7 mol dm-3 p. H 0. 33 7. 48 0. 82 7. 08 1. 0 7 1. 27 6. 92 2. 37 6. 63 What does this tell you about the equation H 2 O(l) H+(aq) + OH¯(aq) ?

Ionic product of water - Kw The value of Kw varies with temperature because it is based on an equilibrium. Temperature / °C 0 20 25 30 60 Kw / 1 x 10 -14 mol 2 dm-6 0. 11 0. 68 1. 0 1. 47 5. 6 H+ / x 10 -7 mol dm-3 p. H 0. 33 7. 48 0. 82 7. 08 1. 0 7 1. 27 6. 92 2. 37 6. 63 What does this tell you about the equation H 2 O(l) H+(aq) + OH¯(aq) ? • Kw gets larger as the temperature increases • this means the concentration of H+ and OH¯ ions gets greater • this means the equilibrium has moved to the right • if the concentration of H+ increases then the p. H decreases • p. H decreases as the temperature increases

Ionic product of water - Kw The value of Kw varies with temperature because it is based on an equilibrium. Temperature / °C 0 20 25 30 60 Kw / 1 x 10 -14 mol 2 dm-6 0. 11 0. 68 1. 0 1. 47 5. 6 H+ / x 10 -7 mol dm-3 p. H 0. 33 7. 48 0. 82 7. 08 1. 0 7 1. 27 6. 92 2. 37 6. 63 What does this tell you about the equation H 2 O(l) H+(aq) + OH¯(aq) ? • Kw gets larger as the temperature increases • this means the concentration of H+ and OH¯ ions gets greater • this means the equilibrium has moved to the right • if the concentration of H+ increases then the p. H decreases • p. H decreases as the temperature increases Because the equation moves to the right as the temperature goes up, it must be an ENDOTHERMIC process

Relationship between p. H and p. OH Because H+ and OH¯ ions are produced in equal amounts when water dissociates their concentrations will be the same. [H+] = [OH¯] = 1 x 10 -7 mol dm-3 Kw = [H+] [OH¯] = 1 x 10 -14 mol 2 dm-6 take logs of both sides log[H+] + log[OH¯] = -14 multiply by minus - log[H+] - log[OH¯] = 14 change to p. H and p. OH p. H + p. OH = 14 (at 25°C) [H+] 100 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 OH¯ 10 -14 10 -13 10 -12 10 -11 10 -10 10 -9 10 -8 10 -7 10 -6 10 -5 10 -4 10 -3 10 -2 10 -1 10 -0 p. H 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 p. OH 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 STRONGLY ACIDIC WEAKLY ACIDIC NEUTRAL WEAKLY ALKALINE STRONGLY ALKALINE

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

Calculating p. H - strong acids and alkalis WORKED EXAMPLE Strong acids and alkalis completely dissociate in aqueous solution It is easy to calculate the p. H; you only need to know the concentration

Calculating p. H - strong acids and alkalis WORKED EXAMPLE Strong acids and alkalis completely dissociate in aqueous solution It is easy to calculate the p. H; you only need to know the concentration Calculate the p. H of 0. 02 M HCl completely dissociates in aqueous solution HCl H+ + Cl¯ One H+ is produced for each HCl dissociating so [H+] = 0. 02 M = 2 x 10 -2 mol dm-3 p. H = - log [H+] = 1. 7

Calculating p. H - strong acids and alkalis WORKED EXAMPLE Strong acids and alkalis completely dissociate in aqueous solution It is easy to calculate the p. H; you only need to know the concentration Calculate the p. H of 0. 02 M HCl completely dissociates in aqueous solution HCl H+ + Cl¯ One H+ is produced for each HCl dissociating so [H+] = 0. 02 M = 2 x 10 -2 mol dm-3 p. H = - log [H+] = 1. 7 Calculate the p. H of 0. 1 M Na. OH completely dissociates in aqueous solution Na. OH One OH¯ is produced for each Na. OH dissociating [OH¯] = 0. 1 M = 1 x 10 -1 mol dm-3 Kw = [H+][OH¯] = 1 x 10 -14 mol 2 dm-6 The ionic product of water (at 25°C) therefore Na+ + OH¯ [H+] = Kw / [OH¯] = 1 x 10 -13 mol dm-3 p. H = - log [H+] = 13

