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CHAPTER 10
2
Gravimetric Analysis and
Precipitation Equilibrium
GRAVIMETRIC ANALYSIS
Key Concepts
 Gravimetric analysis is the quantitative isolation of a
substance by precipitation and weighing of the
precipitate.
 An analyte is the substance to be analysed.
 A precipitating reagent is the reactant used to precipitate
the analyte.
Analyte
+
2ClReagent

PbCl2(s)
Solid Product
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Pb+2
GRAVIMETRIC ANALYSIS
Procedure:
o Weigh the sample to be analysed
o Dissolve the sample in a suitable solvent, eg, water
o Add an excess of the precipitating reagent to precipitate the
analyte
o Filter the mixture to separate the precipitate from the solution
o Wash the precipitate to remove any impurities
o Dry the precipitate by heating to remove water
o Cool the precipitate in a dessicator to prevent the precipitate
absorbing moisture from the air
o Weigh the cooled precipitate
o Repeat the drying and weighing process until a constant mass
for the precipitate is achieved
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o Calculate the percent by mass of analyte in the sample
GRAVIMETRIC ANALYSIS
Key steps in a
precipitation reactionwhat this looks like
Forming a precipitate
Filtering the solution
Weighing the dry precipitate
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Weighing the sample
GRAVIMETRIC ANALYSIS
General calculation of the percent by mass of analyte in a sample:
 Weight the balanced chemical equation for the precipitation reaction
 Calculate the moles of precipitate:
moles = mass ÷ molecular mass
 Calculate moles of analyte from the balanced chemical equation using
the mole ratio of (analyte : precipitate)
 Calculate mass of analyte:
mass = moles x molecular mass
 Calculate percent by mass of analyte in sample:
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= (mass analyte ÷ mass sample) x 100
EXAMPLE
CALCULATION
What is the %KCl in a solid if 5.1367 g of solid gives
rise to 0.8246g AgCl?
Cl- +
Ag+ 
AgCl(s)
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Moles of precipitate (AgCl) = 0.8246/143.3 = 0.00575 mole
Moles of analyte (Cl or KCl) = 0.00575 x (1 mole Cl / 1 mole AgCl)
= 0.00575 mole
Mass of analyte (KCl) = 0.00575 x 74.55 = 0.429 g
Percentage by mass of analyte = (0.429/5.1367)x100
= 8.35%
GRAVIMETRIC ANALYSIS
•
Principals:
Reagent + Analyte ➝ Solid Product (collect and measure mass)
Desired Properties of Solid Product

Should be very insoluble

Easily filterable (i.e., large crystals)

Very Pure

Known and constant composition
Solubility:
The solubility of a precipitate can be decreased by:

Decreasing temperature of solution

Using a different solvent
- usually a less polar or organic solvent (like dissolves like)
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•
GRAVIMETRIC ANALYSIS
Gravimetric Analysis:
Governed by equilibrium: AgCl Ksp = 1.8 x 10-10
Cl- +
Ag+ 
AgCl(ag)
ion pair formation
AgCl(aq)

AgCl(s)
intrinsic solubility
AgCl +Cl-

AgCl2-
complex ion formation
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Solubility of AgCl = [Ag+] + [AgCl] + [AgCl2-]
GRAVIMETRIC ANALYSIS
Filterability:
(i) Want product to be large enough to collect on filter:

Doesn’t clog filter

Doesn’t pass through filter
(ii) Best Case: Pure Crystals
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(iii) Worst Case: Colloidal suspension

Difficult to filter due to small size

Tend to stay in solution indefinitely  suspended by
Brownian motion
(usually 1-100 nm in size )
GRAVIMETRIC ANALYSIS
Process of Crystal Growth: Two Phases in Crystal Growth
Nucleation – molecules in solution come together randomly
and form small aggregates
Particle growth – addition of molecules to a nucleus to form a
crystal
Crystal Impurities:
Impurities are undesirable (known as co-precipitation)

