pH and IONIC EQUILIBRIA A guide for A level students KNOCKHARDY PUBLISHING 2015 SPECIFICATIONS

pH and Ionic Equilibria INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 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 A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at... Navigation is achieved by... either clicking on the grey arrows at the foot of each page orusing the left and right arrow keys on the keyboard

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 pH Ionic product of water K w Relationship between pH and pOH Calculating the pH of strong acids Calculating the pH of weak acids Calculating the pH of mixtures A brief introduction to buffer solutions Check list pH and Ionic Equilibria

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

BRØNSTED-LOWRY THEORY ACIDproton donorHC l ——> H + (aq) + C l ¯ (aq) BASEproton acceptor NH 3 (aq) + H + (aq) ——> NH 4 + (aq) ACIDS AND BASES

BRØNSTED-LOWRY THEORY ACIDproton donorHC l ——> H + (aq) + C l ¯ (aq) BASEproton acceptor NH 3 (aq) + H + (aq) ——> NH 4 + (aq) ACIDS AND BASES Conjugate systems Acids are related to bases ACID PROTON + CONJUGATE BASE Bases are related to acids BASE + PROTON CONJUGATE ACID

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

LEWIS THEORY ACIDlone pair acceptorBF 3 H + AlC l 3 BASElone pair donorNH 3 H 2 O ACIDS AND BASES LONE PAIR DONOR LONE PAIR ACCEPTOR LONE PAIR DONOR LONE PAIR ACCEPTOR

STRONG ACIDS completely dissociate (split up) into ions in aqueous solution e.g. HC l ——> H + (aq) + C l ¯ (aq) MONOPROTIC1 replaceable H HNO 3 ——> H + (aq) + NO 3 ¯ (aq) H 2 SO 4 ——> 2H + (aq) + SO 4 2- (aq) DIPROTIC 2 replaceable H’s STRONG ACIDS AND BASES

STRONG ACIDS completely dissociate (split up) into ions in aqueous solution e.g. HC l ——> H + (aq) + C l ¯ (aq) MONOPROTIC1 replaceable H HNO 3 ——> H + (aq) + NO 3 ¯ (aq) H 2 SO 4 ——> 2H + (aq) + SO 4 2- (aq) DIPROTIC 2 replaceable H’s STRONG BASES completely dissociate into ions in aqueous solution e.g. NaOH (s) ——> Na + (aq) + OH¯ (aq) STRONG ACIDS AND BASES

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

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

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

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) + H + (aq) NH 4 + (aq) The weaker the basethe less it dissociates the more the equilibrium lies to the left The relative strengths of bases can be expressed as K b or pK b values. WEAK BASES

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

Ionic product of water - K w 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) H 3 O + (aq) + OH¯(aq) or, more simply H 2 O(l) H + (aq) + OH¯(aq)

Ionic product of water - K w 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) H 3 O + (aq) + OH¯(aq) or, more simply H 2 O(l) H + (aq) + OH¯(aq) Applying the equilibrium law to the second equation gives K c = [H + (aq)] [OH¯(aq)] [ ] is the equilibrium concentration in mol dm -3 [H 2 O(l)]

Ionic product of water - K w 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) H 3 O + (aq) + OH¯(aq) or, more simply H 2 O(l) H + (aq) + OH¯(aq) Applying the equilibrium law to the second equation gives K c = [H + (aq)] [OH¯(aq)] [ ] is the equilibrium concentration in mol dm -3 [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 - K w 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) H 3 O + (aq) + OH¯(aq) or, more simply H 2 O(l) H + (aq) + OH¯(aq) Applying the equilibrium law to the second equation gives K c = [H + (aq)] [OH¯(aq)] [ ] is the equilibrium concentration in mol dm -3 [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 (K c ) to get a new constant (K w ). K w = [H + (aq)] [OH¯(aq)] mol 2 dm -6 = 1 x mol 2 dm -6 (at 25°C) Because the constant is based on an equilibrium, K w VARIES WITH TEMPERATURE

Ionic product of water - K w The value of K w varies with temperature because it is based on an equilibrium. Temperature / °C K w / 1 x mol 2 dm H + / x mol dm pH What does this tell you about the equation H 2 O(l) H + (aq) + OH¯(aq) ?