Calculating p. H - weak acids • can’t be calculated by just knowing the concentration • need to know. . . the original concentration the extent of the ionisation The extent of the ionisation is given by … Ka The dissociation constant for a weak acid and

The dissociation constant for a weak acid - Ka A weak monobasic acid (HA) dissociates in water thus. HA(aq) + H 2 O(l) Applying the equilibrium law we get [ ] is the equilibrium concentration in mol dm-3 H 3 O+(aq) + A¯(aq) Kc = [H 3 O+(aq)] [A¯(aq)] [HA(aq)] [H 2 O(aq)]

The dissociation constant for a weak acid - Ka A weak monobasic acid (HA) dissociates in water thus. HA(aq) + H 2 O(l) Applying the equilibrium law we get [ ] is the equilibrium concentration in mol dm-3 Assumptions For a weak acid there is little dissociation H 3 O+(aq) + A¯(aq) Kc = [H 3 O+(aq)] [A¯(aq)] [HA(aq)] [H 2 O(aq)] [HA(aq)]equil ~ [HA(aq)]undis

The dissociation constant for a weak acid - Ka A weak monobasic acid (HA) dissociates in water thus. HA(aq) + H 2 O(l) Applying the equilibrium law we get [ ] is the equilibrium concentration in mol dm-3 Assumptions This assumption becomes less valid for stronger weak acids where there is more dissociation. For a weak acid there is little dissociation H 3 O+(aq) + A¯(aq) Kc = [H 3 O+(aq)] [A¯(aq)] [HA(aq)] [H 2 O(aq)] [HA(aq)]equil ~ [HA(aq)]undis

The dissociation constant for a weak acid - Ka A weak monobasic acid (HA) dissociates in water thus. HA(aq) + H 2 O(l) Applying the equilibrium law we get H 3 O+(aq) + A¯(aq) Kc = [ ] is the equilibrium concentration in mol dm-3 Assumptions For a weak acid there is little dissociation In dilute solution, the concentration of water is large compared with the dissociated ions and any changes to its value are insignificant; its concentration can be regarded as ‘constant’. [H 3 O+(aq)] [A¯(aq)] [HA(aq)] [H 2 O(aq)] [HA(aq)]equil ~ [HA(aq)]undis [H 2 O(l)] is ‘constant’

The dissociation constant for a weak acid - Ka A weak monobasic acid (HA) dissociates in water thus. HA(aq) + H 2 O(l) Applying the equilibrium law we get H 3 O+(aq) + A¯(aq) Kc = [ ] is the equilibrium concentration in mol dm-3 Assumptions [HA(aq)] [H 2 O(aq)] For a weak acid there is little dissociation In dilute solution, the concentration of water is large compared with the dissociated ions and any changes to its value are insignificant; its concentration can be regarded as ‘constant’. Combine this ‘constant’ with Kc to get a new constant (Ka) where Ka = Kc [H 2 O(l)] [H 3 O+(aq)] [A¯(aq)] [HA(aq)]equil ~ [HA(aq)]undis [H 2 O(l)] is ‘constant’ Ka = [H 3 O+(aq) ] [A¯(aq)] mol dm-3 [HA(aq)]

The dissociation constant for a weak acid - Ka A weak monobasic acid (HA) dissociates in water thus. HA(aq) + H 2 O(l) Applying the equilibrium law we get H 3 O+(aq) + A¯(aq) Kc = [ ] is the equilibrium concentration in mol dm-3 Assumptions [HA(aq)] [H 2 O(aq)] For a weak acid there is little dissociation In dilute solution, the concentration of water is large compared with the dissociated ions and any changes to its value are insignificant; its concentration can be regarded as ‘constant’. Combine this ‘constant’ with Kc to get a new constant (Ka) where Ka = Kc [H 2 O(l)] [H 3 O+(aq)] [A¯(aq)] [HA(aq)]equil ~ [HA(aq)]undis [H 2 O(l)] is ‘constant’ Ka = [H 3 O+(aq) ] [A¯(aq)] mol dm-3 A simpler way to express the dissociation is [HA(aq)] HA(aq) H+(aq) + A¯(aq)

