Transcript Acids and Bases - EARJ Chemistry [licensed for non

Acids and Bases
http://www.shodor.org/unchem/basic/ab/
http://www.chemtutor.com/acid.htm
Strong Acids
Weak Acids (Carboxylic
Acids)
Strong acids completely dissociate in
water,
A weak acid only partially
dissociates in water
HCl(aq) =>H+1 (aq) + Cl-1 (aq)
1moldm-3
1moldm-3
1moldm-3
HCl - hydrochloric acid
HNO3 - nitric acid
H2SO4 - sulfuric acid
HBr - hydrobromic acid
HI - hydroiodic acid
HClO4 - perchloric acid
CH3COOH(aq) =>H+1 (aq) + CH3COO(aq)
1moldm-3
moldm-3
x
moldm-3
x
Examples of weak acids include
hydrofluoric acid, HF, and
acetic acid, CH3COOH.
2

Strong Bases

The hydroxides of the Group I and
Group II metals usually are
considered to be strong bases.

Ionize completely

NaOH(aq) => Na+(aq) + OH- (aq)

LiOH - lithium hydroxide
NaOH - sodium hydroxide
KOH - potassium hydroxide
RbOH - rubidium hydroxide
Ca(OH)2 - calcium hydroxide
Ba(OH)2 - barium hydroxide






Weak Bases ( AMINES)

Examples of weak bases
include ammonia, NH3,

Ionize partially
NH3 + H2 O=> NH4+1 + OH-

Methylamine and
diethylamine,
(CH3CH2)2NH.
3

Which is not a strong acid?
A. Nitric acid
B. Sulfuric acid
C. Carbonic acid
D. Hydrochloric acid
4
Properties of Acids (page 145)
1.Produce H+ (as H3O+) ions in water
HCl (aq) =>
H+ (aq) + Cl+ (aq)
2.Taste sour
3.Corrode metals:
Zn(s) + HCl(aq)=> ZnCl2 (aq) + H2 (g)
4.Electrolytes
5.React with bases to form a salt and water:
HCl + NaOH => NaCl + H2 O
6.pH is less than 7
7.Turns blue litmus paper to red
8. React with carbonates and bicarbonates to produce carbon dioxide.
CaCO3 (s) + 2HCl(aq) => CaCl2 (aq) + H2 O(l) + CO2 (g)
5
Properties of Bases (page 146)
 Generally produce OH- ions in water:
 NaOH => Na+ + OH Taste bitter, chalky,soapy,slippery
 Are electrolytes
 Displacement of ammonia from ammonium
salts
 React with acids to form salts and water
 pH greater than 7
 Turns red litmus paper to blue
6
Which substance, when dissolved in
water, to give a 0.1 mol dm− solution,
has the highest pH?
 A. HCl
 B. NaCl
 C. NH3
 D. NaOH

7
Definitions
Arhenius
 Bronsted Lowry
 Lewis

8
Arrhenius Definition
Arrhenius
Acid - Substances in water that increase the
concentration of hydrogen ions (H+).
HCl(g) => H+1 (aq)
+ Cl-1 (aq)
Base - Substances in water that increase
concentration of hydroxide ions (OH-).
NaOH(s) => Na+1 (aq) + OH-1 (aq)
9
Bronsted-Lowry Definition
Acids are species that donate a proton (H+).
HNO3 (aq) + H2O(l) => NO3-(aq) + H3O+(aq)
NO3- is called the conjugate base of the acid HNO3, and H3O+ is the
conjugate acid of the base H2O
10

Bases are species that accept a proton.
NH3 (aq) + H2O(l) => NH4+(aq) + OH-(aq)

NH3 is a base and H2O is acting as an acid. NH4+ is the
conjugate acid of the base NH3, and OH- is the
conjugate base of the acid H2O.

