ph and ionic equilibria a guide for a level students knockhardy publishing 2015 specifications
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pH and IONIC pH and IONIC EQUILIBRIAEQUILIBRIA
A guide for A level studentsA guide for A level students
KNOCKHARDY PUBLISHINGKNOCKHARDY PUBLISHING20152015
SPECIFICATIONSSPECIFICATIONS
pH and Ionic EquilibriapH 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...
www.knockhardy.org.uk/sci.htm
Navigation is achieved by...
either clicking on the grey arrows at the foot of each page
or using 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 Kw
• 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 EquilibriapH and Ionic Equilibria
Before you start it would be helpful to…
• know the simple properties of acids, bases and alkalis
Acid & BasesAcid & Bases
BRØNSTED-LOWRY THEORY
ACID proton donor HCl ——> H+(aq) + Cl¯(aq)
BASE proton acceptor NH3(aq) + H+(aq) ——> NH4+(aq)
ACIDS AND BASESACIDS AND BASES
BRØNSTED-LOWRY THEORY
ACID proton donor HCl ——> H+(aq) + Cl¯(aq)
BASE proton acceptor NH3(aq) + H+(aq) ——> NH4+(aq)
ACIDS AND BASESACIDS AND BASES
Conjugate systemsAcids are related to bases ACID PROTON + CONJUGATE BASE
Bases are related to acids BASE + PROTON CONJUGATE ACID
ACIDS AND BASESACIDS AND BASES
Conjugate systemsAcids 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 CH3COO¯ + H2O CH3COOH + OH¯
base acid acid base
ACID-BASEPAIR
BRØNSTED-LOWRY THEORY
ACID proton donor HCl ——> H+(aq) + Cl¯(aq)
BASE proton acceptor NH3(aq) + H+(aq) ——> NH4+(aq)
LEWIS THEORY
ACID lone pair acceptor BF3 H+ AlCl3
BASE lone pair donor NH3 H2O
ACIDS AND BASESACIDS AND BASES
LONE PAIR DONOR LONE PAIRACCEPTOR
LONE PAIR DONOR LONE PAIR ACCEPTOR
STRONGACIDS completely dissociate (split up) into ions in aqueous solution
e.g. HCl ——> H+(aq) + Cl¯(aq) MONOPROTIC 1 replaceable H
HNO3 ——> H+(aq) + NO3¯(aq)
H2SO4 ——> 2H+(aq) + SO42-(aq) DIPROTIC 2 replaceable H’s
STRONG ACIDS AND BASESSTRONG ACIDS AND BASES
STRONGACIDS completely dissociate (split up) into ions in aqueous solution
e.g. HCl ——> H+(aq) + Cl¯(aq) MONOPROTIC 1 replaceable H
HNO3 ——> H+(aq) + NO3¯(aq)
H2SO4 ——> 2H+(aq) + SO42-(aq) DIPROTIC 2 replaceable H’s
STRONGBASES completely dissociate into ions in aqueous solution
e.g. NaOH(s) ——> Na+(aq) + OH¯(aq)
STRONG ACIDS AND BASESSTRONG ACIDS AND BASES
Weak acids partially dissociate into ions in aqueous solution
e.g. ethanoic acid CH3COOH(aq) CH3COO¯(aq) + H+(aq)
When a weak acid dissolves inwater an equilibrium is set up HA(aq) + H2O(l) A¯(aq) + H3O+(aq)
The water stabilises the ions
To make calculations easier the dissociation can be written... HA(aq) A¯(aq) + H+(aq)
WEAK ACIDSWEAK ACIDS
Weak acids partially dissociate into ions in aqueous solution
e.g. ethanoic acid CH3COOH(aq) CH3COO¯(aq) + H+(aq)
When a weak acid dissolves inwater an equilibrium is set up HA(aq) + H2O(l) A¯(aq) + H3O+(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 ACIDSWEAK ACIDS
Weak acids partially dissociate into ions in aqueous solution
e.g. ethanoic acid CH3COOH(aq) CH3COO¯(aq) + H+(aq)
When a weak acid dissolves inwater an equilibrium is set up HA(aq) + H2O(l) A¯(aq) + H3O+(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 Ka or pKa values
WEAK ACIDSWEAK ACIDS
Weak acids partially dissociate into ions in aqueous solution
e.g. ethanoic acid CH3COOH(aq) CH3COO¯(aq) + H+(aq)
When a weak acid dissolves inwater an equilibrium is set up HA(aq) + H2O(l) A¯(aq) + H3O+(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 Ka or pKa values
The dissociation constant for the weak acid HA is Ka = [H+(aq)] [A¯(aq)] mol dm-3
[HA(aq)]
WEAK ACIDSWEAK 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
NH3 (aq) + H2O (l) NH4+ (aq) + OH¯ (aq)
as in the case of acids it is more simply written
NH3 (aq) + H+ (aq) NH4+ (aq)
WEAK BASESWEAK 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
NH3 (aq) + H2O (l) NH4+ (aq) + OH¯ (aq)
as in the case of acids it is more simply written
NH3 (aq) + H+ (aq) NH4+ (aq)
The weaker the base the less it dissociatesthe more the equilibrium lies to the left
The relative strengths of bases can be expressed as Kb or pKb values.
