CONTENTS Brønsted-Lowry theory of acids and bases Lewis theory of acids and bases Strong acids and...

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

• Relation between pH and pOH

• Introduction to buffer solutions

• Check list

Acid & BasesAcid & Bases

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


Conjugate Pairs

HCl (aq)

+ H2O (aq) H3O+(aq) + Cl-

(aq)

Acid Acid BaseBase

Conjugate pair

Conjugate pair

The negative ion is referred to as the conjugate base of the acid

HCl (aq)

+ H2O (aq) H3O+(aq) + Cl-

(aq)

Acid Conjugate Acid

Conjugate Base

Base

The positive ion is referred to as the conjugate acid of the base

CH3COOH (aq)

+ NH3 (aq) NH4+

(aq) + CH3COO-(aq)

Acid Conjugate Acid

Conjugate Base

Base

H2O (aq)

+ NH3 (aq) NH4+

(aq) + OH-(aq)

Acid Conjugate Acid

Conjugate Base

Base

HCOOH (aq)

+ H2O (aq) H3O+(aq) + HCOO-

(aq)

Acid Conjugate Acid

Conjugate Base

Base

• Water can act as an acid and a base – it can be described as amphoteric

HCOOH (aq)

+ H2O (aq) H3O+(aq) + HCOO-

(aq)

H2O (aq)

+ NH3 (aq) NH4+

(aq) + OH-(aq)

Pure water will always contain some H+ & OH- ions it is described as self protonating

2H2O (aq) H3O+

(aq) + OH-(aq)

HNO3 (aq)

+ HF (aq) H2F+(aq) + NO3

-(aq)

Acid Conjugate Acid

Conjugate Base

Base

H2SO4 (aq)

+ HNO3 (aq) H2NO3+

(aq) + HSO4-(aq)

Acid Conjugate Acid

Conjugate Base

Base

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 ——> 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.

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 solutionhydroxide 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

STRONGLY ACIDIC

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+]

WEAKLY ACIDIC

NEUTRAL STRONGLY ALKALINE

WEAKLY ALKALINE

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)]





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


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)

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

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

CONTENTS

• What is pH? - a reminder

• Calculating the pH of strong acids and bases

• Calculating the pH of weak acids

• Calculating the pH of mixtures - strong acid and strong alkali

pH calculationspH calculations


• know the differences between strong and weak acid and bases

• be able to calculate pH from hydrogen ion concentration

• be able to calculate hydrogen ion concentration from pH

• know the formula for the ionic product of water and its value at 25°C

pH calculationspH calculations

What is pH or pOH?What is pH or pOH?

pH = - log10 [H+(aq)]

pOH = - log10 [OH-(aq)]

where [H+] is the concentration of hydrogen ions in mol dm-3

where [OH-] is the concentration of hydroxide ions in mol dm-3

to convert pH intohydrogen ion concentration [H+(aq)] = antilog (-pH) to convert pOH into [OH-(aq)] = antilog (-pOH)OH- concentration

IONIC PRODUCT OF WATER Kw = [H+(aq)] [OH¯(aq)] mol2 dm-6

= 1 x 10-14 mol2 dm-6 (at 25°C)

pKw = 14 = -log(1 x 10-14 )

pKw = pH + pOH

pH + pOH = 14

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 or pOH ; 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

WORKEDEXAMPLEWORKEDEXAMPLE

pH = - log [H+]

= - log [0.02]

= 1.7

Note that negative pH vales are also possibleFor example : The pH of 1.5 mol dm–3 HCl is -0.18

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 or pOH ; you only need to know the concentrationonly need to know the concentration.

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


pH + pOH = 14

pH + 1 = 14

pH = 13

To find pH of a strong base use pH + pOH = 14

pOH = -log [OH¯]

= -log [ 1 ] =

1

Calculating pH - Calculating pH - weak acidsweak acids

A weak acid, HA, dissociates as follows HA(aq) H+(aq) + A¯(aq)(1)

A weak acid is one which only partially dissociates in aqueous solution



Applying the Equilibrium Law Ka = [H+(aq)] [A¯(aq)] mol dm-3 (2)

[HA(aq)]





[HA(aq)]

The ions are formed in equal amounts, so [H+(aq)] = [A¯(aq)]

therefore Ka = [H+(aq)]2 (3)

[HA(aq)]





[HA(aq)]



[HA(aq)]

Rearranging (3) gives [H+(aq)]2

= [HA(aq)] Ka

therefore [H+(aq)] = [HA(aq)] Ka





[HA(aq)]



[HA(aq)]


= [HA(aq)] Ka


pH = [H+(aq)]





[HA(aq)]



[HA(aq)]


= [HA(aq)] Ka


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.