Change the chemical composition of the precipitate

Creates errors in gravimetric analysis
Adsorbed to
crystal surface
Occlusion,
Inclusion
Absorbed or
trapped within
crystal pockets
Impurity placed in crystal
instead of analyte
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Types of Impurities
Adsorption,
GRAVIMETRIC ANALYSIS
Washing Precipitates
Precipitates from ionic compounds
- need electrolyte in wash solution
- keep precipitate from breaking up and redissolving (peptization)
Electrolyte should be volatile
- removed by drying
- HNO3, HCl, NH4, NO3, etc.
Illustration:
- AgCl(s) should not be washed with H2O, instead wash with dilute HNO3
Drying/Igniting Precipitates
Many precipitates contain varying amounts of H2O
- adsorbed from the air (i.e. hygroscopic)
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Precipitates are dried for accurate, stable mass measurements
Precipitates are also ignited to convert to a given chemical form
GRAVIMETRIC ANALYSIS
Example
A 2.00g sample of limestone was dissolved in hydrochloric acid and all the
calcium present in the sample was converted to Ca2+. Excess ammonium
oxalate solution, (NH4)2C2O4 was added to the solution to precipitate the
calcium ions as calcium oxalate, CaC2O4. The precipitate was filtered,
dried and weighed to a constant mass of 2.43g.
Determine the percentage by mass of calcium in the limestone sample.
a. Write the balanced chemical equation for the precipitation reaction:
Ca2+(aq) + C2O42-(aq) -----> CaC2O4(s)
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b. Calculate the moles of calcium oxalate precipitated.
n(CaC2O4(s)) = mass ÷ MM
n(CaC2O4(s)) = 2.43 ÷ (40.08 + 2 x 12.01 + 4 x 16.00)
n(CaC2O4(s)) = 2.43 ÷ 128.10
n(CaC2O4(s)) = 0.019 mol
GRAVIMETRIC ANALYSIS
c.Find the moles of Ca2+(aq).
From the balanced chemical equation, the mole ratio of
Ca2+ : CaC2O4(s) is 1 : 1. So, n(Ca2+(aq)) = n(CaC2O4(s)) = 0.019mol
d.Calculate the mass of calcium in grams
mass (Ca) = n x MM
mass (Ca) = 0.019 x 40.08 = 0.76g
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e.Calculate the percentage by mass of calcium in the original sample:
%Ca = (mass Ca ÷ mass sample) x 100
%Ca = (0.76 ÷ 2.00) x 100 = 38%
PRECIPITATION EQUILIBRIA:
THE SOLUBILITY PRODUCT
“Insoluble” substances still have slight solubility
the solid does not appear in Ksp
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For Ag2CrO4
PRECIPITATION EQUILIBRIA:
THE SOLUBILITY PRODUCT
“Insoluble” substances still have slight solubility
the solid does not appear in Ksp
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For Ag2CrO4
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PRECIPITATION EQUILIBRIA:
THE SOLUBILITY PRODUCT
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PRECIPITATION EQUILIBRIA:
THE SOLUBILITY PRODUCT
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CHAPTER 11
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Precipitation
Titration
PRECIPITATION
TITRATION
Key Concepts
 A titration in which the reaction between the analyte and
titrant involves a precipitation.