WEAK BASESWEAK BASES
Hydrogen ion concentration [HHydrogen ion concentration [H++(aq)(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 muchlarger than their weaker counterparts which only partially dissociate.
Hydrogen ion concentration [HHydrogen ion concentration [H++(aq)(aq)]]
pH hydrogen ion concentration can be converted to pH pH = - log10 [H+(aq)]
to convert pH into hydrogen ion concentration [H+(aq)] = antilog (-pH)
pOH An equivalent calculation for bases convertsthe hydroxide ion concentration to pOH pOH = - log10 [OH¯(aq)]
in both the above, [ ] represents the concentration in mol dm-3
Ionic product of water - KIonic product of water - Kww
Despite being covalent, water conducts electricity to a very small extent.
This is due to the slight ionisation ... H2O(l) + H2O(l) H3O+(aq) + OH¯(aq)
or, more simply H2O(l) H+(aq) + OH¯(aq)
Ionic product of water - KIonic product of water - Kww
Despite being covalent, water conducts electricity to a very small extent.
This is due to the slight ionisation ... H2O(l) + H2O(l) H3O+(aq) + OH¯(aq)
or, more simply H2O(l) H+(aq) + OH¯(aq)
Applying the equilibrium lawto the second equation gives Kc = [H+(aq)] [OH¯(aq)]
[ ] is the equilibrium concentration in mol dm-3 [H2O(l)]
Ionic product of water - KIonic product of water - Kww
Despite being covalent, water conducts electricity to a very small extent.
This is due to the slight ionisation ... H2O(l) + H2O(l) H3O+(aq) + OH¯(aq)
or, more simply H2O(l) H+(aq) + OH¯(aq)
Applying the equilibrium lawto the second equation gives Kc = [H+(aq)] [OH¯(aq)]
[ ] is the equilibrium concentration in mol dm-3 [H2O(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 - KIonic product of water - Kww
Despite being covalent, water conducts electricity to a very small extent.
This is due to the slight ionisation ... H2O(l) + H2O(l) H3O+(aq) + OH¯(aq)
or, more simply H2O(l) H+(aq) + OH¯(aq)
Applying the equilibrium lawto the second equation gives Kc = [H+(aq)] [OH¯(aq)]
[ ] is the equilibrium concentration in mol dm-3 [H2O(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)] mol2 dm-6
= 1 x 10-14 mol2 dm-6 (at 25°C)
Because the constant is based on an equilibrium, Kw VARIES WITH TEMPERATURE
Ionic product of water - KIonic product of water - Kww
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 mol2 dm-6 0.11 0.68 1.0 1.47 5.6
H+ / x 10-7 mol dm-3 0.33 0.82 1.0 1.27 2.37 pH 7.48 7.08 7 6.92 6.63
What does this tell you about the equation H2O(l) H+(aq) + OH¯(aq) ?
