REVISIONACIDS & BASES

MODELS

ARRHENIUS THEORY An acid is a substance that produces hydrogen ions

(H+)/hydronium ions (H3O+ ) when it dissolves in water. A base is a substance that produces hydroxide ions (OH- )

when it dissolves in water.

LOWRY-BRØNSTED THEORYAn acid is a proton (H+ ion) donor.

A base is a proton (H+ ion) acceptor

ACIDSSTRONG ACIDSionise completely in water to form a high concentration of H3O+ ions. Examples of strong acids are hydrochloric acid, sulphuric acid and nitric acid.

WEAK ACIDSionise incompletely in water to form a low concentration of H3O+ ions. Examples of weak acids are ethanoic acid and oxalic acid.

POLYPROTIC ACIDSAcids that can donate more than one proton

AMPHOLYTECan be an acid OR a base

BASES

STRONG BASES dissociate completely in water to form a high

concentration of OH- ions. Examples of strong bases are sodium hydroxide and

potassium hydroxideWEAK BASES

dissociate/ionise incompletely in water to form a low concentration of OH ions.

Examples of weak bases are ammonia, calcium carbonate, potassium carbonate, calcium carbonate and sodium

hydrogen carbonate.

CONCENTRATED AND DILUTE

CONCENTRATED ACIDS/BASES contain a large amount (number of moles) of acid/base in

proportion to the volume of water.

DILUTE ACIDS/BASES contain a small amount (number of moles) of acid/base in

proportion to the volume of water

CONCENTRATED AND DILUTE

3

3

concentration (mol.dm )

number mole

volume (dm )

ncV

c

n

V

1 1 2 2

1

1

2

2

concentration before dilution

V volume before dilution

concentration after dilution

V volume after dilution

cV c V

c

c

IONISATION

HCℓ(g)+H2O(ℓ) →H3O+(aq) + Cℓ-(aq)

NH3(g) + H2O(ℓ) → NH+4(aq)+ OH-(aq)

H2SO4(aq) + 2H2O(ℓ) →2H3O+(aq) +SO24−(aq)

CONJUGATE ACID-BASE PAIRS

HCℓ(g)+H2O(ℓ) →H3O+(aq) + Cℓ-(aq)

NH3(g) + H2O(ℓ) → NH+4(aq)+ OH-(aq)

H2SO4(aq) + 2H2O(ℓ) →2H3O+(aq) +SO24−(aq)

NEUTRALISATION REACTIONSHCℓ(aq)+NaOH(aq)→NaCℓ(aq)+ H2O(ℓ)

HCℓ(aq)+KOH(aq)→KCℓ(aq)+ H2O(ℓ)

HCℓ(aq)+Na2CO3(aq)→NaCℓ(aq) +H2O(ℓ) + CO2(g)

HNO3(aq) +NaOH(aq)→NaNO3(aq) +H2O(ℓ)

H2SO4(aq) +2NaOH(aq)→Na2SO4(aq) +2H2O(ℓ)

(COOH)2(aq) +NaOH(aq)→(COO)2Na2(aq)+ H2O(ℓ)

CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(ℓ)

HYDROLYSIS & NEUTRALISATIONHYDROLYSIS as the reaction of a salt with water

Hydrolysis of the salt of a weak acid and a strong base results in an alkaline solution, i.e. the pH > 7. Examples of such salts are sodium ethanoate, sodium oxalate and sodium carbonate. O

Hydrolysis of the salt of a strong acid and a weak base results in an acidic solution, i.e. the pH < 7. An example of such a salt is ammonium chloride.

The salt of a strong acid and a strong bases does not undergo hydrolysis and the solution of the salt will be neutral, i.e. pH = 7.

PHENOLPHTHALEIN STRONG BASE WITH WEAK ACIDSTRONG BASE WITH STRONG ACID

BROMOTHYMOL STRONG BASE WITH STRONG ACID

METHYL ORANGEWEAK BASE WITH STRONG ACID

TITRATION

EQUIVALENCE POINT of a titration is the point at which the acid /base has

completely reacted with the base/acid.

ENDPOINT of a titration is the point where the indicator changes

colour.

EXPERIMENTOXALIC ACID WITH SODIUM HYDROXIDE

• List the apparatus needed or identify the apparatus from a diagram.

• Describe the procedure to prepare a standard oxalic acid solution.

