Transcript Slide 1

REDOX REACTIONS
FUNDAMENTALS OF ELECTROCHEMISTRY
ELECTROCHEMISTRY IS THE BRANCH OF CHEMISTRY THAT
DEALS WITH THE RELATIONSHIP BETWEEN ELECTRICITY AND
CHEMICAL REACTIONS.
 PRODUCTION OF ELECTRICAL CURRENT BY CHEMICAL
REACTIONS (Batteries, Fuel cells)
 CHEMICAL CHANGES PRODUCED BY ELECTRIC CURRENT
(Electrolysis, Electroplating and refining of metals)
 ELECTRIC CURRENT = Transfer of charge per unit time
 BIOELECTROCHEMISTRY – study of electron transfer in
biological regulations of organisms
BASIC CONCEPTS
 A redox reaction involves transfer
of electrons from one species to
another.
Oxidation: loss electrons Reducing agent
Reduction : gain electrons Oxidizing agent
Electric charge (q) = n x F
= Coulombs
F – Faraday constant = 96 485.3415 C/mol of eElectric current (I) – is the quantity of charge flowing
each second through the circuit. Unit – Amperes (A)
Eg. Calculate the mass of aluminum produced in 1 hour
by electrolysis of molten AlCl3 if the electrical current is
20 A.
(Answer: 33 g)
Potential difference (E) between 2 points is the work
needed when moving an electric charge from 1 point to
another, unit is Volt
Work = E. q OR J = CV
1 Joule is the energy gained or lost when 1 coulomb of
charge moves between points whose potentials differ by 1V
The greater the potential difference between 2 points the
stronger will be the “push” on a charged particle travelling
between those points. A 12V battery pushes e- 8X harder
than a 1.5V dry cell
 The free energy of change, ∆G, represents the maximum
electrical work that can be done by the reaction on its
surroundings.
Work done on surroundings = ∆G = - work = -E.q
∆G = - nFE
(-ve ∆G = spontaneous rxn)
Ohm’s Law, I = E/R
Unit of resistance is ohms or Greek symbol Ω (omega)
Power , P = work/time
= E.q = E. q = E. I = I2. R
s
s
BALANCING REDOX REACTIONS
 In an acidic medium
 In a basic medium
Split the reaction into 2 components or half reactions by identifying
which species are oxidised and which ones are reduced. Add
electrons appropriately to match the change in oxidation state of
each element
 Introduce H2O to balance the oxygen atoms
 Introduce H+ to balance the charges as well as the hydrogen atoms
 Multiply the half reactions so as to have an equal no. of e- and add
the half reactions
Whichever case, balance for the acid (H+) and then for basic media add OH- to
the side where there is H+ to eliminate it as a H2O molecule (this is a 1:1
reaction)
STANDARD POTENTIALS
LHS – negative terminal
Reference electrode
The voltmeter tells how
much work is done by
e- flowing from one
side to the other, +ve
V means e- flow into
negative terminal
RHS
connected
to the
positive
terminal
Line notation
for cell
 STANDARD REDUCTION POTENTIAL (Eo)
for each half cell is measured or setup by the above experiment.
‘Standard’ means the activities ( A ) of all species are unity.
The half reaction of interest is : 2Ag+ + 2e-
2Ag(s)
AAg+ = 1 by definition activity of Ag (s) = unity