Ionic product of water - KIonic product of water - Kww
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 mol2 dm-6 0.11 0.68 1.0 1.47 5.6
H+ / x 10-7 mol dm-3 0.33 0.82 1.0 1.27 2.37 pH 7.48 7.08 7 6.92 6.63
What does this tell you about the equation H2O(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 pH decreases
• pH decreases as the temperature increases
Ionic product of water - KIonic product of water - Kww
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 mol2 dm-6 0.11 0.68 1.0 1.47 5.6
H+ / x 10-7 mol dm-3 0.33 0.82 1.0 1.27 2.37 pH 7.48 7.08 7 6.92 6.63
What does this tell you about the equation H2O(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 pH decreases
• pH decreases as the temperature increases
Because the equation moves to the right as thetemperature goes up, it must be an ENDOTHERMIC process
Relationship between pH and pOHRelationship between pH and pOH
Because H+ and OH¯ ions are producedin equal amounts when water dissociates [H+] = [OH¯] = 1 x 10-7 mol dm-3
their concentrations will be the same.
Kw = [H+] [OH¯] = 1 x 10-14 mol2 dm-6
take logs of both sides log[H+] + log[OH¯] = -14
multiply by minus - log[H+] - log[OH¯] = 14
change to pH and pOH pH + pOH = 14 (at 25°C)
STRONGLY ACIDIC
WEAKLY ACIDIC
NEUTRAL STRONGLY ALKALINE
WEAKLY ALKALINE
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
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
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14pH
OH¯
[H+]
14 13 12 11 10 9 8 7 6 5 4 3 2 1 0pOH
Relationship between pH and pOHRelationship between pH and pOH
Because H+ and OH¯ ions are producedin equal amounts when water dissociates [H+] = [OH¯] = 1 x 10-7 mol dm-3
their concentrations will be the same.
Kw = [H+] [OH¯] = 1 x 10-14 mol2 dm-6
take logs of both sides log[H+] + log[OH¯] = -14
multiply by minus - log[H+] - log[OH¯] = 14
change to pH and pOH pH + pOH = 14 (at 25°C)
N.B. As they are based on the position of equilibrium and that varies withtemperature, 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)
Kw is constant for any aqueous solution at the stated temperature
Calculating pH - Calculating pH - strong acids and alkalisstrong acids and alkalis
Strong acids and alkalis completely dissociate in aqueous solution
It is easy to calculate the pH; you only need to know the concentrationonly need to know the concentration.
WORKEDEXAMPLEWORKEDEXAMPLE
Calculating pH - Calculating pH - strong acids and alkalisstrong acids and alkalis
Strong acids and alkalis completely dissociate in aqueous solution
It is easy to calculate the pH; you only need to know the concentrationonly need to know the concentration.
Calculate the pH of 0.02M HCl
HCl completely dissociates in aqueous solution HCl H+ + Cl¯
One H+ is produced for each HCl dissociating so [H+] = 0.02M = 2 x 10-2 mol dm-3
pH = - log [H+] = 1.7
WORKEDEXAMPLEWORKEDEXAMPLE
Strong acids and alkalis completely dissociate in aqueous solution
It is easy to calculate the pH; you only need to know the concentrationonly need to know the concentration.