A compound that can act as either an acid or a base,
such as the H2O in the above examples, is called
amphoteric
11
Conjugate Acid Base Pairs
Conjugate Base - The species remaining after an acid
has transferred its proton.
Conjugate Acid - The species produced after base
has accepted a proton.
12
Bronsted-Lowry Acid Base
Systems
Acid
Conjugate Acid
Base
Conjugate Base
+
H2O

H3O+ +
Cl-
H2PO4- +
H2O

H3O+
+
HPO42-
NH4+
+
H2O 
H3O+
+
NH3
Base
:NH3
+
Acid
Conjugate Acid Conjugate Base
H2O 
NH4+ +
OH-
PO43-
+
HCl
H2O 

HPO42- +
OH-
13
Lewis Definition



A Lewis acid is defined to be any species that accepts
lone pair electrons.
A Lewis base is any species that donates lone pair
electrons.
H+
+
: OH-1
=> H2O
14
Auto Ionization of Water
http://www.wwnorton.com/college/chemistry/gilbert2/tutorials/interface.asp?chapter=chapter_16&
folder=self_ionization


self-ionization of water
H2O (l) => H+ (aq) + OH− (aq)
or

2 H2O (l) => H3O+ (aq) + OH− (aq)
hydronium ion

It is an example of autoprotolysis, and relies on the amphoteric
nature of water.
15
Kw
The autoionization of water in equilibrium:
H2O(l)

H+(aq) + OH-(aq) ΔH> 0
The equilibrium constant expression for this reaction is given by:
Keq = [H+] [OH-]/[H2 O]
Kc x [H2O]= Kw = 1.0 x 10-14
ion product for water
The value for Kw is for room temperature, 25 °C, and 1.0 atm
The equilibrium constant expression applies not only to pure (distilled)
water but to any aqueous solution. It can be used to calculate either
[H+] or [OH-] provided one of them is known.
16
Kw and T:
http://mmsphyschem.com/autoIon.htm
H2O(l)  H+(aq) + OH-(aq) ΔH> 0
T
0
5
10
15
20
25
Kw
1.14 x 10-15
1.85 x 10-15
2.92 x 10-15
4.53 x 10-15
6.81 x 10-15
1.01 x 10-14

The ionization of
water is endo so,

as the
temperature
increases,so does
the Kw
17
Formulas
pH = -log [H]
pKa = -log Ka
pOH = -log [OH]
pKb = -log Kb
pH + pOH = 14
pKb + pKa = 14
[H] [OH] = 1 x 10 -14 = Ka x Kb = Kw
Kw = 1.0 x 10-14
pKw = 14
pKw = pH + pOH
18
pH Formulas
19
20

Lime(calcium hydroxide) is added to a lake to neutralize
the effects of acid rain. The pH value of the lake water
rises from 4 to 7. What is the change in concentration of
H+ in the lake water?

A. An increase by a factor of 3
B. An increase by a factor of 1000
C. A decrease by a factor of 3
D. A decrease by a factor of 1000



21
The pH Scale
22
Calculating the pH
pH = - log [H3O+]
Example 1: If [H3O+] = 1 X 10-10
pH = - log 1 X 10-10
pH = - (- 10)
pH = 10
Example 2: If [H3O+] = 1.8 X 10-5
pH = - log 1.8 X 10-5
pH = - (- 4.74)
pH = 4.74
23
pH and acidity
The pH values of several
common substances are
shown at the right.
Many common foods are
weak acids
Some medicines and many
household cleaners are
bases.
24
Indicators: Substances that change color when
the concentration of hydrogen changes.
25
Neutralization
An acid will neutralize a base,
giving a salt and water as products
Examples:
HCl
H2SO4
H3PO4
2 HCl
+ NaOH
+ 2 NaOH
+ 3 KOH
+ Ca(OH) 2