Standard Hydrogen Electrode (SHE), consists of a catalytic Pt
surface in contact with an acid solution H+ (aq,1M), AH+ =1
By convention the LHS electrode (Pt) is attached to the
negative terminal of the potentiometer
 equilibrium rxn at SHE : 2H+ (aq, A = 1) + 2e-
H2 (g, A = 1)
 half reaction – always written as reduction reactions
BY INTERNATIONAL AGREEMENT THE SHE IS ARBITARILY
ASSIGNED Eo = 0.00V at 25oC.
 E0cell = + 0.799V
 E0cell = Eored (cathode) – Eored (anode) or
 E0cell = RHS electrode potential – LHS electrode potential
E0cell = E0red (reduction process) – E0red (oxidation process)
 A positive E0 = spontaneous process
 A negative E0 = non spontaneous process
 Sketch the cell construction : SHE ll Cd2+(aq, A=1) l Cd(s)
The half reaction with a more positive E0 is more reduced and
that with a less positive E0 is less reduced or more oxidised
NERNST EQUATION
Le Chatelier`s principle tells us that
conc. of
reactants or products drives rxn to the right or left
respectively.
The net driving force is expressed by the NERNST
EQUATION
For the half reaction : a A + ne- ↔ b B
Nernst Equation: for the half cell potential, E
Q = Reaction
b
B
a
A
RT A
E=E ln
nF A
o
quotient
where
Eo = standard reduction potential ( AA = AB = 1)
Ai = activity of species i
R = gas constant ( 8.314 J/K.mol)
T = Temperature (K)
N = number of electrons in the half reaction
F = Faraday constant (9.6485 x 104C/mol)
[concentration in Kc can be replaced by activities to
account for ionic strength ]
Ac = [C] γc
γ = activity co efficient – measures
the deviation from ideality , γ = 1
behavior is ideal, low ionic strength
Pure solids and liquids are omitted from Q, because their
activities are (close to) unity.
Concentration mol/L and pressure in bars
When all activities are unity Q = 1, ln Q = 0 then E = Eo
Log form of the Nernst equation, at 25oC T = 298K
b
B
a
A
0.05916V
A
E=E log
n
A
o
NERNST EQUATION FOR A COMPLETE CELL
E = E+ - E-
Find the voltage for
the cell if the right half
cell contains 0.50M
AgNO3 (aq) and the left
half cell 0.010M
Cd(NO3)2 (aq).
STEP 1 Right electrode : 2 Ag+ + 2e- ↔
Left electrode : Cd2+
+ 2e- ↔
2Ag(s)
Eo = 0.799V
Cd(s)
Eo = - 0.402V
STEP 2 Nernst equation for the right electrode
0.05916V
1
E = 0.799log
2
[0.50]2
= 0.781V
STEP 3 Nernst equation for the left electrode
0.05916V
1 = - 0.461V
E = -0.402log
2
[0.010]
STEP 4 Cell Voltage E = E+ - E- = 0.781 - (-0.461) = + 1.242V
STEP 5 Net cell reaction: Eqn Right electrode – Eqn Left
electrode
Cd (s) + 2 Ag+ ↔ Cd2+ + 2 Ag (s)
Note : Multiplying a half reaction by a number does not
change E0 nor E
To figure half cell reactions look for the element in two
different oxidations states
Eo and the Equilibrium Constant
A galvanic cell produces electricity because the cell is not at
equilibrium.
Relating E to the reaction quotient Q :
Consider:
Right electrode: aA + ne- ↔ cC
Left electrode: dD + ne- ↔ bB
E o+
E o-
The Nernst equation will be:
E = E+ - E=
=
ACc
0.05916V
E log a
n
AA
o