Calculate the pH of 0.02M HCl
HCl completely dissociates in aqueous solution HCl H+ + Cl¯
One H+ is produced for each HCl dissociating so [H+] = 0.02M = 2 x 10-2 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 10-1 mol dm-3
The ionic product of water (at 25°C) Kw = [H+][OH¯] = 1 x 10-14 mol2 dm-6
therefore [H+] = Kw / [OH¯] = 1 x 10-13 mol dm-3
pH = - log [H+] = 13
WORKEDEXAMPLEWORKEDEXAMPLECalculating pH - Calculating pH - strong acids and alkalisstrong 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 … Ka The dissociation constant for a weak acid
Calculating pH - Calculating pH - weakweak acidsacids
A weak monobasic acid (HA)dissociates in water thus. HA(aq) + H2O(l) H3O+(aq) + A¯(aq)
Applying the equilibrium law we get Kc = [H3O+(aq)] [A¯(aq)]
[ ] is the equilibrium concentration in mol dm-3 [HA(aq)] [H2O(aq)]
The dissociation constant for a weak acid - KThe dissociation constant for a weak acid - Kaa
A weak monobasic acid (HA)dissociates in water thus. HA(aq) + H2O(l) H3O+(aq) + A¯(aq)
Applying the equilibrium law we get Kc = [H3O+(aq)] [A¯(aq)]
[ ] is the equilibrium concentration in mol dm-3 [HA(aq)] [H2O(aq)]
Assumptions For a weak acid there is little dissociation
The dissociation constant for a weak acid - KThe dissociation constant for a weak acid - Kaa
[HA(aq)]equil ~ [HA(aq)]undis
A weak monobasic acid (HA)dissociates in water thus. HA(aq) + H2O(l) H3O+(aq) + A¯(aq)
Applying the equilibrium law we get Kc = [H3O+(aq)] [A¯(aq)]
[ ] is the equilibrium concentration in mol dm-3 [HA(aq)] [H2O(aq)]
Assumptions For a weak acid there is little dissociation
The dissociation constant for a weak acid - KThe dissociation constant for a weak acid - Kaa
[HA(aq)]equil ~ [HA(aq)]undis
This assumption becomes less
valid for strongerweak acids where
there is more dissociation.
This assumption becomes less
valid for strongerweak acids where
there is more dissociation.
A weak monobasic acid (HA)dissociates in water thus. HA(aq) + H2O(l) H3O+(aq) + A¯(aq)
Applying the equilibrium law we get Kc = [H3O+(aq)] [A¯(aq)]
[ ] is the equilibrium concentration in mol dm-3 [HA(aq)] [H2O(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 - KThe dissociation constant for a weak acid - Kaa
[HA(aq)]equil ~ [HA(aq)]undis
[H2O(l)] is ‘constant’
A weak monobasic acid (HA)dissociates in water thus. HA(aq) + H2O(l) H3O+(aq) + A¯(aq)
Applying the equilibrium law we get Kc = [H3O+(aq)] [A¯(aq)]
[ ] is the equilibrium concentration in mol dm-3 [HA(aq)] [H2O(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 Kc
to get a new constant (Ka) where Ka = Kc [H2O(l)]
The dissociation constant for a weak acid - KThe dissociation constant for a weak acid - Kaa
[HA(aq)]equil ~ [HA(aq)]undis
[H2O(l)] is ‘constant’
Ka = [H3O+(aq) ] [A¯(aq)] mol dm-3
[HA(aq)]
A weak monobasic acid (HA)dissociates in water thus. HA(aq) + H2O(l) H3O+(aq) + A¯(aq)
Applying the equilibrium law we get Kc = [H3O+(aq)] [A¯(aq)]
[ ] is the equilibrium concentration in mol dm-3 [HA(aq)] [H2O(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 Kc
to get a new constant (Ka) where Ka = Kc [H2O(l)]
A simpler way to express the dissociation is HA(aq) H+(aq) + A¯(aq)
The dissociation constant for a weak acid - KThe dissociation constant for a weak acid - Kaa
[HA(aq)]equil ~ [HA(aq)]undis
[H2O(l)] is ‘constant’
Ka = [H3O+(aq) ] [A¯(aq)] mol dm-3
[HA(aq)]
A weak monobasic acid (HA)dissociates in water thus. HA(aq) + H2O(l) H3O+(aq) + A¯(aq)
Applying the equilibrium law we get Kc = [H3O+(aq)] [A¯(aq)]
[ ] is the equilibrium concentration in mol dm-3 [HA(aq)] [H2O(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 Kc
to get a new constant (Ka) where Ka = Kc [H2O(l)]
A simpler way to express the dissociation is HA(aq) H+(aq) + A¯(aq)
The dissociation constant Ka is then Ka = [H+(aq)] [A¯(aq)] mol dm-3
[HA(aq)]
The dissociation constant for a weak acid - KThe dissociation constant for a weak acid - Kaa
[HA(aq)]equil ~ [HA(aq)]undis
[H2O(l)] is ‘constant’
Ka = [H3O+(aq) ] [A¯(aq)] mol dm-3
[HA(aq)]
A weak monobasic acid (HA)dissociates in water thus. HA(aq) + H2O(l) H3O+(aq) + A¯(aq)
Applying the equilibrium law we get Kc = [H3O+(aq)] [A¯(aq)]
[ ] is the equilibrium concentration in mol dm-3 [HA(aq)] [H2O(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 Kc
to get a new constant (Ka) where Ka = Kc [H2O(l)]
A simpler way to express the dissociation is HA(aq) H+(aq) + A¯(aq)
The dissociation constant Ka is then Ka = [H+(aq)] [A¯(aq)] mol dm-3
[HA(aq)]
The dissociation constant for a weak acid - KThe dissociation constant for a weak acid - Kaa
[HA(aq)]equil ~ [HA(aq)]undis
[H2O(l)] is ‘constant’
Ka = [H3O+(aq) ] [A¯(aq)] mol dm-3
[HA(aq)]
This assumption becomes less
valid for strongerweak acids where
there is more dissociation.