NaCl
Na2SO4
K3PO4
CaCl2
+ H2O
+ 2 H2O
+ 3 H2O
+ 2 H 2O
26
Neutralization Problems
# moles acid = # moles base
If an acid and a base combine in a 1 to 1 ratio, then
the volume of the acid multiplied by the
concentration of the acid is equal to the volume of
the base multiplied by the concentration of the
base
Vacid M acid = V base M base
27
Neutralization Problems
Example 1:
HCl + KOH  KCl + H2O
If 15.00 cm3 of 0.500 M HCl exactly neutralizes 24.00 cm3 of
KOH solution, what is the concentration of the KOH
solution?
28
Neutralization Problems
Whenever an acid and a base do not combine in a 1 to 1
ratio, a mole factor must be added to the neutralization
equation
n Vacid C acid = V base C base
The mole factor (n) is the number of times the moles the
acid side of the above equation must be multiplied so as to
equal the base side. (or vice versa)
Example
H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
moles base = 2 x moles acid
29
Neutralization Problems
Example 3:
H3PO4 + 3 KOH  K3PO4 + 3 H2O
If 30.00 cm3 of 0.300 M KOH exactly neutralizes 15.00 cm3 of
H3PO4 solution, what is the concentration of the H3PO4
solution?
Solution:
30
Neutralization Problems
Example 2: Sulfuric acid reacts with sodium hydroxide
according to the following reaction:
H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
If 20.00 cm3 of 0.400 M H2SO4 exactly neutralizes 32.00 cm3 of
NaOH solution, what is the concentration of the NaOH
solution?
Solution:
In this case the mole factor is 2 and it goes on the acid side,
since the mole ratio of acid to base is 1 to 2. Therefore
2 Vacid Cacid = Vbase Cbase
2 (20.00 cm3 )(0.400 M) = (32.00 cm3 ) Cbase
Cbase = (2) (20.00 cm3 )(0.400 M)
(32.00 cm3 )
Cbase = 0.500 M
31
Neutralization Problems
Example 4:
2 HCl + Ca(OH)2  CaCl2 + 2 H2O
If 25.00 cm3 of 0.400 M HCl exactly neutralizes 20.00 cm3 of
Ca(OH)2 solution, what is the concentration of the Ca(OH)2
solution?
32

YEAR 2
33
Acid Base Dissociation
Acid-base reactions are equilibrium processes.
The relationship between the relative concentrations of the
reactants and products is a constant for a given temperature. It
is known as the Acid or Base Dissociation Constant: Ka and
Kb
The stronger the acid or base, the larger the value of the
dissociation constant.
For an acid in w ate r
K eq
[: A - ][H3 O  ]

[HA] [H2 O]
For a bas e in w ate r
K eq
[HB ][OH - ]

[: B - ] [H2 O]
Note :
H3 O   [H ]

[H2 O] in dilute s olutions is cons tant.
K eq [H2 O]  K a
[: A - ] [H ]

[HA]
K eq [H2 O]  K b
[HB] [OH- ]