The dissociation constant for a weak acid - Ka A weak monobasic acid (HA) dissociates in water thus. HA(aq) + H 2 O(l) Applying the equilibrium law we get H 3 O+(aq) + A¯(aq) Kc = [ ] is the equilibrium concentration in mol dm-3 Assumptions [HA(aq)] [H 2 O(aq)] [HA(aq)]equil ~ [HA(aq)]undis For a weak acid there is little dissociation In dilute solution, the concentration of water is large compared with the dissociated ions and any changes to its value are insignificant; its concentration can be regarded as ‘constant’. Combine this ‘constant’ with Kc to get a new constant (Ka) where Ka = Kc [H 2 O(l)] is ‘constant’ Ka = [H 3 O+(aq) ] [A¯(aq)] mol dm-3 [HA(aq)] A simpler way to express the dissociation is The dissociation constant Ka is then [H 3 O+(aq)] [A¯(aq)] Ka = HA(aq) H+(aq) + A¯(aq) [H+(aq)] [A¯(aq)] mol dm-3 [HA(aq)]

The dissociation constant for a weak acid - Ka A weak monobasic acid (HA) dissociates in water thus. HA(aq) + H 2 O(l) Applying the equilibrium law we get H 3 O+(aq) + A¯(aq) Kc = [ ] is the equilibrium concentration in mol dm-3 Assumptions This assumption becomes less valid for stronger weak acids where there is more dissociation. [HA(aq)] [H 2 O(aq)] [HA(aq)]equil ~ [HA(aq)]undis For a weak acid there is little dissociation In dilute solution, the concentration of water is large compared with the dissociated ions and any changes to its value are insignificant; its concentration can be regarded as ‘constant’. Combine this ‘constant’ with Kc to get a new constant (Ka) where Ka = Kc [H 2 O(l)] is ‘constant’ Ka = [H 3 O+(aq) ] [A¯(aq)] mol dm-3 [HA(aq)] A simpler way to express the dissociation is The dissociation constant Ka is then [H 3 O+(aq)] [A¯(aq)] Ka = HA(aq) H+(aq) + A¯(aq) [H+(aq)] [A¯(aq)] mol dm-3 [HA(aq)]

The dissociation constant for a weak acid - Ka A weak monobasic acid (HA) dissociates in water thus. HA(aq) + H 2 O(l) The dissociation constant Ka is then Ka = H 3 O+(aq) + A¯(aq) [H+(aq)] [A¯(aq)] mol dm-3 [HA(aq)] The weaker the acid • the less it dissociates • the fewer ions you get • the smaller Ka The stronger the acid • the more the equilibrium lies to the right • the larger Ka p. Ka • very weak acids have very small Ka values • it is easier to compare the strength as p. Ka values The conversion is carried out thus. . . To convert p. Ka into Ka p. Ka = - log 10 Ka Ka = antilog (-p. Ka) or 10 -Ka

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

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

WORKED EXAMPLE Calculating p. H – weak acids Calculate the p. H of a weak acid HX of concentration 0. 1 M ( Ka = 4 x 10 -5 mol dm-3 ) HX dissociates as follows HX(aq) H+(aq) + X¯(aq)

WORKED EXAMPLE Calculating p. H – weak acids Calculate the p. H of a weak acid HX of concentration 0. 1 M ( Ka = 4 x 10 -5 mol dm-3 ) HX dissociates as follows Dissociation constant for a weak acid HX(aq) Ka H+(aq) = + X¯(aq) [H+(aq)] [X¯(aq)] [HX(aq)] mol dm-3

WORKED EXAMPLE Calculating p. H – weak acids Calculate the p. H of a weak acid HX of concentration 0. 1 M ( Ka = 4 x 10 -5 mol dm-3 ) HX dissociates as follows Dissociation constant for a weak acid HX(aq) Ka H+(aq) = + X¯(aq) [H+(aq)] [X¯(aq)] mol dm-3 [HX(aq)] Substitute for X¯ as ions are formed in equal amounts and then rearrange equation [H+(aq)] = [HX(aq)] Ka mol dm-3