This assumption becomes less
valid for strongerweak acids where
there is more dissociation.
A weak monobasic acid (HA)dissociates in water thus. HA(aq) + H2O(l) H3O+(aq) + A¯(aq)
The dissociation constant Ka is then Ka = [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
pKa • very weak acids have very small Ka values• it is easier to compare the strength as pKa values
The conversion is carried out thus... pKa = - log10 Ka
To convert pKa into Ka Ka = antilog (-pKa) or 10-Ka
The dissociation constant for a weak acid - KThe dissociation constant for a weak acid - Kaa
A weak acid, HA, dissociates as follows HA(aq) H+(aq) + A¯(aq)(1)
Applying the Equilibrium Law Ka = [H+(aq)] [A¯(aq)] mol dm-3 (2)
[HA(aq)]
The ions are formed in equal amounts, so [H+(aq)] = [A¯(aq)]
therefore Ka = [H+(aq)]2 (3)
[HA(aq)]
Rearranging (3) gives [H+(aq)]2
= [HA(aq)] Ka
therefore [H+(aq)] = [HA(aq)] Ka mol dm-3
pH = [H+(aq)]
A weak acid is one which only partially dissociates in aqueous solution
Calculating pH – Calculating pH – weak acidsweak acids
A weak acid, HA, dissociates as follows HA(aq) H+(aq) + A¯(aq)(1)
Applying the Equilibrium Law Ka = [H+(aq)] [A¯(aq)] mol dm-3 (2)
[HA(aq)]
The ions are formed in equal amounts, so [H+(aq)] = [A¯(aq)]
therefore Ka = [H+(aq)]2 (3)
[HA(aq)]
Rearranging (3) gives [H+(aq)]2
= [HA(aq)] Ka
therefore [H+(aq)] = [HA(aq)] Ka 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 – Calculating pH – weak acidsweak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows HX(aq) H+(aq) + X¯(aq)
WORKEDEXAMPLEWORKEDEXAMPLECalculating pH – Calculating pH – weak acidsweak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows HX(aq) H+(aq) + X¯(aq)
Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
WORKEDEXAMPLEWORKEDEXAMPLECalculating pH – Calculating pH – weak acidsweak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows HX(aq) H+(aq) + X¯(aq)
Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
Substitute for X¯ as ions are formed in [H+(aq)] = [HX(aq)] Ka mol dm-3
equal amounts and then rearrange equation
WORKEDEXAMPLEWORKEDEXAMPLECalculating pH – Calculating pH – weak acidsweak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows HX(aq) H+(aq) + X¯(aq)
Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
Substitute for X¯ as ions are formed in [H+(aq)] = [HX(aq)] Ka mol dm-3
equal amounts and the rearrange equation
ASSUMPTIONHA 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
WORKEDEXAMPLEWORKEDEXAMPLECalculating pH – Calculating pH – weak acidsweak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows HX(aq) H+(aq) + X¯(aq)
Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
Substitute for X¯ as ions are formed in [H+(aq)] = [HX(aq)] Ka mol dm-3
equal amounts and the rearrange equation
ASSUMPTIONHA 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 10-5 mol dm-3
= 4.00 x 10-6 mol dm-3
= 2.00 x 10-3 mol dm-3
ANSWER pH = - log [H+(aq)] = 2.699
WORKEDEXAMPLEWORKEDEXAMPLECalculating pH – Calculating pH – weak acidsweak acids
CALCULATING THE pH OF MIXTURESCALCULATING 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 EXCESSSTRONG ACID and STRONG BASE - EITHER IN EXCESS
pH of mixturespH of mixtures
Strong acids and strong alkalis (either in excess)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 dm3 (divide cm3 by 1000)
5. Divide moles by volume to find concentration of excess the ion in mol dm-3
6. Convert concentration to pH
If the excess is H+ pH = - log[H+]