[: B - ]
34
35
Acid Strength
Strong Acid
- Transfers all of its protons to water;
- Completely ionized;
- Strong electrolyte;
- The conjugate base is weaker and has a
negligible tendency to be protonated.
Weak Acid
- Transfers only a fraction of its protons to
water;
- Partly ionized;
- Weak electrolyte;
- The conjugate base is stronger, readily
accepting protons from water
 As acid strength decreases, base strength increases.
 The stronger the acid, the weaker its conjugate base
 The weaker the acid, the stronger its conjugate base
36
Acid Dissociation Constants
Dissociation constants for some weak acids
37
Base Strength
Strong Base - all molecules accept a proton;
- completely ionizes;
- strong electrolyte;
- conjugate acid is very weak, negligible
tendency to donate protons.
Weak Base
- fraction of molecules accept proton;
- partly ionized;
- weak electrolyte;
- the conjugate acid is stronger. It more
readily donates protons.
 As base strength decreases, acid strength increases.
 The stronger the base, the weaker its conjugate acid.
 The weaker the base the stronger its conjugate acid.
38
Weak Acid Equilibria
A weak acid is only partially ionized.
Both the ion form and the unionized form exist at
equilibrium
HA + H2O  H3O+ + AThe acid equilibrium constant is
Ka = [H3O+ ] [A-]
[HA]
Ka values are relatively small for most weak acids.
The greatest part of the weak acid is in the
unionized form
39
Weak Acid Equilibrium
Constants
Sample problem . A certain weak acid dissociates in water as
follows:
HF + H2O  H3O+ + F-
If the initial concentration of HF is 1.5 M and the equilibrium
concentration of H3O+ is 0.0014 M. Calculate Ka for this acid
40
Weak Base Equilibria
Weak bases, like weak acids, are partially
ionized. The degree to which ionization
occurs depends on the value of the base
dissociation constant
General form: B + H2O  BH+ + OHKb
= [BH+][OH-]
[B]
Example
NH3 + H2O  NH4+ + OHKb
= [NH4+][ OH-]
[NH3]
41
Weak Base Equilibrium
Constants
Sample problem . A certain weak base dissociates in water as
follows:
B + H2O  BH+ + OHIf the initial concentration of B is 1.2 M and the equilibrium
concentration of OH- is 0.0011 M. Calculate Kb for this base
Solution
Kb = [BH+ ] [OH-]
[B]
I
C
E
[B]
1.2 -x 1.2-x
[OH-]
0 +x
x
[BH+ ] 0 +x
x
x = 0.0011
1.2-x = 1.1989
Substituting
Kb =
(0.0011)2 = 1.01 x 10-6
1.1989
42
Weak Acid Equilibria
Concentration Problems
Problem 1. A certain weak acid dissociates in water as follows:
HA + H2O  H3O+ + AThe Ka for this acid is 2.0 x 10-6. Calculate the [HA] [A-], [H3O+ ]
and pH of a 2.0 M solution
Solution
Ka = [H3O+ ] [A-] = 2.0 x 10-6
[HA]
I
C
E
Substituting
[HA] 2.0 -x 2.0-x
Ka =
x2
= 2.0 x 10-6
[A-]
0 +x
x
2.0-x
[H3O+ ] 0 +x
x
If x <<< 2.0 it can be dropped
from the denominator
The x2 = (2.0 x10-6)(2.0) = 4.0 x10-6 x = 2.0 x 10-3
[A-] = [H3O+ ] = 2.0 x10-3 [HA] = 2.0 - 0.002= 1.998
pH = - log [H3O+ ] =-log (2.0 x 10-3) = 2.7
43
Weak Acid Equilibria
Concentration Problems
Problem 2. Acetic acid is a weak acid that dissociates in water as
follows: CH3COOH + H2O  H3O+ + CH3COOThe Ka for this acid is 1.8 x 10-5. Calculate the [CH3COOH],[CH3COO-]
[H3O+ ] and pH of a 0.100 M solution
Solution
Ka = [H3O+ ] [CH3COO- ] = 1.8 x 10-5
[CH3COOH]
I
C
E
Substituting
[CH3COOH] 0.100 -x 0.100-x
Ka =
x2
= 1.8 x 10-5
[CH3COO- ]
0
+x
x
0.100-x
+
[H3O ]
0
+x
x
If x <<< 0.100 it can be dropped
from the denominator
The x2 = (1.8 x10-5)(0.100) = 1.8 x10-6 x = 1.3 x 10-3
[CH3COO--] = [H3O+ ] = 1.3 x10-3 [CH3COOH ] = 0.100 - 0.0013 = 0.0987
pH = - log [H3O+ ] =-log (1.3 x10-3) = 2.88
44
Weak Base Equilibria
Example1. Ammonia dissociates in water according to the
following equilibrium
NH3 + H2O  NH4+ + OHKb = [NH4+][ OH-] = 1.8 x 10-5
[NH3]
Calculate the concentration of [NH4+][ OH-] [NH3 ]and the pH of
a 2.0M solution.
I
C
E
Substituting
[NH3] 2.0 -x 2.0-x
Kb =
x2
= 1.8x 10-5
[OH-]
0 +x
x
2.0-x
[NH4+]
0 +x
x
If x <<< 2.0 it can be dropped
from the denominator
The x2 = (1.8 x10-5)(2.0) = 3.6 x10-5 x = 6.0 x 10-3
[OH-] = [NH4+] = 6.0 x10-3 [NH3] = 2.0- 0.006= 1.994
pOH = - log [OH-] =-log (6.0 x10-3) = 2.22
pH = 14-pOH = 14-2.22 = 11.78
45
Amphoteric Solutions