c d
A
0
.
05916
V
o
o
C AD
E  E- log a b
n
AA AB
E0
=
b
0
.
05916
V
A
- ( E-o log Bd )
n
AD
Q
0.05916 V
E log Q
n
o
When the cell is at equilibrium E = 0 and Q = K
K
nE 0 / 0.05916
= 10
Eo can be calculated and used to find K for 2 half reactions!
Question 10 in Tutorial on Complexation rxns
Use the following standard-state cell potentials to calculate the complex
formation equilibrium constant for the Zn(NH3)42+ complex ion.
Zn(NH3)42+ + 2e- ⇌ Zn + 4NH3
Zn2+ + 2e- ⇌ Zn
SOLUTION: E0cell
Eored = - 0.7628 V
= E0red (reduction process) – E0red (oxidation process)
(i) Zn + 4NH3 ⇌ Zn(NH3)42+ + 2e(ii) Zn2+ + 2e- ⇌ Zn
(i) + (ii): Zn2+ + 4NH3 ⇌ Zn(NH3)42+
K
nE 0 / 0.05916
= 10
Eored = -1.04 V
= 10
E0 = + 1.04 V
E0 = - 0.7628 V
E0 = + 0.28 V
(2)(0.28)/0.05916 =
2.92 x 109
REDOX TITRATIONS
 Theory of redox titrations and common titrants.
A redox titration is based on an oxidation- reduction reaction
between an analyte and titrant.
Environmental and biological analytes can be measured by
redox titrations. Other analytes include laser and
superconductor materials.
REDOX TITRATION CURVE
Consider the potentiometric titration of Fe(II) and cerium (IV).
Titration rxn:
Ce4+ + Fe2+
Ce3+ + Fe3+ (1)
Each mole of Ce4+ ion oxidizes 1 mole of ferrous ion, the
titration creates a mixture of Ce4+, Fe2+, Ce3+ and Fe3+ ions.
e- flow from the anode to
the cathode, the circuit
measures the potential for
Fe3+ /Ce4+ reduction at the
Pt surface by e- from the
calomel electrode
Calomel reference electrode : 2Hg(l) + 2 Cl-
Hg2Cl2 (s) + 2e-
Pt indicator electrode: 2 rxns coming to equilibrium:
Fe3+ + e- ↔ Fe2+
Eo = + 0.767V
(2)
Ce4+ + e- ↔ Ce3+
Eo = + 1.70V
(3)
Cell rxns
2Fe3+ + 2Hg(l) + 2Cl-
↔ Fe2+ + Hg2Cl2 (s)
(4)
2Ce4+ + 2Hg(l) + 2Cl-
↔ Ce3+ + Hg2Cl2 (s)
(5)
At equilibrium the potential driving rxns 1 and 2 must be the same.
THE CELL RXNS ARE NOT THE SAME AS THE TITRATION RXN!
The titration reaction goes to completion and is an oxidation of Fe2+
and reduction of Ce4+
Cell rxns proceed to negligible extent. The cell is used to measure
activities, not to change them.
HOW CELL VOLTAGE CHANGES AS Fe2+ IS TITRATED WITH Ce4+
REGION 1 : BEFORE THE EQUIVALENCE POINT
As each aliquot of Ce4+ is added , it is consumed (eqn 1) and
creates an equal number of moles of Ce3+ and Fe3+
Prior to the equivalence point excess unreacted Fe remains in
solution
Since the amounts of Fe2+ and Fe3+ are known, cell voltage can
be calculated from eqn 2 rather than 3 E = E+ - E- (calomel)
[ Fe2 ]
E = 0.767- 0.05916log
- 0.241
3
[ Fe ]
When volume titrant is half the equivalence point ( V = 1/2Ve)
the concentration of Fe2+ and Fe3+ are equal, thus E+ = Eo
For an acid-base titration pH = pKa when V = 1/2Ve
SHAPES OF TITRATION CURVES
E ~ Eo(Ce4+I Ce3+)
- 0.241V = 1.46V
1:1 stoichiometry symmetric
about equivalence point & same
curve for diluted sample
Not symmetric about the equivalence
point – 2:1
REGION 2 : AT THE EQUIVALENCE POINT
All cerium is in the form of Ce3+
[Ce3+] = [Fe3+]
Thus the equilibrium form of eqn 1 Ce 4+ + Fe2+ ↔ Ce3+ + Fe3+
If a little Fe3+ goes back to Fe2+, an equal no. of moles Ce4+
must be made and [Ce4+] = [Fe2+]
Eqns 2 and 3 are in equilibrium at the Pt electrode, it is
convenient to use both these eqns to calculate the cell voltage
At equilibrium in a redox titration, Ecell = (Eo1 + Eo2)/2
REGION 3 : AFTER THE EQUIVALENCE POINT
Almost all the iron atoms are Fe3+, the moles of Ce3+ = moles
Fe3+, and there is a known excess of unreacted Ce4+. We know
[Ce3+] and [Ce4+] and so we can use eqn 3 to calc E .
FINDING THE ENDPOINT
As in acid-base titration, indicators and electrodes are
commonly used to find the endpoints of a redox titration.
REDOX INDICATORS
A redox indicator is a compound that changes colour when it
goes from oxidized to a reduced state, eg. Ferroin changes
from pale blue to red.
By writing the Nernst equation we can predict the potential
range over which the indicator will change :
In (oxidised) + ne- ↔ In (reduced)
0.05916V
[ In(reduced)]
E=E log
n
[ In(oxidised)]
o
(6)
As with acid base indicators, the colour of In (reduced) will be
observed when:
[In( reduced)]
[In(oxidised)]