WORKED EXAMPLE Calculating p. H – weak acids Calculate the p. H of a weak acid HX of concentration 0. 1 M ( Ka = 4 x 10 -5 mol dm-3 ) HX dissociates as follows Dissociation constant for a weak acid HX(aq) Ka H+(aq) = + X¯(aq) [H+(aq)] [X¯(aq)] mol dm-3 [HX(aq)] Substitute for X¯ as ions are formed in equal amounts and the rearrange equation [H+(aq)] = [HX(aq)] Ka mol dm-3 ASSUMPTION HA is a weak acid so it will not have dissociated very much. You can assume that its equilibrium concentration is approximately that of the original concentration

WORKED EXAMPLE Calculating p. H – weak acids Calculate the p. H of a weak acid HX of concentration 0. 1 M ( Ka = 4 x 10 -5 mol dm-3 ) HX dissociates as follows HX(aq) Dissociation constant for a weak acid Ka H+(aq) = + X¯(aq) [H+(aq)] [X¯(aq)] mol dm-3 [HX(aq)] Substitute for X¯ as ions are formed in equal amounts and the rearrange equation [H+(aq)] = [HX(aq)] Ka mol dm-3 ASSUMPTION HA is a weak acid so it will not have dissociated very much. You can assume that its equilibrium concentration is approximately that of the original concentration [H+(aq)] ANSWER = 0. 1 x 4 x 10 -5 mol dm-3 = 4. 00 x 10 -6 mol dm-3 = 2. 00 x 10 -3 mol dm-3 p. H = - log [H+(aq)] = 2. 699

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

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

p. H of mixtures WORKED EXAMPLE Strong acids and alkalis (either in excess) Calculate the p. H of a mixture of 25 cm 3 of 0. 1 M Na. OH is added to 20 cm 3 of 0. 1 M HCl

p. H of mixtures WORKED EXAMPLE Strong acids and alkalis (either in excess) Calculate the p. H of a mixture of 25 cm 3 of 0. 1 M Na. OH is added to 20 cm 3 of 0. 1 M HCl 1. Calculate the number of moles of H+ and OH¯ ions present 25 cm 3 of 20 cm 3 of 0. 1 M Na. OH 0. 1 M HCl 2. 5 x 10 -3 moles 2. 0 x 10 -3 moles of OH ¯ = 0. 1 x 25/1000 = 2. 5 x 10 -3 moles of H+ = 20 x 20/1000 = 2. 0 x 10 -3

WORKED EXAMPLE p. H of mixtures Strong acids and alkalis (either in excess) Calculate the p. H of a mixture of 25 cm 3 of 0. 1 M Na. OH is added to 20 cm 3 of 0. 1 M HCl 1. Calculate the number of moles of H+ and OH¯ ions present 2. As the ions react in a 1: 1 ratio, calculate the unreacted moles of the excess species 25 cm 3 of 20 cm 3 of 0. 1 M Na. OH 0. 1 M HCl 2. 5 x 10 -3 moles 2. 0 x 10 -3 moles The reaction taking place is… or in its ionic form HCl + Na. OH H+ + OH¯ Na. Cl H 2 O + H 2 O (1: 1 molar ratio)

WORKED EXAMPLE p. H of mixtures Strong acids and alkalis (either in excess) Calculate the p. H of a mixture of 25 cm 3 of 0. 1 M Na. OH is added to 20 cm 3 of 0. 1 M HCl 1. Calculate the number of moles of H+ and OH¯ ions present 2. As the ions react in a 1: 1 ratio, calculate the unreacted moles of the excess species 25 cm 3 of 20 cm 3 of 0. 1 M Na. OH 0. 1 M HCl 2. 5 x 10 -3 moles 5. 0 x 10 -4 moles of OH¯ 2. 0 x 10 -3 moles The reaction taking place is… or in its ionic form UNREACTED HCl + Na. OH H+ + OH¯ Na. Cl H 2 O + H 2 O (1: 1 molar ratio) 2. 0 x 10 -3 moles of H+ will react with the same number of moles of OH¯ this leaves 2. 5 x 10 -3 - 2. 0 x 10 -3 = 5. 0 x 10 -4 moles of OH¯ in excess