If the excess is OH¯ pOH = - log[OH¯] then
pH + pOH = 14
or use Kw = [H+] [OH¯] = 1 x 10-14 at 25°C therefore
[H+] = Kw / [OH¯] then
pH = - log[H+]
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
pH of mixturespH of mixtures
Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)
WORKEDEXAMPLEWORKEDEXAMPLE
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
pH of mixturespH of mixtures
Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)
25cm3 of
0.1M NaOH
20cm3 of
0.1M HCl
2.5 x 10-3 moles
2.0 x 10-3 moles
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
WORKEDEXAMPLEWORKEDEXAMPLE
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M 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
pH of mixturespH of mixtures
Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)
The reaction taking place is… HCl + NaOH NaCl + H2O
or in its ionic form H+ + OH¯ H2O (1:1 molar ratio)
25cm3 of
0.1M NaOH
20cm3 of
0.1M HCl
2.5 x 10-3 moles
2.0 x 10-3 moles
WORKEDEXAMPLEWORKEDEXAMPLE
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M 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
pH of mixturespH of mixtures
Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)
The reaction taking place is… HCl + NaOH NaCl + H2O
or in its ionic form H+ + OH¯ H2O (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
5.0 x 10-4
moles of OH¯
UNREACTED
25cm3 of
0.1M NaOH
20cm3 of
0.1M HCl
2.5 x 10-3 moles
2.0 x 10-3 moles
WORKEDEXAMPLEWORKEDEXAMPLE
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M 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
pH of mixturespH of mixtures
Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)
the volume of the solution is 25 + 20 = 45cm3
WORKEDEXAMPLEWORKEDEXAMPLE
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M 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 dm3 (divide cm3 by 1000)
pH of mixturespH of mixtures
Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)
the volume of the solution is 25 + 20 = 45cm3
there are 1000 cm3 in 1 dm3
volume = 45/1000 = 0.045dm3
WORKEDEXAMPLEWORKEDEXAMPLE
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M 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 dm3 (divide cm3 by 1000)
5. Divide moles by volume to find concentration of excess ion in mol dm-3
pH of mixturespH of mixtures
Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)
WORKEDEXAMPLEWORKEDEXAMPLE
[OH¯] = 5.0 x 10-4 / 0.045 = 1.11 x 10-2 mol dm-3
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M 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 dm3 (divide cm3 by 1000)
5. Divide moles by volume to find concentration of excess ion in mol dm-3
6. As the excess is OH¯ use pOH = - log[OH¯] then pH + pOH = 14
or Kw = [H+][OH¯] so [H+] = Kw / [OH¯]
pH of mixturespH of mixtures
Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)
Kw = 1 x 10-14
mol2 dm-6 (at 25°C)
Kw = 1 x 10-14
mol2 dm-6 (at 25°C)
WORKEDEXAMPLEWORKEDEXAMPLE
[OH¯] = 5.0 x 10-4 / 0.045 = 1.11 x 10-2 mol dm-3
[H+] = Kw / [OH¯] = 9.00 x 10-13 mol dm-3
pH = - log[H+] = 12.05
Buffer solutions - Buffer solutions - BriefBrief introductionintroduction
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
Uses Standardising pH metersBuffering biological systems (eg in blood)Maintaining the pH of shampoos
REVISION CHECKREVISION 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 Kw
CAN YOU DO ALL OF THESE? CAN YOU DO ALL OF THESE? YES YES NONO
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