10
1
And the colour of the In(oxidised) will be observed when:
[In( reduced)]
[In(oxidised)]

1
10
Using these 2 quotients in eqn 6, tells us the colour
range will occur over the range
0.05916
E = (E 
)volts
n
o
Eo= 1.147V, we expect the
colour change to occur ~ 1.088
– 1.206V wrt SHE
STARCH IODINE COMPLEX
Many analytical procedures use redox titrations involving
iodine. Starch is used as the indicator, since it forms an intense
blue complex with iodine. Starch is not a redox indicator
because it responds to I2, not to a change in potential.
Starch is readily biodegradable and must be freshly prepared.
ADJUSTMENT OF ANALYTE OXIDATION STATES
Before titration we adjust the oxidation state of the analyte, eg
Mn2+ can be pre-oxidised to MnO4- and then titrated with Fe2+
Pre-adjustment must be quantitative and all excess reagent
must be destroyed.
Pre-oxidation :
Persulphate S2O82- is a powerful oxidant that requires Ag+ as a
catalyst: S2 O82- + Ag+
SO42- + SO4- + Ag2+
Two powerful oxidants
Excess reagent is destroyed after by boiling the solution
after oxidation is complete
2S2O82- + 2 H2O
boiling
4SO42- + O2 + 4H+
H2O2 is a good oxidant in basic solution and reductant in
acidic solution. The excess spontaneously disproportionate
in boiling water.
H2O2
H2O + O2
PRE-REDUCTION
Stannous & chromous chloride, SO2 , H2S are used to prereduce analytes to a lower oxidation state.
An important pre-reduction technique uses a packed column
to pre-reduce analyte to a lower oxidation state (analyte is
drawn by suction).
Jones reductor, which contains Zn coated with
Zn amalgam. Zn is a powerful reducing agent (Eo
= -0.764V) making the Jones reductor
unselective, other species eg, Cr3+ are reduced
and may interfere with the titration analysis.
OXIDATION WITH POTASSIUM PERMANGANATE
KMnO4 is a strong oxidant with an intense violet colour. In
strongly acidic solutions (pH< 1) it is reduced to colourless
Mn2+ (manganous).
In neutral or alkaline the product is a brown solid – MnO2
In strongly alkaline solution, green manganate MnO42- is
produced.
KMnO4 is not a primary
standard, and can contain
traces of MnO2, thus it must
be standardized with pure Fe
wire or sodium oxalate (pink
end-point) for greater
accuracy,
The KMnO4 solution is
unstable :
4MnO4- + 2H2O
4MnO2 + 3O2 + 4OH-
OXIDATION WITH Ce4+
Reduction of Ce4+ (yellow) to Ce3+ (colorless) can be used in
place of KMnO4. Ce4+ is used for the quantitative determination
of malonic acid as well as alcohol, ketones and carboxylic
acids.
The primary standard is prepared by dissolving the salt in 1M
H2SO4 and is stable indefinitely.
OXIDATION WITH K2Cr2O7
Powerful oxidant – in acidic solution orange dichromate iron is
reduced to green chromic ion.
In 1M HCl the formal potential is 1.00V and 2M H2SO4 it is 1.11V,
thus less powerful than Ce 4+ and MnO4-
Stable primary standard that is employed to determine
Fe2+
Also used in environmental analysis of oxygen demand.
COD or chemical oxygen demand is defined as the oxygen
that is equivalent to the Cr2O72- consumed by the oxidation
of organics in water.
METHODS INVOLVING IODINE
