ph and ionic equilibria a guide for a level students knockhardy publishing 2015 specifications

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

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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





Conjugate systemsAcids are related to bases ACID PROTON + CONJUGATE BASE

Bases are related to acids BASE + PROTON CONJUGATE ACID


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




LEWIS THEORY

ACID lone pair acceptor BF3 H+ AlCl3

BASE lone pair donor NH3 H2O


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






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 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)]


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)





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.







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


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 / 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


• 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




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


• 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



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 0.02M 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




Assumptions For a weak acid there is little dissociation


[HA(aq)]equil ~ [HA(aq)]undis







This assumption becomes less

valid for strongerweak acids where

there is more 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’.



[H2O(l)] is ‘constant’






Combine this ‘constant’ with Kc

to get a new constant (Ka) where Ka = Kc [H2O(l)]




Ka = [H3O+(aq) ] [A¯(aq)] mol dm-3

[HA(aq)]








A simpler way to express the dissociation is HA(aq) H+(aq) + A¯(aq)





[HA(aq)]









The dissociation constant Ka is then Ka = [H+(aq)] [A¯(aq)] mol dm-3

[HA(aq)]





[HA(aq)]










[HA(aq)]





[HA(aq)]









[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


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



Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3

[HX(aq)]





[HX(aq)]

Substitute for X¯ as ions are formed in [H+(aq)] = [HX(aq)] Ka mol dm-3

equal amounts and then rearrange equation





[HX(aq)]


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





[HX(aq)]


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


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


Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)



1. Calculate the number of moles of H+ and OH¯ ions present



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




2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species



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







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





3. Calculate the volume of the solution by adding the two individual volumes



the volume of the solution is 25 + 20 = 45cm3









the volume of the solution is 25 + 20 = 45cm3

there are 1000 cm3 in 1 dm3

volume = 45/1000 = 0.045dm3







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






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¯]



Kw = 1 x 10-14

mol2 dm-6 (at 25°C)

Kw = 1 x 10-14

mol2 dm-6 (at 25°C)


[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

You need to go over the You need to go over the relevant topic(s) againrelevant topic(s) again

Click on the button toClick on the button toreturn to the menureturn to the menu

WELL DONE!WELL DONE!Try some past paper questionsTry some past paper questions

© 2015 JONATHAN HOPTON & KNOCKHARDY PUBLISHING© 2015 JONATHAN HOPTON & KNOCKHARDY PUBLISHING

pH and IONIC pH and IONIC EQUILIBRIAEQUILIBRIA

THE ENDTHE END

ph and ionic equilibria a guide for a level students knockhardy publishing 2015 specifications

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