A chemical compound able to react with both an
acid or a base is amphoteric.
Water is amphoteric. The two acid-base couples
of water are H3O+/H2O and H2O/OHIt behaves sometimes like an acid, for example

And sometimes like a base :

Hydrogen carbonate ion HCO3- is also amphoteric,
it belongs to the two acid-base couples
H2CO3/HCO3- and HCO3-/CO32-

46
Common Ion Effect
The common ion effect is a consequence of Le
Chatelier’s Principle
When the salt with the anion (i.e. the conjugate
base) of a weak acid is added to that acid,
It reverses the dissociation of the acid.
 Lowers the percent dissociation of the
acid.

A similar process happens when the salt with the
cation (i.e, conjugate acid) is added to a weak
base.
These solutions are known as Buffer Solutions.
47
Buffers

A buffer solution is na aqueous solution that resists a
change in pH when a small amount of acid, base or
water is added to it.
Acidic Buffers: Weak acid and its salt
Can be prepared by using excess of a weak acid and a strong base .
NaOH(aq) + CH3 COOH(aq) => CH3 COONa(aq) + CH3 COOH(aq) + H2 O (l)
salt
excess weak acid
CH3COOH (aq) => CH3COO- (aq) + H+ (aq)
A
CB
pH = pKa + log [ CB] / [ A ]
Ka = 1.74 x 10-5
Henderson Hasselbach Equation
48
Formulas
pH = -log [H]
pKa = -log Ka
pOH = -log [OH]
pKb = -log Kb
pH + pOH = 14
pKb + pKa = 14
[H] [OH] = 1 x 10 -14 = Ka x Kb
Kw = 1.0 x 10-14
pKw = 14
49

1. Given 30 cm3
of 0.1 M ethanoic acid and 10 cm3
NaOH, find the pH of the buffer solution.
CH3COOH (aq) => CH3COO- (aq) + H+ (aq)
A
CB
of 0.1 M
Ka = 1.74 x 10-5
pKa = -log (1.74 x 10-5 )=4.76
n CH3COOH = 0.1 x 0.03 = 0.003
n NaOH = 0.1 x 0.01 = 0.001
excess CH3COOH after titration = 0.002
pH = pKa + log [ CB] / [ A ]
pH = 4.76 + log(0.001/0.002)= 4.5
50
2. Calculate the pH of a solution that is 0.50 M CH3COOH and 0.25
M NaCH3COO.
CH3COOH + H2O  H3O+ + CH3COO- (Ka = 1.8 x 10-5)
51
Buffer Solution Calculations
Calculate the pH of a solution that is 0.50 M CH3COOH and 0.25
M NaCH3COO.
CH3COOH + H2O  H3O+ + CH3COO- (Ka = 1.8 x 10-5)
Solution
Ka = [H3O+ ] [CH3COO- ] = 1.8 x 10-5
[CH3COOH]
I
C
E
.
Substituting
[CH3COOH] 0.50 -x 0.50-x
Ka = x (0.25+x) = 1.8 x 10-5
[CH3COO-]
0.25 +x 0.25+x
(0.50-x)
+
[H3O ]
0
+x
x
If x <<< 0.25 it can be dropped from both
expressions in ( ) since adding or
subtracting a small amount will not
significantly change the value of the ratio
Then the expression becomes x(0.25)/(0.50) = 1.8 x 10-5
x = 3.6 x 10-5 = [H3O+]
pH = - log [H3O+] =-log(3.6 x 10-5 ) = 4.44
52
Using the Henderson Hasselbach Equation
pH = pKa + log([A-]/[HA])
Example
Calculate the pH of the following of a mixture that contains
0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka =
1.4 x 10-4)
HC3H5O3 + H2O  H3O+ + C3H5O3Solution
Using the Henderson-Hasselbach equation
pH = - log (1.4 x 10-4) + log ( 0.25/0.75 )
= 3.85 + (-0.477) = 3.37
53