Ag (aq)
 Cl(aq)
 AgCl (s)
K sp  1.82 1010


Ag (aq)
 I (aq)
 AgI (s)
K sp  8.3 10-17

2Ag (aq)
 CrO 24 (aq)  Ag 2CrO 4 (s)
K sp  1.2 1012
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The solids produced from the titration reactions should
have very low Ksp values.
PRECIPITATION
TITRATION
 For the reaction
A+(aq) + B-(aq) ⇌
AB(s)
The concentrations of A+ and B- are varying through the titration
process. When titrating B- (analyte) with A+ (titrant).
a) in the beginning of titration, concentration of A+ is zero and
concentration of B- is calculated from its salt.
b) After addition of few milliliters of A+, B- will react with A+ forming
AB precipitate, concentrations of A+ will be calculated according
to Ksp. B- will be excess.
c) At the end point, concentrations A+, B- will be calculated from Ksp.
d) After the end point A+ is the excess and B- will be calculated from
Ksp.
Titration curve in the precipitation titration is a plot of
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pA and/or pB versus volume of titrant (A)
PRECIPITATION
TITRATION
Example: Calculate pCl for the titration of 100.0mL of 0.1000 M Cl− with 0.1000
M AgNO3 for the addition of 0.00, 20.00, 99.00, 99.50, 100.00, 100.50,
and 110.00mL AgNO3 (assume Ksp = 1.0x10-10)
a) at 0.0mL addition
pCl = -log(0.1000) = 1.000
b) after 20.0mL addition
mmol Cl− = 100.0 mL × 0.1000 mmol/mL = 10.00 mmol
mmol Ag+ = 20.00 mL × 0.1000 mmol/mL = 2.000 mmol
Cl− left = 10.00 − 2.00 = 8.00 mmol/120.0 mL = 0.0667 M
pCl = −log 0.0667 = 1.18
c) after 99.00 mL addition
mmol Ag + = 99.00 mL × 0.1000 mmol/mL = 9.900 mmol
pCl = −log 5.0×10−4 = 3.26
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Cl − left = 10.00 − 9.90 = 0.10 mmol/199.0 mL = 5.0×10−4 M
PRECIPITATION
TITRATION
d) After 99.50mL addition
mmol Ag+ = 99.50 mL × 0.1000 mmol/mL = 9.950 mmol
Cl− left = 10.00 − 9.95 = 0.05 mmol/199.5 mL = 2.5×10−4M
pCl = −log 2.5×10−4 = 3.60
e) after 100.00 mL addition
at 100.00 mL, all the Cl− is reacted with Ag+
[Cl−] =
K sp  11010  1105
pCl = −log 1.0×10−5 = 5.00
f) after 100.50 mL addition
mmol Ag+ = 100.50 mL × 0.1000 mmol/mL = 10.05 mmol
Ag+ left = 10.05 − 10.00 = 0.05 mmol/200.5 mL = 2.5×10−4 M
[Cl−] = Ksp/[Ag+] = 1.0×10−10/2.5×10−4 = 4.0×10−7 M
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pCl = −log 4.0×10−7 = 6.40
PRECIPITATION
TITRATION
g) after 110.00 mL addition
mmol Ag+ = 110.00 mL × 0.1000 mmol/mL = 11.00 mmol
Ag+ left = 11.00 − 10.00 = 1.00 mmol/210 mL = 4.76×10−3 M
[Cl−] = Ksp/[Ag+] = 1.0×10−10/4.76×10−3 = 2.1×10−8 M
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pCl = −log 2.1×10−8 = 7.67
PRECIPITATION
TITRATION
Example: 20 mL of 0.05M CaCl2 was titrated with 0.10M AgNO3. Calculate the
concentration of Cl- and Ag+ in the solution after addition of 0.0mL
AgNO3, 5.0mL AgNO3 , at the equivalence point and after addition of 30mL
AgNO3. (assume Ksp = 1.82x10-10)
a) at 0.0mL addition
b) after 5.0mL addition
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Firstly, calculate the new rough concentrations after addition
PRECIPITATION
TITRATION
Secondly, calculate the exact concentrations from Ksp
c) at the equivalence point.
at the equivalence point, volume of titrant will be calculated as following
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mole Ag = mole Cl ⇨ (M.V)Ag = (M.V)Cl ⇨ 0.1xVAg = 0.1x20 ⇨ Vag = 20mL
at this point [Ag+] and [Cl] only comes from dissociation of AgCl
PRECIPITATION
TITRATION
d) after addition of 30mL AgNO3 , Ag is excess now
Firstly, calculate the new rough concentrations after addition
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Secondly, calculate the exact concentrations from Ksp
SUMMARY OF
CALCULATIONS
 For the reaction
A+(aq) + B-(aq) ⇌
AB(s)
When titrating B (analyte) with A+ (titrant).