• Describe the procedure to conduct the titration.• Describe safety precautions. o • Describe measures that need to be in place to ensure reliable

results. • Interpret given results to determine the unknown

concentration.

pH values of strong acids&bases

pH = -log[H3O+]

Kw as the equilibrium constant for the ionisation of water or the ionic product of water or the ionisation constant of water

Kw = [H3O+][OH-] = 1 x 1014 at 298 K

auto-ionisation of water, i.e. the reaction of water with itself to form H3O+ ions and OH- ions

pH values of strong acids&bases

2 4

22 4 4

2 4

3

3

Determine the pH of a 0.1M H solution

2

:

1 : 2

0.1 :

0.12 0.2 .

1

log

log 0.2

0.7

SO

H SO H SO

H SO H

x

x mol dm

pH H O

3

3 2 4

3

3

13

3

14 3

3

3

3

1

Determine the pH of a 0.1M

:

1 : 1

0.1 :

0.11 0.1 .

1

0.1

1 10

log

log 1

1 1  

1

0

 

10

3

wK

NH solution

NH H O NH OH

NH OH

x

x mol dm

pH H O

H O OH

H O

H O

Concentration from pH3

3

3

4 33

3 3

3

4

44 3

determine the concentration of a solution

with pH=4

log

4 log

1 10 .

:

1 : 1

: 1 10

1 101 1 10 .

1

CH COOH

pH H O

H O

H O mol dm

CH COOH CH COO H

CH COOH H

x

x mol dm

3

3

9 33

9

5

3

14 

3

determine the concentration of a solution

with pH=9

log

9 log

1 10 .

1 10

1 10 .

:

1 : 1

: 1 1

1 10  

w

NaOH

pH H O

H O

H O mo

K H O OH

OH

O

l dm

mol dm

NaOH Na OH

NaOH

H

OH

x

5

55 3

0

1 101 1 10 .

1x mol dm

-3A learner accidentally spills some sulphuric acid of concentration 6

from a flask on the laboratory bench. Her teacher tells her to neutralise

the spilled acid by sprinkling sodium hydrog

mo

e

l.dm

n ca

2 4 3 2 4 2

2

2

4

rbonate powder onto it.

The reaction that takes place is: (Assume that the H SO ionises completely.)

The fizzing, due

H SO a

to t

q + 2NaHCO s Na SO aq +

he formation of carbon

2H O l + 2CO g

dioxide, s

tops after the learner has

added 27 g sodium hydrogen carbonate to the spilled acid.

Calculate the volume of sulphuric acid that spilled.

Assume that all the sodium hydrogen carbonate reacts wit

a

h all

-3 3

3-3

the acid

The learner now dilutes some of the 6 sulphuric acid solution in the flask to 0,1 mol.

Calculate the volume of the 6 sulphuric acid solution needed to prep

mol.dm dm

b mol. are 1 dm odm f

th

3

3

3

e dilute acid.

During a titration 25 cm of the 0,1mol. sulphuric acid solution is added to an Erlenmeyer

flask and titrated with a 0,1mol. sodium hydroxide solution.

The learner uses bromo

dm

dm

c thymol

3

blue as indicator. What is the purpose of this indicator?

Calculate the pH of the solution in the flask after the addition of 30 cm of sodium hydroxide.

The endpoint of the titration is not ye

d

t reached at this point

3

2 4 2

3

270.321

48from balance equation:

:

1 : 2

: 0.321

0.3211 0.16

2convert mole to volume

c

0.166

0.16

6

NaHCO

a

mn mol

M

H SO CO

x

x mol

n

V

V

V dm

31 1 2 21

11

3 32 1

32

6 .

6 0.1 1?

0.1 . 0.02

1

cV c Vb c mol dm

VV

c mol dm V dm

V dm

2 4

2 4 2 2 4

2 4

determine how much H is neutralised

H 2 2

1

2

0.10.1 1

2 0.1 30?

0.1 15 is neutral

30 10 H influences pH

a a a a

b b bb

aa

a

b a

b

SO

SO NaOH H O Na SO

n n c V

n c Vn

Vc

V

c V ml

V ml ml SO

2 4

31 1 2 21

21

32 2

2

Determine the new concentration of 10ml H

0.1 .

0.1 10 5510

? 0.018 .25 30

SO

cV c Vc mol dm

cV ml

c c mol dmV ml

2 4

22 4 4

2 4

3

Determine the pH of a 0.018M H solution

2

:

1 : 2

0.018 :

0.0182 0.036 .

1

SO

H SO H SO

H SO H

x

x mol dm

3log

log 0.036

1.44

pH H O