Basic Buffers: Weak base and its salt
Can be prepared by using excess of a weak base and a strong acid .
HCl(aq) + CH3 NH2(aq) => CH3 NH2Cl(aq) + CH3 NH3 +(aq)
salt
excess base
CH3NH2 (aq) + H2 O => CH3NH3+ (aq) + OH- (aq)
B
CA
pOH = pKb + log [ CA] / [ B ]
+ H2 O (l)
Kb = 4.37 x 10-4
Henderson Hasselbach Equation
54

Find the pH of a buffer made with 0.025 M
methylamine and 0.010 M HCl.
55
Methyl amine is a weak base with a Kb or 4.38 x 10-4
CH3NH2 + H2O  CH3NH3+ + OHCalculate the pH of a solution that is 0.10 M in methyl
amine and 0.20 M in methylamine hydrochloride.
56
Henderson-Hasselbach
Equation & Base Buffers
Methyl amine is a weak base with a Kb or 4.38 x 10-4
CH3NH2 + H2O  CH3NH3+ + OHCalculate the pH of a solution that is 0.10 M in methyl
amine and 0.20 M in methylamine hydrochloride.
pOH = pKb + log ([BH+] / [B])
Solution
pOH = -log (4.38 x 10-4) + log (0.20 / 0.10)
= 3.36 + 0.30 = 3.66
pH = 14- 3.66 = 10.34
57
Additional Buffer Problems
How many grams of sodium formate, NaCHOO, would have to
be dissolved in 1.0 dm3 of 0.12 M formic acid, CHOOH, to
make the solution a buffer of pH 3.80? Ka= 1.78 x 10-4
pH = pKa + Log ([A-]/[HA])
Solution
3.80 = -log (1.78 x 10-4) + Log [A-] - Log [0.12]
3.80 = 3.75 + Log [A-] - (-0.92)
Log [A-] = 3.80 - 3.75 - 0.92 = - 0.87
[A-] = 10-0.87 = 0.135 mol dm-3
The molar mass of NaCHOO = 23+12+1+2(16) = 58.0 gmol-1
So (0.135 mol dm-3)(58.0 gmol-1 ) = 7.8 grams per dm-3
58
Relationship of Ka, Kb & Kw

HA weak acid. Its acid ionization is

A- is the conjugate base Its base ionization is

Multiplying Ka and Kb and canceling like terms
59
Titration Curves

Simulations: http://chemilp.net/labTechniques/AcidBaseIdicatorSimulation.htm

A graph showing pH vs volume of acid or base added

The pH shows a sudden change near the equivalence
point

The equivalence point is the point at which the moles of
OH- are equal to the moles of H3O+

The end point is when the indicatorchanges color
acurate to the addition of one dro.
60
Equivalence Point x End Point
http://www.chem.ubc.ca/courseware/pH/section15/index.html

The endpoint of a titration is NOT the same thing as the
equivalence point.

The equivalence point is a single point defined by the reaction
stoichiometry as the point at which the base (or acid) added
exactly neutralizes the acid (or base) being titrated.

The endpoint is defined by the choice of indicator as the point at
which the colour changes. Depending on how quickly the colour
changes, the endpoint can occur almost instantaneously or be
quite wide.

