Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

CHE027 - 2018

ACIDS AND BASES
Objectives:
 Theories of acidity
 The ionization of water
 Measurement of acidity – the pH scale
 Strong and weak acids and bases
 Calculating pH of solutions of weak acids and bases
 pH changes during acid-base titrations
 Indicators
 Buffer Solution

Properties of acids and Bases
acids

Bases

Taste sour
Acids change litmus (a blue vegetable dye) from
blue to red
Their aqueous (water) solutions conduct electric
current (are electrolytes)
reacts with bases to form salts and water
Evolve hydrogen gas (H2) upon reaction with most
metals
Common examples: Lemons, oranges, vinegar,
urine, sulfuric acid, hydrochloric acid

Taste bitter
Feels slippery or soapy
Turn red (acidified) litmus to blue
Their aqueous (water) solutions conduct electric
current (are electrolytes)
Reacts with acids to form salt and water

Common Examples: Soap, toothpaste, bleach,
cleaning agents, limewater, ammonia water,
sodium hydroxide
...
In 1884 a Swedish chemist by the name of
Arrhenius suggested that:


Acids are substances which produce hydrogen ions (H+) in solution
e
...

HCl(l) +
aq

H+(aq)
+
Cl-(aq)



Bases are substances which produce hydroxide ions (OH¯) in solution
e
...

NaOH(s)
+
aq

Na+(aq) +

OH-(aq)

Neutralization
HCl(aq)
1|Page

+

NaOH(aq)



NaCl(aq) +

H2O(l)
MNU
...

Hydrogen chloride is neutralized by both sodium hydroxide as well as ammonia
...

As a result it was suggested that the H+ ion exists in association with the water molecule as the H3O+ ion
called the oxonium ion (also known as hydronium ion)
...

 An acid is a proton (H+) donor
 A base is a proton (H+) acceptor

This proton will be donated to the
ammonia molecule

Dative covalent is formed in
between NH3 and H+ where N
donates the lone pair of electrons

An acid is a substance which can behave as a proton (hydrogen ion) donor
...

HCl(aq)
+
H2O(l)

H3O+(aq) +
Cl-(aq)
2|Page

MNU
...

A base has a lone pair of electrons which can form dative covalent bond with the proton
...

NH4+ +

NH3(aq) + H2O(l)

OH-

Conjugate acids and bases
Above reactions can be regarded as equilibrium
...

The proton donor and proton acceptor on the right hand side of the equation are called the conjugate acid
and conjugate base
...


Try these
...
In pure ethanol, C2H5OH, the following equilibrium can exist with ammonium ions:
NH4+

+

C2H5OH

NH3

+

C2H5OH2+

Does the ethanol behave as a Bronsted acid?
3|Page

MNU
...
Identify the conjugate acid and base pairs in the following
...

pH is defined as the negative logarithm to the base 10 of hydrogen/ hydroxonium ion in moldm-3
+

pH = -log[H ]
Examples:
-3
0
...
1 = 1
-3
0
...
2 = 0
...
01 mol dm HNO3 -log 0
...
001 mol dm HI -log 0
...
1 mol dm to 0
...

 Some acids, such as sulphuric are dibasic; this means they can release two hydrogen ions from each
-3
-3
molecule
...

Examples:
-3
+
-3
0
...
2 mol dm pH = -log 0
...
7
-3
+
-3
0
...
02 mol dm pH = -log 0
...
7

4|Page

MNU
...

Auto-ionisation of water
Water molecules dissociate to very small extent:
2H2O(l)

H3O+(aq)

+

OH-(aq)

H2O(l)

H+(aq) +

OH-(aq)

Or

This is known as the auto-ionisation of water
...

We can simplify this equilibrium expression by including the constant [H2O] term with the value of Kc to
obtain a new equilibrium constant, the ionic product constant for water, Kw (which is a constant at a
given temperature)
Kc[H2O] =

Kw = [H+][OH-]

=

1
...