a) In the beginning of titration, concentration of A+ is zero and
concentration of B- is calculated from its salt.
b) Before the equivalence point
CB= (original moles of B – moles of A) / total volume
CA= Ksp/CB
c) At the equivalence point, CB=CA=
K sp
d) After the equivalence point
CB=Ksp/CA
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CA= (moles of A added – original moles of B) / total volume
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TITRATION CURVE
END POINT AND
INDICATORS
Three methods are well known for precipitation titrations involving Ag+ as
titrant. These are:
 Mohr method (direct method)
in this method, excess Ag cations after the equivalence point will react with the
indicator (CrO42-) forming red precipitate. This method carried at pH 7-10, to
prevent formation of chromic acid (H2CrO4).
 Fajans method (direct method)
in this method, the indicator (Fluorescein) adsorbed on the surface of the
colloidal particles of the precipitate. This will change its color from yellow-green
to red. The used of this method is limited because of the relatively few
precipitation reactions in which a colloidal precipitate is formed rapidly.
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 Volhard method (back titration method)
in this method, the analyte is treated with excess amount of AgNO3, and the
remaining AgNO3 is titrated then with SCN-. At the end point, the titrant SCN- will
react with the indicator Fe3+ forming red color complex. pH in this method
should be acidic to prevent the formation of Fe(OH)3.
EXAMPLE
(VOLHARD)
25.0mL of 0.10M AgNO3 were added to 30 mL of unknown chloride
solution. The resulted mixture was then titrated with 19.2mL of 0.10M
thiocyanate solution (SCN-). Calculate the concentration of Cl- in the
unknown solution.
moleAg = moleCl + moleSCN
(M.V)Ag = (M.V)Cl + (M.V)SCN
25x0.1 =
MClx30 + 19.2x0.1
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2.5 -1.92 = MClx30 ⇨ MCl = 0.0193M
EXAMPLE
(MOHR)
35.7mL of 0.10M AgNO3 were needed to reach the end point for titration
of 75.0mL of unknown chloride solution using mohr method. Calculate the
concentration of Cl- in the unknown solution.
moleAg = moleCl
(M.V)Ag = (M.V)Cl
35.7x0.1 =
MClx75
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3.57 = MClx75 ⇨ MCl = 0.0467M
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CHAPTER 14
Redox Reactions and
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Titrations
OXIDATION REDUCTION
REACTION
Oxidation is defined as a loss of electrons.
e.g. iron rusting: When iron metal is exposed to oxygen in the air the iron combines with the oxygen to
form iron (II) oxide and further reacts to form iron(III) oxide
Fe  Fe2  2e 
Fe2  Fe3  e 
Reduction is defined as a gain of electrons.
e.g. Formation of copper from copper (II) oxide: When hydrogen gas is passed over heated copper (II)
oxide, the copper (II) oxide is reduced to copper metal.
Cu 2  2e   Cu
Redox reactions are a combination of reduction and oxidation, which
occur simultaneously.
Mg  2H   Mg 2  H 2
Redox
2H   2e   H 2
Reduction
Mg  Mg 2  2e 
Oxidation
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e.g. Displacement of hydrogen from acidic solution with magnesium.
OXIDATION NUMBERS
Oxidation numbers are hypothetical numbers assigned to an individual atom or
ion present in a substance using a set of rules. Oxidation numbers can be positive,
negative, or zero.
It is VERY IMPORTANT to remember that oxidation numbers are always reported for
one individual atom or ion and not for groups of atoms or ions.
atom/ion
Fe3+
H in H2
Oxidation #
3+
Mg
0
0
Cl in NaCl
C in CO2
1-
4+
N in NH4+
3+
When the oxidation number for an atom or ion increased then we said it was oxidized,
and we can also consider it as reducing agent
ox.#
Fe2
2