Ionic product of water - K w The value of K w varies with temperature because it is based on an equilibrium. Temperature / °C K w / 1 x mol 2 dm H + / x mol dm pH What does this tell you about the equation H 2 O(l) H + (aq) + OH¯(aq) ? K w 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 pH decreases pH decreases as the temperature increases

Ionic product of water - K w The value of K w varies with temperature because it is based on an equilibrium. Temperature / °C K w / 1 x mol 2 dm H + / x mol dm pH What does this tell you about the equation H 2 O(l) H + (aq) + OH¯(aq) ? K w 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 pH decreases pH 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 pH and pOH Because H + and OH¯ ions are produced in equal amounts when water dissociates[H + ] = [OH¯] = 1 x mol dm -3 their concentrations will be the same. K w = [H + ] [OH¯] = 1 x mol 2 dm -6 take logs of both sideslog[H + ] + log[OH¯] = -14 multiply by minus- log[H + ] - log[OH¯] = 14 change to pH and pOHpH + pOH = 14 (at 25°C) STRONGLY ACIDIC WEAKLY ACIDIC NEUTRAL STRONGLY ALKALINE WEAKLY ALKALINE pH OH¯ [H + ] pOH

Relationship between pH and pOH Because H + and OH¯ ions are produced in equal amounts when water dissociates[H + ] = [OH¯] = 1 x mol dm -3 their concentrations will be the same. K w = [H + ] [OH¯] = 1 x mol 2 dm -6 take logs of both sideslog[H + ] + log[OH¯] = -14 multiply by minus- log[H + ] - log[OH¯] = 14 change to pH and pOHpH + pOH = 14 (at 25°C) N.B.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 (298K) Neutral solutions may be regarded as those where [H + ] = [OH¯]. Therefore a neutral solution is pH 7 only at a temperature of 25°C (298K) K w is constant for any aqueous solution at the stated temperature

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

Calculating pH - strong acids and alkalis Strong acids and alkalis completely dissociate in aqueous solution only need to know the concentration It is easy to calculate the pH; you only need to know the concentration. Calculate the pH of 0.02M HCl HC l completely dissociates in aqueous solution HC l H + + C l ¯ One H + is produced for each HCl dissociating so [H + ] = 0.02M = 2 x mol dm -3 pH = - log [H + ] = 1.7 WORKED EXAMPLE WORKED EXAMPLE

Strong acids and alkalis completely dissociate in aqueous solution only need to know the concentration It is easy to calculate the pH; you only need to know the concentration. Calculate the pH of 0.02M HC l HC l completely dissociates in aqueous solution HC l H + + C l ¯ One H + is produced for each HCl dissociating so [H + ] = 0.02M = 2 x mol dm -3 pH = - log [H + ] = 1.7 Calculate the pH of 0.1M NaOH NaOH completely dissociates in aqueous solution NaOH Na + + OH¯ One OH¯ is produced for each NaOH dissociating [OH¯] = 0.1M = 1 x mol dm -3 The ionic product of water (at 25°C) K w = [H + ][OH¯] = 1 x mol 2 dm -6 therefore [H + ] = K w / [OH¯] = 1 x mol dm -3 pH = - log [H + ] = 13 WORKED EXAMPLE WORKED EXAMPLE Calculating pH - strong acids and alkalis

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

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

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

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

A weak monobasic acid (HA) dissociates in water thus. HA (aq) + H 2 O (l ) H 3 O + (aq) + A¯ (aq) Applying the equilibrium law we getK c = [H 3 O + (aq) ] [A¯ (aq) ] [ ] is the equilibrium concentration in mol dm -3 [HA (aq) ] [H 2 O (aq) ] 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’. The dissociation constant for a weak acid - K a [HA (aq) ] equil ~ [HA (aq) ] undis [H 2 O (l) ] is ‘constant’

A weak monobasic acid (HA) dissociates in water thus. HA (aq) + H 2 O (l ) H 3 O + (aq) + A¯ (aq) Applying the equilibrium law we getK c = [H 3 O + (aq) ] [A¯ (aq) ] [ ] is the equilibrium concentration in mol dm -3 [HA (aq) ] [H 2 O (aq) ] 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’. Combine this ‘constant’ with K c to get a new constant (K a ) where K a = K c [H 2 O(l)] The dissociation constant for a weak acid - K a [HA (aq) ] equil ~ [HA (aq) ] undis [H 2 O (l) ] is ‘constant’ K a = [H 3 O + (aq) ] [A¯ (aq) ] mol dm -3 [HA (aq) ]