When a reducing analyte is titrated with iodine to produce Ithe method is called Iodimetry (titration with I3-).
In the iodimetric determination of vitamin C-starch is added to
give an intense blue end-point.
Iodometry – oxidizing analyte added to I- to produce I2 which
is then titrated with thiosulfate standard, starch is added only
before the endpoint.
When we speak of using iodine as a titrant we mean a
solution of I2 plus excess II2 (aq) + I-
I3 -
K= 7x102
A 0.05M solution of I3- is prepared by dissolving0.12M KI plus
0.05M I2 in water.
Reducing agent + I3Oxidizing agent + 3I-
3II3 -
Precipitation Reactions
Gravimetric Analysis: Solid product formed
Relatively insoluble
Easy to filter
High purity
Known Chemical composition
Precipitation Conditions: Particle Size
Small Particles: Clog & pass through filter paper
Large Particles: Less surface area for attachment of
foreign particles.
1. Nucleation
Molecules form small
Aggregates randomly
Crystallization
2. Particle
Growth
Addition of more
molecules to a
nucleus.
Supersaturated Solution: More solute than should be
present at equilibrium.
Supersaturated Solution: Nucleation faster; Suspension
(colloid) Formed.
Less Supersaturated Solution: Nucleation slower, larger
particles formed.
How to promote Crystal Growth
1. Raise the temperature
Increase solubility
Decrease supersaturation
2. Precipitant added slowly with
vigorous stirring.
3. Keep low concentrations of precipitant
and analyte (large solution volume).
Homogeneous Precipitation
Precipitant generated slowly by a chemical reaction
O
Heat
+
C
CO2
3H2O
2NH4+
+
2OH-
NH2
H2N
O
OH-
+
C
H
+
HCO2Formate
+
H2O
OH
pH gradually
increases
Formic Acid
Large particle size
3HCO2-
+
Fe3+
Fe(HCO2)3.nH2O(s)
Fe(III)formate
Net +ve charge on colloidal particle because of
adsorbed Ag+
Precipitation in the Presence of an Electrolyte
Consider titration of Ag+ with Cl- in the presence of
0.1 M HNO3.
Colloidal particles of ppt: Surface is +vely charged
Adsorption of excess Ag+
on surface (exposed Cl-)
Colloidal particles need enough kinetic energy to
collide and coagulate.
Addition of electrolyte (0.1 M HNO3) causes
neutralisation of the surface charges.
Decrease in ionic atmosphere (less electrostatic
repulsion)
Digestion and Purity
Digestion: Period of standing in hot mother liquor.
Promotion of recrystallisation
Crystal particle size increases and expulsion
of impurities.
Purity:
Adsorbed impurities: Surface-bound
Absorbed impurities: Within the crystal Inclusions &
Occlusions
Inclusion: Impurity ions
occupying crystal lattice sites.
Occlusion: Pockets of
impurities trapped within
a growing crystal.
Coprecipitation: Adsorption, Inclusion and Occlusion
Colloidal precipitates: Large surface area
BaSO4; Al(OH)3; and Fe(OH)3
How to Minimise Coprecipitation:
1. Wash mother liquor, redissolve, and reprecipitate.
2. Addition of a masking agent:
Gravimetric analysis of Be2+, Mg2+, Ca2+, or Ba2+
with N-p-chlorophenylcinnamohydroxamic acid.
Impurities are Ag+, Mn2+, Zn2+, Cd2+, Hg2+, Fe2+,
and Ga2+. Add complexing KCN.
Ca 2+
+
2RH
Analyte
Mn2+
Impurity