Fe3
3
 increase in oxidation number  oxidation occures  Fe2 was oxidized, it is reducing agent
If the oxidation number decreased, then the atom (or ion) reduced and it is considered
as oxidizing agent

Cu
0
 decrease in oxidation number  reduction occures  Cu 2 was reduced, it is oxidizing agent
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ox.#
Cu 2
2
OXIDATION NUMBERS
RULES
The oxidation number for an atom in its elemental form is always zero.
• A substance is elemental if both of the following are true:
• only one kind of atom is present
• charge = 0
• Examples:
• S8: The oxidation number of S = 0
• Fe: The oxidation number of Fe = 0
The oxidation number of a monoatomic ion = charge of the monatomic ion.
• Examples:
• Oxidation number of S2- is -2.
• Oxidation number of Al3+ is +3.
The oxidation number of all Group 1A metals = +1 (unless elemental). (Li, Na, K,…)
The oxidation number of all Group 2A metals = +2 (unless elemental). (Mg, Ca, Ba,…)
Hydrogen (H) has two possible oxidation numbers (unless elemental) :
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• +1 when bonded to a nonmetal
• -1 when bonded to a metal
OXIDATION NUMBERS
RULES
Oxygen (O) has two possible oxidation numbers (unless elemental):
• -2 in all other compounds...most common
• -1 in peroxides (O22-)....pretty uncommon
The oxidation number of fluorine (F) is always -1 (unless elemental).
The sum of the oxidation numbers of all atoms (or ions) in a neutral compound = 0.
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The sum of the oxidation numbers of all atoms in a polyatomic ion = charge on the
polyatomic ion.
OXIDATION NUMBERS
RULES
Examples: Find oxidation numbers for each atom in A) Na2SO4, B) Cr2O72A) Na2SO4 : no elemental form, no monoatomic ion, Na from the 1A group so its oxidation
number is 1+. No group 2A, no hydrogen, O is in its most common form 2-.
The sum of the oxidation numbers of all atoms (or ions) in a neutral compound = 0.
2xNa + 1xS + 4xO = 0
2x(1+) + S + 4x(2-) = 0 ⇨ 2 + S – 8 = 0 ⇨ S = 6+
B) Cr2O72- : no elemental form, no monoatomic ion, no group 1A, no group 2A, no hydrogen,
O is in its most common form 2-.
The sum of the oxidation numbers of all atoms in a polyatomic ion = charge on the
polyatomic ion.
2xCr + 7xO = 2-
Find the oxidation number for each atom in the following: Ba(NO3)2, NF3, (NH4)2SO4
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2xCr + 7x(2-) = 2- ⇨ 2Cr – 14 = 2- ⇨ Cr = 6+
OXIDATION REDUCTION
REACTION AND
OXIDATION NUMBERS
Example:
In the reactions given below, identify the species undergoing oxidation and
reduction and balance the reaction:
Cu(s) + Ag+(aq) ➝ Cu2+(aq) + Ag(s)
The reaction could be divided to two reactions
1- Cu(s) ➝ Cu2+(aq). In this reaction ox.# of Cu was changed from 0➝2+
2- Ag+(aq) ➝ Ag(s). In this reaction ox.# of Ag was changed from 1+➝0
Reaction 1 is oxidation, Cu was oxidized and it is called reducing agent, while reaction
2 is reduction, Ag was reduced and it is called oxidizing agent.