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)]






[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

CONTENTS

• What is a buffer solution?

• Uses of buffer solutions

• Acidic buffer solutions

• Alkaline buffer solutions

• Buffer solutions - ideal concentration

• Calculating the pH of a buffer solution

• Salt hydrolysis

• Check list

Buffer solutionsBuffer solutions


• know that weak acids and bases are only partly ionised in solution

• be able to calculate pH from hydrogen ion concentration

• be able to construct an equation for the dissociation constant of a weak acid

Buffer solutionsBuffer solutions

Buffers

• A buffer is a solution whose pH is resistant to change on the addition of relatively small quantities of an acid or base.

• Buffers have the ability to absorb added H+

or OH- ions.

Buffer solutions - usesBuffer solutions - uses

Definition “Solutions which resist changes in pH when small quantities of acid or alkali are added.”

Biological Uses

In biological systems (saliva, stomach, and blood) it is essential thatthe pH stays ‘constant’ in order for any processes to work properly.e.g. If the pH of blood varies by 0.5 it can lead to unconsciousness and coma

Most enzymes work best at particular pH values.

Other Uses Many household and cosmetic products need to control their pH values.

Shampoo Buffer solutions counteract the alkalinity of the soap and prevent irritation

Baby lotion Buffer solutions maintain a pH of about 6 to prevent bacteria multiplying

Others Washing powder, eye drops, fizzy lemonade

Acidic Buffers

• Contain a weak acid and its salt formed with a strong base

• Ethanoic acid & Sodium ethanoate

CH3COOH & CH3COO-Na+

Buffer Action

• The important species present in the buffer solution are the

• undissociated weak acid CH3COOH

• and its conjugate base CH3COO-

• Undissociated ethanoic acid can remove any added OH- ions.

• CH3COOH + OH- → CH3COO- + H2O

• The conjugate base can remove any added H+ ions

• CH3COO- + H+ → CH3COOH

• For a buffer to work effectively it must contain a large reservoir of the weak acid and its conjugate base.

• CH3COOH + H2O CH3COO- + H3O+

• CH3COOH CH3COO- + H+

• Write an expression for Ka state units

Ka = [CH3COO-] [H+] [CH3COOH]

CH3COOH CH3COO- + H+

[H+] = Ka x [CH3COOH] [CH3COO-]

If both [CH3COOH] and [CH3COO-] are large………..


• If both [CH3COOH] and [CH3COO-] are large

• small changes in their concentrations will not affect the overall ratio significantly

• so [H+] remains almost constant

• So very small change in pH

Basic Buffers

• Contain a weak base and its salt formed with a strong acid

• Ammonia & Ammonium chloride

NH3 & NH4+

Cl-

Buffer Action

• The important species present in the buffer solution are the

• undissociated weak base NH3

• and its conjugate acid NH4+

• Undissociated ammonia can remove any added H+ ions.

• NH3 + H+ → NH4+

• The conjugate acid can remove any added OH- ions

• NH4+ + OH- → NH3 + H2O

Calculating the pH of a Buffer

• When performing acidic buffer calculations it is assumed that

• The acid is completely undissociated

• The A- ions are formed solely from the salt

Ka = [H+] [A-] [HA]

HA H+ + A-

[H+] = Ka x [HA] [A-]

From saltWeak acid completely undissociated

Calculate the pH of a buffer solution produced by adding 3.28g of sodium ethanoate to 1dm3 of 0.01M ethanoic acid. The Ka of ethanoic acid is 1.84 x 10-5 at 300K

moles = mass/Mr

1. Find the number of moles of sodium ethanoate

Moles = 3.28/82 = 0.04

2. Find the [CH3COOH] and [CH3COO-]

[CH3COOH] = 0.01M [CH3COO-] = 0.04M

3. Find [H+]


[H+] = 1.84 x 10-5 x 0.01 0.04

= 4.6 x 10-6

3. Find pH

pH = -log[H+]

pH = -log 4.6 x 10-6 = 5.34

Calculate the pH of this buffer if 10cm3 of 0.1M HCl are now added

1. Find the number of moles of acid added

moles = conc x volume

moles of acid = 0.1 x 10/1000 = 1 x 10-3

CH3COO- + H+ → CH3COOH

1 x 10-3 + 1 x 10-3 → 1 x 10-3

Buffer solutions - Buffer solutions - ideal concentrationideal concentration

The concentration of a buffer solution is also important

If the concentration is too low, there won’t be enough CH3COOH and CH3COO¯

to cope with the ions added.