CaR2(s)
+
2H+
Precipitate
6CN-
+
Masking agent

Mn(CN)64Stays in solution
Postprecipitation: Collection of impurities on ppt during
Digestion: A supersaturated impurity
e.g., MgC2O4 on CaC2O4.
Peptization:
Breaking up of charged solid particles
when ppt is washed with water.
AgCl is washed with volatile electrolyte (0.1 M HNO3).
Other electrolytes: HCl; NH4NO3; and (NH4)2CO3.
Product Composition
Hygroscopic substances:
Difficult to weigh accurately
Some ppts: Variable water quantity as water of
Crystallisation.
Drying
Change final composition by ignition:
Fe(HCO2)3.nH2O
850 oC
Fe2O3
+
CO2(g)
+ xH2O(g)
(1 Hour)
1100oC
2Mg(NH4)PO4.6H2O
(1 Hour)
Mg2P2O7
+
2NH3 +
13H2O
Thermogravimetric Analysis
Heating a substance and measuring its mass as a
function of temperature.
OH
OH
OH
OH
H2O
200oC
CO2CaO2C
CO2CaO2C
300oC
Calcium salicylate monohydrate
O
CaO
Calcium
oxide
700oC
CaCO3
Calcium
carbonate
Ca
500oC
O
O
Example
In the determination of magnesium in a sample, 0.352 g
of this sample is dissolved and precipitated as
Mg(NH4)PO4.6H2O. The precipitate is washed and
filtered. The precipitate is then ignited at for 1 hour
1100 oC and weighed as Mg2P2O7.The mass of Mg2P2O7
is 0.2168 g.
1100oC
2Mg(NH4)PO4.6H2O
Mg2P2O7
+
2NH3 +
13H2O
(1 Hour)
Calculate the percentage of magnesium in the sample.
Solution:
Relative atomic mass
Of Mg
The gravimetric factor is:
Grams of Mg in analyte
Grams of Mg2P2O7
=
2 x (24.305)
222.553
FM of Mg2P2O7
Note: 2 mol Mg2+ in 1 mol Mg2P2O7.
Grams of Mg in analyte = Grams of Mg2P2O7 formed
2 x (24.3050)
222.553
2 x(24.3050 )
= 0.2168 g
222 .553
Mass of Mg 2+ = 0.0471 g
% Mg =
Mass of Mg2+ (100)
=
sample Mass
= 13.45 %
0.0474 (100)
0.352
Combustion Analysis
Determination of the carbon and Hydrogen content of
organic compounds burned in excess oxygen.
H2O absorption
CO2 Absorption
Prevention of
entrance of
atmospheric O2
and CO2.
Note: Mass increase
in each tube.
C, H, N, and S Analyser:
Modern Technique
Thermal Conductivity, IR,or Coulometry for
Measuring products.
2 mg sample in tin or silver capsule.
Capsule melts and sample is oxidised in excess of O2.
Dynamic Flash combustion: Short burst of gaseous products
C, H,N, S
1050 oC; O2
CO2(g) + H2O(g)
+
N2(g) +
SO2(g)
+
SO3(g)
(95 % SO2)
Products
Hot WO3 catalyst: Carbon
Then, metallic Cu at 850 oC:
Cu + SO3
2Cu + O2
850 oC
850 oC
SO2 + CuO(s)
2CuO(s)
Heat
CrO3 Cat.
CO2
Oxygen Analysis:
Pyrolysis or thermal decomposition in absence of oxygen.
Carbon
Gaseous products: Nickelised
1075 oC
CO formed
Halogen-containing compounds:
CO2, H2O, N2, and HX products
HX(aq) titration with Ag+ coulometrically.
Silicon Compounds (SiC, Si3N4, & Silicates from rocks):
Combustion with F2 in nickel vessel
Volatile SiF4 & other fluorinated products
Mass
Spectrometry
Example 1:
Write a balanced equation for the
combustion of benzoic acid, C6H5CO2H, to give CO2
and H2O. How many milligrams of CO2 and H2O will be
produced by the combustion of 4.635 mg of C6H5CO2H?
Solution:
C6H5CO2H
+
FW = 122.123
15/
4.635 mg of C6H5CO2H
2O2
7CO2
44.010
+
3H2O
18.015
4.634m g
= 0.03795m m ol
=
122.123m g / m m ol
1 mole C6H5CO2H yields 7 moles CO2 and 3 moles H2O
Mass CO2 = 7 x 0.03795 mmol x 44.010 mg/mmol = 11.69 mg CO2
Mass H2O = 3 x 0.03795 mmol x 18.015 mg/mmol = 11.69 mg H2O
Example 2: A 7.290 mg mixture of cyclohexane, C6H12
(FW 84.159), and Oxirane, C2H4O (FW 44.053) was
analysed by combustion, and 21.999 mg CO2
(FW 44.010) were produced. Find the % weight of
oxirane in the sample mixture.
Solution:
C6H12 + C2H4O + 23/2O2
8CO2 + 8H2O
Let x = mg of C6H12 and y = mg of C2H4O.
X
+
y
=
7.290 mg
Also,CO2 = 6(moles of C6H12) + 2(moles of C2H4O)
21.999m g
 x   y 
6
  2
=
 84.161  44.053 44.010m g / m m ol
21.999
g
 x   y 
COm
2
6
  2
=
 84.161  44.053 44.010m g / m m ol
X
+
y
=
7.290 mg