We can re-write the reactions as following:
Cu(s) ➝ Cu2+(aq) + 2eAg+(aq) + e- ➝ Ag(s)
To balance the reaction we should equalize number of electrons, this will be happened
when we multiply reaction 2 by 2
2x
Cu2+(aq) + 2e-
(Ag+(aq) + e- ➝
Ag(s))
⇨
⇨
Cu(s) ➝
Cu2+(aq) + 2e-
2Ag+(aq) + 2e- ➝ 2Ag(s)
Cu(s)+ 2Ag+(aq) ➝ Cu2+(aq)+ 2Ag(s)
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Cu(s) ➝
OXIDATION REDUCTION
REACTION AND
OXIDATION NUMBERS
Example:
In the reactions given below, identify the species undergoing oxidation and reduction and
balance the reaction:
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Al(s)  CuCl 2(aq)  AlCl 3(aq)  Cu(s)
OXIDATION REDUCTION
REACTION AND
OXIDATION NUMBERS
Questions:
a)
Mg (s)  HCl (aq)  MgCl 2(aq)  H 2(aq)
b)
Fe(s)  H 2 O (l)  Fe2 O3(S)  H 2 (g)
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In the reactions given below, identify the species undergoing oxidation and reduction and
balance the reaction:
BALANCE REDOX EQUATIONS
IN ACIDIC AND BASIC MEDIUM
Acidic Conditions
3
Cr2O72 (aq)  HNO2 (aq)  Cr(aq)
 NO3 (aq)
1-Separate the half-reactions.
3
Cr2 O 72 (aq)  Cr(aq)
HNO 2 (aq)  NO3 (aq)
2-Balance elements other than O and H.
3
Cr2 O 72 (aq)  2Cr(aq)
HNO 2 (aq)  NO3 (aq)
3-Add H2O to balance oxygen.
3
Cr2O 72 (aq)  2Cr(aq)
 7H 2O(l)
HNO 2 (aq)  H 2O (l)  NO3 (aq)
4-Balance hydrogen with protons.
3
Cr2O72 (aq)  14H  (aq)  2Cr(aq)
 7H 2O (l)
HNO 2 (aq)  H 2O(l)  NO3 (aq)  3H  (aq)
(aq)
3
 14H  (aq)  6e   2Cr(aq)
 7H 2O(l)
HNO 2 (aq)  H 2O(l)  NO3 (aq)  3H  (aq)  2e 
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2
5-Balance the charge with e. Cr2 O 7
BALANCE REDOX EQUATIONS IN
ACIDIC AND BASIC MEDIUM
6-Scale the reactions so that they have an equal amount of electrons.
3
3
 Cr2O72 (aq)  14H  (aq)  6e   2Cr(aq)
 7H 2O(l)   Cr2O72 (aq)  14H  (aq)  6e   2Cr(aq)
 7H 2O(l)
3   HNO 2 (aq)  H 2O(l)  NO3 (aq)  3H  (aq)  2e    3HNO 2 (aq)  3H 2O(l)  3NO3 (aq)  9H  (aq)  6e 
1
7-Add the reactions and cancel out the electrons and common terms.
3
Cr2O72 (aq)  14H  (aq)  3HNO 2 (aq)  3H 2O(l)  2Cr(aq)
 7H 2O(l)  3NO3 (aq)  9H  (aq)
3
 Cr2O72 (aq)  5H  (aq)  3HNO 2 (aq)  2Cr(aq)
 4H 2O(l)  3NO3 (aq)
Basic Conditions
Same as in acidic conditions but complete by
8-Add OH- to balance H+.
3
Cr2O72 (aq)  5H  (aq)  3HNO2 (aq)  5OH ( aq)  2Cr(aq)
 4H 2O(l)  3NO3 (aq)  5OH ( aq)
9-Combine OH- ions and H+ ions that are present on the same side to form water.
10-Cancel common terms.
3
Cr2O72 (aq)  H 2O(l)  3HNO2 (aq)  2Cr(aq)
 3NO3 (aq)  5OH (aq)
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3
Cr2O72 (aq)  5H 2O(l)  3HNO2 (aq)  2Cr(aq)
 4H 2O(l)  3NO3 (aq)  5OH (aq)
BALANCE REDOX EQUATIONS
IN ACIDIC AND BASIC MEDIUM
Example: Balance the following reaction in acidic medium
2
MnO 4 (aq)  C2O24 (aq)  Mn (aq)
 CO2 (g)
C 2 O 24 (aq)  CO2 (g)
2
MnO 4 (aq)  Mn (aq)
C 2 O 24 (aq)  2CO2 (g)
2
MnO 4 (aq)  Mn (aq)
 4H 2 O
C 2 O 24 (aq)  2CO2 (g)
2
MnO 4 (aq)  8H   Mn (aq)
 4H 2 O
C 2 O 24 (aq)  2CO2 (g)
2
MnO 4 (aq)  8H   5e   Mn (aq)
 4H 2 O
C 2 O 24 (aq)  2CO2 (g)  2e 