p. H of mixtures WORKED EXAMPLE Strong acids and alkalis (either in excess) Calculate the p. H of a mixture of 25 cm 3 of 0. 1 M Na. OH is added to 20 cm 3 of 0. 1 M HCl 1. Calculate the number of moles of H+ and OH¯ ions present 2. As the ions react in a 1: 1 ratio, calculate the unreacted moles of the excess species 3. Calculate the volume of the solution by adding the two individual volumes the volume of the solution is 25 + 20 = 45 cm 3

p. H of mixtures WORKED EXAMPLE Strong acids and alkalis (either in excess) Calculate the p. H of a mixture of 25 cm 3 of 0. 1 M Na. OH is added to 20 cm 3 of 0. 1 M HCl 1. Calculate the number of moles of H+ and OH¯ ions present 2. As the ions react in a 1: 1 ratio, calculate the unreacted moles of the excess species 3. Calculate the volume of the solution by adding the two individual volumes 4. Convert volume to dm 3 (divide cm 3 by 1000) the volume of the solution is 25 + 20 = 45 cm 3 there are 1000 cm 3 in 1 dm 3 volume = 45/1000 = 0. 045 dm 3

p. H of mixtures WORKED EXAMPLE Strong acids and alkalis (either in excess) Calculate the p. H of a mixture of 25 cm 3 of 0. 1 M Na. OH is added to 20 cm 3 of 0. 1 M HCl 1. Calculate the number of moles of H+ and OH¯ ions present 2. As the ions react in a 1: 1 ratio, calculate the unreacted moles of the excess species 3. Calculate the volume of the solution by adding the two individual volumes 4. Convert volume to dm 3 (divide cm 3 by 1000) 5. Divide moles by volume to find concentration of excess ion in mol dm -3 [OH¯] = 5. 0 x 10 -4 / 0. 045 = 1. 11 x 10 -2 mol dm-3

WORKED EXAMPLE p. H of mixtures Strong acids and alkalis (either in excess) Calculate the p. H of a mixture of 25 cm 3 of 0. 1 M Na. OH is added to 20 cm 3 of 0. 1 M HCl 1. Calculate the number of moles of H+ and OH¯ ions present 2. As the ions react in a 1: 1 ratio, calculate the unreacted moles of the excess species 3. Calculate the volume of the solution by adding the two individual volumes 4. Convert volume to dm 3 (divide cm 3 by 1000) 5. Divide moles by volume to find concentration of excess ion in mol dm -3 6. As the excess is OH¯ use or [OH¯] [H+] p. H = 5. 0 x 10 -4 / 0. 045 = Kw / [OH¯] = - log[H+] p. OH = - log[OH¯] Kw = [H+][OH¯] then p. H + p. OH = 14 so [H+] = Kw / [OH¯] = 1. 11 x 10 -2 mol dm-3 = 9. 00 x 10 -13 mol dm-3 = 12. 05 Kw = 1 x 10 -14 mol 2 dm-6 (at 25°C)

Buffer solutions - Brief introduction Definition “Solutions which resist changes in p. H when small quantities of acid or alkali are added. ” Acidic Buffer (p. H < 7) made from Alkaline Buffer (p. H > 7) made from a Uses a weak acid + its sodium or potassium salt ethanoic acid sodium ethanoate weak base ammonia Standardising p. H meters Buffering biological systems (eg in blood) Maintaining the p. H of shampoos + its chloride ammonium chloride

REVISION CHECK What should you be able to do? Recall the definition of acids and bases in the Brønsted-Lowry system Recall the definition of acids and bases in the Lewis system Recall and explain the difference between strong and weak acids Recall and explain the difference between strong and weak bases Recall the definition of p. H Recall the definition of the ionic product of water Explain how and why p. H varies with temperature Recall the relationship between p. H, [H+], [OH¯], p. OH and Kw CAN YOU DO ALL OF THESE? YES NO

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