x = 7.290 - y
21.999m g
 7.290 - y   y 
6
  2
=
 84.161   44.053 44.010m g / m m ol
 y = mass of C2H4O = 0.767 mg
Therefore, % Weight Oxirane = 0.767m g (100)
7.294m g
= 10.52 %
The Precipitation Titration Curve
Reasons for calculation of titration curves:
1. Understand the chemistry occurring.
2. How to exert experimental control to influence the
quality of analytical titration.
In precipitation titrations:
1. Analyte concentration
2. Titrant concentration
3. Ksp magnitude
Influence the sharpness
of the end point
Titration Curve
A graph showing variation of concentration of one
reactant with added titrant.
Concentration varies over many orders of magnitude
P function:
pX = -log10[X]
Consider the titration of 25.00 mL of 0.1000 M I- with
0.05000 M Ag+.
I- + Ag+  AgI(s)
There is small solubility of AgI:
AgI(s)  I- + Ag+
Ksp = [Ag+][I-] = 8.3 x 10-17
I- + Ag+  AgI(s)
K =1/Ksp = 1.2 x 1016
Ve = Volume of titrant at the equivalent point:
(0.02500 L)(0.1000 mol I-/L) = (Ve)(0.05000 mol Ag+/L)
mol I-
mol Ag+
 Ve = 0.05000 L = 50.00 mL
Before the Equivalence Point:
Addition of 20 mL of Ag+:
This reaction: I- + Ag+  AgI(s)
goes to completion.
Some AgI redissolves: AgI(s)  I- + Ag+
Ag  = I 

K sp
[I-] due to I- not precipitated by
20.00 mL of Ag+.
-
(20.00m L)
Fraction of I- reacted: (50.00m L)
(30.00m L)
Fraction of I- remaining: (50.00m L)
Original volume
of I-
 30.00m L 
 25.00m L 
Therefore, [ I ] =  50.00m L (0.1000M ) 45.00m L 




-
Fraction
Remaining
Original
Conc.
Dilution
Factor
Total
volume

[I-] = 3.33 x 10-2 M
Ag  = I 

K sp
-


Ag 

8.3x10-17
=
3.33x10-2
[Ag+] = 2.49 x 10-15 M
pAg+ = -log[Ag+] = 14.60
The Equivalence Point:
All AgI is precipitated
Then,
AgI(s)  I- + Ag+
Ksp = [Ag+][I-] = 8.3 x 10-17
And
[Ag+] = [I-] = x
Ksp = (x)(x) = 8.3 x 10-17

 X = 9.1 x 10-9 M
pAg+ = -log x = 8.04
At equivalence point:
pAg+ value is independent of the
original volumes or concentrations.
After the Equivalence Point:
Note: Ve = 50.00 mL
[Ag+] is in excess after the equivalence point.
Suppose that 52.00 mL is added:
Therefore, 2.00 mL excess Ag+
 2.00m L 
[ Ag ] = (0.05000M )