Intuition may suggest that the endpoint of the titration will occur at
the equivalence point if we choose an indicator whose pKa is equal
to the pH of the equivalence point
61
Indicators: Substances that change color when the concentration of
hydrogen changes. The choice of an indicator is determined by the pH of
the solution at the equivalence point
http://www.avogadro.co.uk/chemeqm/acidbase/titration/phcurves.htm
62

An indicator is a solution of a weak acid in which the the
conjugate base has a different color from that of the
undissociated acid

Given na aqueous solution of na indicator HIn(aq)
HIn(aq)  H+(aq) + In-(aq)
Color A
Color B
(in acid solution)
( in basic solution)



The end point of the indicator will be when pH = pkIN
At low pH we see color A
At high pH we see color B
63
Strong acid-strong base

At equivalence point, Veq:
Moles of H3O+ = Moles of OH-

There is a sharp rise in the pH as
one approaches the equivalence
point

With a strong acid and a strong
base, the equivalence point is at
pH =7
Indicators: methyl red and
phenophtalein
Change between 4 and 10
64



a strong acid-strong base has a very sharp equivalence point,
meaning that a very large change in pH occurs due to the addition
of a very small amount of titrant, often a single drop.
. This means that any indicator that starts to change colour in this
range will signal equally well that the equivalence point has been
reached.
Any acid-base indicator that changes colour between pH 4 and pH
10 is suitable to detect the end-point for a strong acid - strong
base titration. Both methyl orange and phenolphthalein could be
used. Just one drop of the added base will bring about a change in
colour of the indicator.
65
66
67
Weak acid-strong base

The increase in pH is more gradual
as one approaches the
equivalence point

With a weak acid and a strong
base, the equivalence point is
higher than pH = 7

CH3COOH + NaOH

Ka

Indicator : phenophtalein
68
WEAK ACID - STRONG BASE
NaOH(aq) + HF(aq)  NaF(aq) + H2O(l)
Strong Acid X Weak Base

The equivalence point is below 7 because the salt (NH4Cl) formed at
the neutralization reacts with water to give H+ ions. The equivalence
point lies at about pH 5.3. It is, therefore necessary to use an indicator
with pH range slightly on the acidic side. Methyl orange can be used.
Phenolphthalein is not suitable because its colour change occurs away
from the equivalence point.
70
Weak Acid Weak Base
There will hardly be a pH change
around the equivalence point
71
Buffered Weak Acid-Strong
Base Titration Curve

The initial pH is higher than
the unbuffered acid

As with a weak acid and a
strong base, the equivalence
point for a buffered weak
acid is higher than pH =7

The conjugate base is strong
enough to affect the pH
72
Polyprotic Weak AcidStrong Base Titration Curve


Phosphoric Acid has three
hydrogen ions.
There are three equivalence
points
H3P04 + H2O  H3O+ + H2PO4H2PO4- +H2O  H3O+ + HPO42HPO42- +H2O  H3O+ + PO43-
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SALTS & Hydrolysis

Salts are ionic and already completely
dissociated. They are also strong
electrolytes.

I. When a salt comes from a strong acid and a
strong base, they form neutral solutions when
they dissolve in water.
NaCl(aq) => Na+(aq) + Cl-(aq)
74

II. When a salt comes from a weak acid and a
strong base, it forms na alkaline solution when it
dissolves in water.
CH3COOH(aq) + NaOH(aq) => CH3COONa(aq) + H2O(l)
CH3COOH(aq) => Na+(aq) + CH3COO-(aq)
+
H2O(l) => OH-(aq) + H+(aq)
↨
CH3COOH(aq)
75

III. When a salt comes from a strong acid
and a weak base, it will be acidic in solution.
HCl(aq) + NH3(aq) => NH4Cl(aq)
NH4+(aq)
+
H2O(l) => OH-(aq)
↨
NH3(aq)
NH4Cl(aq) =>
+
Cl-(aq)
+ H+(aq)
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http://home.clara.net/rod.beavon/AlCl3_and_water.htm


The acidity of a salt also depends on the size and charge of the
anion.
Aluminum chloride reacts vigorously with water to produce a strong
acidic solution.
AlCl3(s) + 3H2O(l) → Al(OH )3 (s) + 3HCl(g)
amphoteric nature
The charge of aluminum is spread over the small ion giving it a high charge density. It will
then attract six pair of electrons forming the hexahydrated ion CC page 155
http://en.wikipedia.org/wiki/Hydrolysis
The greater the charge of nthe ion or
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