5  C 2 O 24 (aq)  2CO2 (g)  2e 
2
2MnO 4 (aq)  16H   10e   2Mn (aq)
 8H 2 O
5C 2 O 24 (aq)  10CO2 (g)  10e 
2
2  MnO 4 (aq)  8H   5e   Mn (aq)
 4H 2 O

2
2MnO 4 (aq)  5C2O24 (aq)  16H   2Mn (aq)
 10CO2 (g)  8H 2O

46
2
MnO 4 (aq)  Mn (aq)
BALANCE REDOX EQUATIONS IN
ACIDIC AND BASIC MEDIUM
Example: Balance the following reaction in basic medium
CrOH3  ClO 3  CrO 24  Cl
Cr(OH) 3  CrO 24
ClO 3  Cl
Cr(OH) 3  H 2 O  CrO 24
ClO 3  Cl  3H 2 O
Cr(OH) 3  H 2 O  CrO 24  5H 
ClO 3  6H   Cl  3H 2 O
Cr(OH) 3  H 2 O  CrO 24  5H   3e 
ClO 3  6H   6e   Cl  3H 2 O

2Cr(OH) 3  2H 2 O  2CrO 24  10H   6e 

1 ClO 3  6H   6e   Cl  3H 2 O

ClO 3  6H   6e   Cl  3H 2 O
2Cr(OH) 3  ClO 3  2CrO 24  4H   Cl  H 2 O
2Cr(OH) 3  ClO 3  4OH   2CrO 24  4H   Cl  H 2 O  4OH 
2Cr(OH) 3  ClO 3  4OH   2CrO 24  Cl  5H 2 O
47