 77.00m L 
Volume of excess
Ag+

Original Ag+
Concentration
Dilution
Factor
Total volume
of solution
[Ag+] = 1.30 x 10-3 M
pAg+ = -log[Ag+] = 2.89
Shape of the Titration Curve:
Equivalence point: point of maximum slope
Steepest slope:
dy
dx
2
Inflection point:
has maximum value
d y
=0
2
dx
Titration Curves: Effect of Diluting the reactants
1. 0.1000 M I- vs
0.05000 M Ag+
2. 0.01000 M I- vs
0.005000 M Ag+
3. 0.001000 M I- vs
0.0005000 M Ag+
Titrations involving 1:1 stoichiometry of reactants
Equiv. Point: Steepest point in titration curve
Other stoichiometric ratios: 2Ag+ + CrO42-  Ag2CrO4(s)
1. Curve not symmetric near equiv. point
2. Equiv. Point: Not at the centre of the steepest section
of titration curve
3. Equiv. Point: not an inflection point
In practice: Conditions chosen such that curves are steep
enough for the steepest point to be a good estimate of
the equiv. point
Effect of Ksp on the Titration Curve
 K = 1/Ksp largest
AgI is least soluble
Sharpest change at
equiv. point
Least sharp, but steep
enough for Equiv.
point location
Titration of a Mixture
Less soluble precipitate forms first.
Titration of KI & KCl solutions with AgNO3
Ksp (AgI) << Ksp (AgCl)
First precipitation of AgI nearly complete before the
second (AgCl) commences.
When AgI pption is almost complete, [Ag+] abruptly
increeases and AgCl begins to precipitate.
Finally, when Cl- is almost completely consumed,
another abrupt change in [Ag+] occurs.
Titration Curve for 40.00 mL of 0.05000 M KI and 0.05000 M KCl
with 0.084 M AgNO3.
I- end point: Intersection of the steep and nearly
horizontal curves.
Note: Precipitation of AgI not quite complete when AgCl
begins to precipitate.
End of steep portion better approximation of the
equivalence point.
AgCl End Point: Midpoint of the second steep section.
The AgI end point is always slightly high for I-/Clmixture than for pure I-.
1. Random experimental error: both +tive and –tive.
2. Coprecipitation: +ve error
Example: Some Cl- attached to AgBr ppt and carries
down an equivalent amount of Ag+.
Coprecipitation error lowers the calculated concentration
of the second precipitated halide.
High nitrate concentration to minimise coprecipitation.
NO3- competes with Cl- for binding sites.
Separation of Cations by Precipitation
Consider a solution of Pb2+ and Hg22+: Each is 0.01 M
PbI2(s) ⇌ Pb2+ + 2I-
Ksp = 7.9 x 10 -9
Hg2I2(s) ⇌ Hg22+ + 2I-
Ksp = 1.1 x 10 -28
Smaller Ksp
Considerably
Less soluble
Is separation of Hg22+ from Pb2+ “complete”?
Is selective precipitation of Hg22+ with I- feasible?
Can we lower [Hg22+ ] to 0.010 % of its original value
without precipitating Pb2+?
From 0.010 M to 1.0 x 10–6 M?
Add enough I- to precipitate 99.990 % Hg22+.
Hg2I2(s)
Initial Concentration:
Final Concentration:
[Hg22 ][I - ]2 = K sp
0
solid
⇌
Hg22+
+
0.010
1.0 x 10-6
2I0
x
 (1.0 x 10-6)(x)2 = 1.1 x 10-28
 X = [I-] = 1.0 x 10 –11 M
Will this [I-] = 1.0 x 10 –11 M precipitate 0.010 M Pb2+?
Q = [ Pb2 ][I - ]2 = (0.010)(1.0x10-11 ) 2
Q = 1.0 x 10-24 << 7.9 x 10–9 = Ksp for PbI2
Therefore, Pb2+ will not precipitate.
Prediction: All Hg22+ will virtually precipitate before any
Pb2+ precipitates on adding I-.