Any solution in which the concentrations of H3O+ and OH- are equal is said to be neutral, pH is 7
So

+

-7

-3

o

[H3O ] =1
...
0 x 10-7 moldm-3 at 25oC

[H3O+]
0
...
1

In a neutral solution,

[H3O+]

In an acidic solution,

[H3O+]

>

[OH-]

In a basic solution,

[H3O+]

<

[OH-]

=

[OH-]

Type of solution
Acidic
Acidic
Acidic
Neutral
Alkaline
Alkaline
Alkaline

MNU
...
0 x 10-14
+
Applying - logarithm:
-log Kw =-log [H ]-log[OH ]
= 14
pKw = p[H+] + p [OH-] = 14
Hence

pH =

14- p [OH-]

Calculating the pH of strong acids and strong bases
Since strong acids dissociate fully, the concentration of H+ will be the same as the concentration of the
acid
 Assumption: H+ ions due to the auto-ionization of water is insignificant (hence ignored)
Example 1:A 1
...
The
H3O+ present due to the auto-ionisation of water (which will be small in comparison) is ignored
...
0 M
+
If [H ] = 1
...
0]
= 0

H+(aq)
1
...

NaOH(aq) Na+(aq) +
OH-(aq)
1
...
0 M
so

[H+]

= Kw/[OH-]
= Kw/0
...


pH

= -log [H3O+]
= -log [1 x 10-13]
= 13
...
1)
= 14-1= 13

MNU
...

1
...
Calculate the pH of H2SO4 of concentration 5
...
If the dissociation only
occurs partially it is classified as a weak acid
...

Note:




Strength of an acid is related to the amount of H3O+ produced per mole of substance dissolved
...
1
0
...
1
0
...
1
0
...
024
0
...
9
1
...
7

HCl is a strong acid and it dissociates completely to product 1 mole of H3O+
While CH3COOH is a weak acid where the equilibrium of this reaction is well over to the left
...

HCl(aq)
CH3COOH(aq)

7|Page

+
+

H2O(l)
H2O(l)



H3O+(aq)
CH3COO-(aq) +

+

Cl-(aq)
H3O+(aq)

MNU
...
Observe the changes in pH brought by changes in concentration
Concentration of HCl
pH
1
0
-1
1
1x10
-2

2

1x10

-3

3
1x10
Strong acid has a weak conjugate base
...

Similarly, weak acids have strong conjugate bases
...

For each dilution of 10x, the pH increases by 1 unit
...

For each dilution of 10x, the pH increases by 0
...
Dilution factor

Concentration of
-3

acid /moldm
0
...
01
0
...
0001

0
10x
100x
1000x

pH of strong acid
(monobasic)

pH of weak acid
(monobasic)

1
2
3
4

2
...
38
3
...
38

Acid Dissociation constant Ka
A strong acid is one that dissociated completely
...
The
amount of dissociation is therefore an indication of the strength of an acid and is measured using the
dissociation constant
...

8|Page

MNU
...

 This equation can then be used to find the pH of a weak acid
...

As Kw and Ka values are very small and inconvenient, pKwand pKa is used for convenience
...

HA(aq)
C-x

+ H2O(l)

H3O+(aq)
x

+

A-(aq)
x

Ka = [H3O+][A-]
[HA]
Example
...
7x10-5 mol dm-3)
CH3COOH(aq) + H2O(l)
1
...
7x10-5= [H+]2
1
...
7x10-5
[H+] = 4
...
18x10-3) = 2
...
FOUNDATION STUDIES

CHE027 - 2018

Smiliar to weak acid, weak bases also dissociate only partially in water
...

Example
...
8x10-5 mol dm-3)


Assumption:

*Since NH3 is a weak base it dissociates partially hence [NH3] does not change significantly (c-x = c)
*OH- ions due to the auto-ionization of water is insignificant- hence ignored
...
8x10-5= [OH-]2
15
...
8x10-5 x 15
...
643x10-2
pH
= 14- p [OH-]

= 14 -1
...
215

Calculating Ka or Kb value of an acid or base (known pH, known concentration)
Example
-3

Nitrous acid of concentration 0
...
17
...