A weak monobasic acid (HA) dissociates in water thus. HA (aq) + H 2 O (l ) H 3 O + (aq) + A¯ (aq) Applying the equilibrium law we getK c = [H 3 O + (aq) ] [A¯ (aq) ] [ ] is the equilibrium concentration in mol dm -3 [HA (aq) ] [H 2 O (aq) ] 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’. Combine this ‘constant’ with K c to get a new constant (K a ) where K a = K c [H 2 O(l)] A simpler way to express the dissociation is HA (aq) H + (aq) + A¯ (aq) The dissociation constant for a weak acid - K a [HA (aq) ] equil ~ [HA (aq) ] undis [H 2 O (l) ] is ‘constant’ K a = [H 3 O + (aq) ] [A¯ (aq) ] mol dm -3 [HA (aq) ]

A weak monobasic acid (HA) dissociates in water thus. HA (aq) + H 2 O (l ) H 3 O + (aq) + A¯ (aq) Applying the equilibrium law we getK c = [H 3 O + (aq) ] [A¯ (aq) ] [ ] is the equilibrium concentration in mol dm -3 [HA (aq) ] [H 2 O (aq) ] 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’. Combine this ‘constant’ with K c to get a new constant (K a ) where K a = K c [H 2 O(l)] A simpler way to express the dissociation is HA (aq) H + (aq) + A¯ (aq) The dissociation constant K a is then K a = [H + (aq) ] [A¯ (aq) ] mol dm -3 [HA (aq) ] The dissociation constant for a weak acid - K a [HA (aq) ] equil ~ [HA (aq) ] undis [H 2 O (l) ] is ‘constant’ K a = [H 3 O + (aq) ] [A¯ (aq) ] mol dm -3 [HA (aq) ]

A weak monobasic acid (HA) dissociates in water thus. HA (aq) + H 2 O (l ) H 3 O + (aq) + A¯ (aq) Applying the equilibrium law we getK c = [H 3 O + (aq) ] [A¯ (aq) ] [ ] is the equilibrium concentration in mol dm -3 [HA (aq) ] [H 2 O (aq) ] 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’. Combine this ‘constant’ with K c to get a new constant (K a ) where K a = K c [H 2 O(l)] A simpler way to express the dissociation is HA (aq) H + (aq) + A¯ (aq) The dissociation constant K a is then K a = [H + (aq) ] [A¯ (aq) ] mol dm -3 [HA (aq) ] The dissociation constant for a weak acid - K a [HA (aq) ] equil ~ [HA (aq) ] undis [H 2 O (l) ] is ‘constant’ K a = [H 3 O + (aq) ] [A¯ (aq) ] mol dm -3 [HA (aq) ] This assumption becomes less valid for stronger weak acids where there is more dissociation. This assumption becomes less valid for stronger weak acids where there is more dissociation.

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

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

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

Calculate the pH of a weak acid HX of concentration 0.1M ( K a = 4x10 -5 mol dm -3 ) HX dissociates as followsHX (aq) H + (aq) + X¯ (aq) WORKED EXAMPLE WORKED EXAMPLE Calculating pH – weak acids

Calculate the pH of a weak acid HX of concentration 0.1M ( K a = 4x10 -5 mol dm -3 ) HX dissociates as followsHX (aq) H + (aq) + X¯ (aq) Dissociation constant for a weak acid K a = [H + (aq) ] [X¯ (aq) ] mol dm -3 [HX (aq) ] WORKED EXAMPLE WORKED EXAMPLE Calculating pH – weak acids

Calculate the pH of a weak acid HX of concentration 0.1M ( K a = 4x10 -5 mol dm -3 ) HX dissociates as followsHX (aq) H + (aq) + X¯ (aq) Dissociation constant for a weak acid K a = [H + (aq) ] [X¯ (aq) ] mol dm -3 [HX (aq) ] Substitute for X¯ as ions are formed in [H + (aq) ] = [HX (aq) ] K a mol dm -3 equal amounts and then rearrange equation WORKED EXAMPLE WORKED EXAMPLE Calculating pH – weak acids

Calculate the pH of a weak acid HX of concentration 0.1M ( K a = 4x10 -5 mol dm -3 ) HX dissociates as followsHX (aq) H + (aq) + X¯ (aq) Dissociation constant for a weak acid K a = [H + (aq) ] [X¯ (aq) ] mol dm -3 [HX (aq) ] Substitute for X¯ as ions are formed in [H + (aq) ] = [HX (aq) ] K a mol dm -3 equal amounts and the rearrange equation 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 WORKED EXAMPLE Calculating pH – weak acids