SummaryFor an acidic buffer solution one needs ...

large [CH3COOH(aq)] - for dissociating into H+(aq) when alkali is added

large [CH3COO¯(aq)] - for removing H+(aq) as it is added

This situation can’t exist if only acid is present; a mixture of the acid and salt is used.

The weak acid provides the equilibrium and the large CH3COOH(aq) concentration.

The sodium salt provides the large CH3COO¯(aq) concentration.

One uses a WEAK ACID + its SODIUM OR POTASSIUM SALT

Calculating the pH of an acidic buffer solutionCalculating the pH of an acidic buffer solution

Ka * concentration of acid

concentration of salt[H+] =

Ka * concentration of acidconcentration of salt

Log [H+] = Log

= Log Ka + Log concentration of acidconcentration of salt

=> (both sides put log)

=>(multiply all by -1) = - Log Ka - Log concentration of acidconcentration of salt

- Log [H+]

pKa - Log concentration of acidconcentration of salt

pH =

Calculating the pH of an acidic buffer solutionCalculating the pH of an acidic buffer solution

Calculate the pH of a buffer whose concentration of acid is 0.40 mol dm-3

Methanoic acid and is 1 mol dm-3 sodium ethanoate

Ka of ethaoic acid = 1.6*10-4

Ka * concentration of acid

concentration of salt[H+] =

1.6*10-4 *

0.40

1=

= 6.4*10-5

pH = -log (6.4*10-5 ) = 4.19

SALT HYDROLYSISSALT HYDROLYSIS

Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems.

All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate.




Salts of strong acids and strong bases - SODIUM CHLORIDE

NaCl dissociates completely in water Na+ Cl¯ ——> Na+ + Cl¯ Water only ionises to a very small extent H2O OH¯ + H+

Na+ and OH¯ are ions of a strong base so remain apart H+and Cl¯ are ions of a strong acid so remain apart

all the OH¯ and H+ ions remain in solution

therefore [H+] = [OH¯] and the solution will be NEUTRAL




Salts of strong acids and weak bases - AMMONIUM CHLORIDE

NH4Cl dissociates completely in waterNH4+ Cl¯ ——> NH4

+ + Cl¯

Water only ionises to a very small extent H2O OH¯ + H+

NH3 + H2O

Na+ and OH¯ are ions of a strong base so tend to be associatedH+and Cl¯ are ions of a strong acid so remain apart

all the H+ ions remain in solution

therefore [H+] > [OH¯] and the solution will be ACIDIC




Salts of weak acids and strong bases - SODIUM ETHANOATE

CH3COONa dissociates completely in water CH3COO ¯ Na + ——> Na+ + CH3COO¯


CH3COOH

Na+ and OH¯ are ions of a strong base so remain apartH+and CH3OO¯ are ions of a weak acid so tend to be associated

all the OH¯ ions remain in solution

therefore [OH¯] > [H+] and the solution will be ALKALINE




Salts of weak acids and weak bases - AMMONIUM ETHANOATE

CH3COONH4 dissociates completely in water CH3COO ¯ NH4 + ——> NH4+ + CH3COO¯


NH3 + H2O CH3COOH

Na+ and OH¯ are ions of a weak base so tend to be associatedH+and CH3OO¯ are ions of a weak acid so tend to be associated

the solution might be alkaline or acidic i.e. APPROXIMATELY NEUTRAL

REVISION CHECKREVISION CHECK

What should you be able to do?

Recall the definition of a buffer solution

Recall the difference between an acidic and an alkaline buffer solution

Recall the uses of buffer solutions

Understand the action of buffer solutions

Calculate the pH of an acidic buffer solution

Recall and understand the reactions due to salt hydrolysis

CAN YOU DO ALL OF THESE? CAN YOU DO ALL OF THESE? YES YES NONO

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REVISION CHECKREVISION CHECK

What should you be able to do?

Calculate pH from hydrogen ion concentration

Calculate hydrogen ion concentration from pH

Write equations to show the ionisation in strong and weak acids

Calculate the pH of strong acids and bases knowing their molar concentration

Calculate the pH of weak acids knowing their Ka and molar concentration

Calculate the pH of mixtures of strong acids and strong bases

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

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

CONTENTS Brønsted-Lowry theory of acids and bases Lewis theory of acids and bases Strong acids and...

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Transcript of CONTENTS Brønsted-Lowry theory of acids and bases Lewis theory of acids and bases Strong acids and...