+

[H ] = 10

-2
...
76 x10 moldm

-3

pKa =[H+]2
[HA]

= (6
...
1

2

-4

-3

= 4
...

-3

3
...
0 moldm has a pH of 4
...
Calculate its Ka value
...
Ka for a weak monobasic acid is 1
...
In a 0
...
FOUNDATION STUDIES

CHE027 - 2018

ii)

The pH

iii)

The concentration of OH- ions (Kw = 1
...
Dimethylamine ((CH3)2NH), a key intermediate in detergent manufacture has a Kb of 5
...

What is the pH of 1
...
Calculate the pH of 0
...
8 x 10-10mol dm-3
...
The pH of 0
...
17 at 25oC
...


8
...

a) A 0
...
0 x 10-4 or one with Ka = 4
...

b) A 0
...
01 M solution of the same acid
...
1 M solution of a weak acid or a 0
...

d) A 0
...
1 M solution of a base
...
FOUNDATION STUDIES

CHE027 - 2018

An indicator is in fact a weak acid, HIn
...

The HIn will be one colour and the In- will be a different colour
...

 In alkali: H+ will be removed →equilibrium will shift to the right →colour 2
...
One
drop of acid or alkali, and the equivalence point should be able to change the colour of the indicator
...

Equivalence point = when the stoichiometric amounts of acid and alkali have been added
...

If the indicator is chosen correctly the end point and the equivalence point are the same
...

The titration curve serves to profile the unknown solution
...


You need to choose an indicator which changes colour as close as possible to that equivalence point
...

12 | P a g e

MNU
...
10 mol dm
–3

of stated acid against 0
...

Indicators:
methyl orange: color changes around 3
...
0
phenolphthalein color changes around around 8
...

Curve (b) CH3COOH/NaOH:

 Initial pH 2
...
5
 Quickly flattens out again, heading towards pH=13
...


(c)

(d)

Curve (c) HCl/NH3:
13 | P a g e

MNU
...

Curve (d) CH3COOH/NH3:
 weak acid and base give no sharp endpoint and there is no suitable indicator
...

Examples;
(i)
A weak acid with its sodium salt (or similar) e
...
ethanoic acid and sodium ethanoate, giving
high concentrations of CH3COOH molecules and CH3COO– ions
...

CH3COOH

H+

+

CH3COO–

When a little additional strong acid is added, →most of the H+ ions react with some of the ethanoate
to form ethanoic acid:
H+ + CH3COO– → CH3COOH
Hence the concentration of H+ in the solution only rises slightly, and there is a very small drop in pH
...

When a little additional alkali is added, → most of the OH– ions react with CH3COOH molecules:
OH– + CH3COOH → H2O + CH3COO–
This time the H+ doesn’t fall by nearly as much as expected, and the pH remains relatively constant
...

(i)

14 | P a g e

A weak base with one of its salts e
...
ammonia and ammonium chloride, giving high
concentrations of NH3 molecules and NH4+ ions
...
FOUNDATION STUDIES

CHE027 - 2018
In the NH3/NH4Cl buffer, the NH3 reacts with H+, and the ammonium ions react with OH–:
H+ + NH3 → NH4+
OH– + NH4+ → H2O + NH3
Buffered solutions do change in pH upon addition of H+ or OH- ions
...

Calculating the pH of a buffer solution
-3
Example: what is the pH of a buffer solution containing 0
...
800
-3
-5
-3
moldm ethanoic acid (Ka = 1
...
500) = 1
...
800

-5

From which [H+] = 2
...
56
Or use Henderson-Hasslebach equation which is a derivation from Ka expression
...

-3

9
...
1 moldm ethanoic acid is
-3

mixed with 20 cm3 of 0
...
76)

10
...
800 moldm-3 ethanoic acid ( pKa=4
...
FOUNDATION STUDIES



---