Calculate the pH of a weak acid HX of concentration 0.1M ( K a = 4x10 -5 mol dm -3 ) HX dissociates as followsHX (aq) H + (aq) + X¯ (aq) Dissociation constant for a weak acid K a = [H + (aq) ] [X¯ (aq) ] mol dm -3 [HX (aq) ] Substitute for X¯ as ions are formed in [H + (aq) ] = [HX (aq) ] K a mol dm -3 equal amounts and the rearrange equation 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) ] = 0.1 x 4 x mol dm -3 = 4.00 x mol dm -3 = 2.00 x mol dm -3 ANSWER pH = - log [H + (aq) ] = WORKED EXAMPLE WORKED EXAMPLE Calculating pH – weak acids

CALCULATING THE pH OF MIXTURES The method used to calculate the pH of a mixture of an acid and 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

pH 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 Convert concentration to pH If the excess isH + pH = - log[H + ] If the excess is OH¯ pOH = - log[OH¯] then pH + pOH = 14 or use K w = [H + ] [OH¯] = 1 x at 25°C therefore [H + ] = K w / [OH¯] then pH = - log[H + ]

Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l pH of mixtures Strong acids and alkalis (either in excess) WORKED EXAMPLE WORKED EXAMPLE

Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 1. Calculate the number of moles of H + and OH¯ ions present pH of mixtures Strong acids and alkalis (either in excess) 25cm 3 of 0.1M NaOH 20cm 3 of 0.1M HC l 2.5 x moles 2.0 x moles moles of OH ¯ = 0.1 x 25/1000 = 2.5 x moles of H + = 20 x 20/1000 = 2.0 x WORKED EXAMPLE WORKED EXAMPLE

Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 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 pH of mixtures Strong acids and alkalis (either in excess) The reaction taking place is… HC l + NaOH NaC l + H 2 O or in its ionic form H + + OH¯ H 2 O (1:1 molar ratio) 25cm 3 of 0.1M NaOH 20cm 3 of 0.1M HC l 2.5 x moles 2.0 x moles WORKED EXAMPLE WORKED EXAMPLE

Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 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 pH of mixtures Strong acids and alkalis (either in excess) The reaction taking place is… HC l + NaOH NaC l + H 2 O or in its ionic form H + + OH¯ H 2 O (1:1 molar ratio) 2.0 x moles of H + will react with the same number of moles of OH¯ this leaves 2.5 x x = 5.0 x moles of OH¯ in excess 5.0 x moles of OH¯ UNREACTED 25cm 3 of 0.1M NaOH 20cm 3 of 0.1M HC l 2.5 x moles 2.0 x moles WORKED EXAMPLE WORKED EXAMPLE

Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 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 pH of mixtures Strong acids and alkalis (either in excess) the volume of the solution is = 45cm 3 WORKED EXAMPLE WORKED EXAMPLE

Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 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) pH of mixtures Strong acids and alkalis (either in excess) the volume of the solution is = 45cm 3 there are 1000 cm 3 in 1 dm 3 volume = 45/1000 = 0.045dm 3 WORKED EXAMPLE WORKED EXAMPLE

Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 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 pH of mixtures Strong acids and alkalis (either in excess) WORKED EXAMPLE WORKED EXAMPLE [OH¯]= 5.0 x / = 1.11 x mol dm -3

Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 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 As the excess is OH¯ usepOH = - log[OH¯] then pH + pOH = 14 or K w = [H + ][OH¯] so [H + ] = K w / [OH¯] pH of mixtures Strong acids and alkalis (either in excess) K w = 1 x mol 2 dm -6 (at 25°C) WORKED EXAMPLE WORKED EXAMPLE [OH¯]= 5.0 x / = 1.11 x mol dm -3 [H + ] = K w / [OH¯] = 9.00 x mol dm -3 pH = - log[H + ] = 12.05

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

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 pH Recall the definition of the ionic product of water Explain how and why pH varies with temperature Recall the relationship between pH, [H + ], [OH¯], pOH and K w CAN YOU DO ALL OF THESE? YES NO

You need to go over the relevant topic(s) again Click on the button to return to the menu

WELL DONE! Try some past paper questions

© 2015 JONATHAN HOPTON & KNOCKHARDY PUBLISHING pH and IONIC EQUILIBRIA THE END