2  Cr(OH) 3  H 2 O  CrO 24  5H   3e 
POPULAR REDUCING AND
OXIDIZING TITRANTS
The two most common reducing agents used as titrants are thiosulfate
ion S2O32- and iron (II) Fe2+
Fe2  Fe3  e 
2S2O32  S4O62  2e 
The most common reducing agents used as titrants are Iodine I2,
Potassium permanganate KMnO4 and potassium dichromate K2Cr2O7
I 2  2e   2I 
MnO 4  8H   5e   Mn 2  4H 2 O
48
Cr2 O 72  14H   6e   2Cr 3  7H 2 O
IODOMETRY AND IODIMETRY
Iodometry involves indirect titration of iodine liberated by reaction with the
analyte, whereas iodimetry involves direct titration using iodine as the
titrant.
Iodine has brawn to pale yellow color, it gives a dark
blue color in the presence of starch. Usually starch is
added to the solution to indicate the appearance or
disappearance of iodine.
In iodometry, excess I react with the analyte (which is usually is an oxidizing
agent), the reaction forms I2, this I2 is then titrated with strong reducing
agent like S2O32. I2 is not soluble in water, therefore it is forming I3. For
example:
2 Cu2+ + 4 I → 2 CuI + I2
I2 + I → I3 (yellow)
I3 + 2 S2O32 → S4O62 + 3 I
49
at the end point mole of Cu2+ = moles of S2O32
IODOMETRY AND IODIMETRY
While titration with S2O32 , the yellow color of the solution start decreasing
and before its disappearance, starch is added before the total
disappearance of I3, the color change to dark blue before resuming titration.
In iodimetry, I2 produced in the reaction vessel (by reaction between IO3and
I) and react with the analyte, the excess amount of I2 is then titrated with
reducing agent like S2O32 . starch is added before the total disappearance of
I3 to indicate the sharp end point.
For example:
HO
HO
O
O
H
HO
HO
IO3 + 8I + 6H+ → 3I3 + 3H2O
C6H8O6 + I3 → C6H6O6 + 3I + 2H+
I3 + 2S2O32 → 3I- + S4O62
OH
HO
O
O
H
O
O
50
at the end point, moles of I3 (3 IO3 moles) = moles of C6H8O6 + ½ moles of S2O32
IODOMETRY AND IODIMETRY
Example:
36.40 mL of 0.3300 M Na2S2O3to titrate the I3 in a 15.00 mL sample, calculate the
molarity of I3 in the sample solution.
I3 + 2S2O32 → 3I + S4O62
Solution:
moles of S2O32 = M.V = 0.3300 M x 36.40 mL = 12.012 mmol
2 moles of S2O32
react with
1 mole of I3
12.012 mmole
react with
???
⇨ moles of I3 = 12.012 x 1 / 2 = 6.006 mmole
51
⇨ molarity of I3 = moles/volume = 6.006 mmol / 15.00 mL = 0.4004 M
IODOMETRY AND IODIMETRY
The amount of ascorbic acid, C6H8O6, in orange juice was determined by oxidizing the
ascorbic acid to dehydroascorbic acid, C6H6O6, with a known excess of I3–, and back
titrating the excess I3– with Na2S2O3.
Example:
A 5.00 mL sample of filtered orange juice was treated with 50.00 mL of excess 0.01023 M I3–. After the
oxidation was complete, 13.82 mL of 0.07203 M Na2S2O3 was needed to reach the starch indicator end
point. Report the concentration of ascorbic acid in grams per 100 mL.
Solution:
0.01023x0.050 = moles C6H8O6 +(0.5 x 0.07203x0.01382)
moles of C6H8O6 = 1.377x10-5
mass of C6H8O6= moles x M.wt = 1.377x10-5 x 176.13 = 0.002425g
concentration of ascorbic acid in grams per 100 mL= 100x0.002425/5 =0.0485g/100mL
52
0.002425g was found in 5 mL, So how much ????? are in 100mL
IODOMETRY AND IODIMETRY
Example:
A 0.20g sample containing copper is analyzed iodometrically. Copper(II) is reduced to
copper(I) by iodide
2Cu2+ + 4I → 2CuI + I2
What is the percent copper in the sample if 20.0 mL of 0.100M Na2S2O3 is required for
titration of the liberated I2.
I2 + I → I3
I3 + 2S2O32 → S4O62 + 3 I
Solution:
Over all reaction is
2Cu2+ + 4I + 2S2O32 → 2CuI + S4O62
moles of S2O32= M.V = 0.1x20 = 2mmol
moles of Cu2+ = moles of S2O32 = 2mmol
mass of copper = moles x atomic mass =(2/1000) x 63.54 = 0.12708g
53
% copper = [mass copper/mass sample] x 100%
=[0.12708/0.2]x100=63.54%
PERMANGANATE AS
TITRANT
KMnO4 was used as an oxidizing titrant in many redox titrations. It is self
indicating material, its color was changed from purple to colorless during
the reaction.
Example:
The concentration of hydrogen peroxide solution can be found by a redox titration with
acidified potassium permanganate solution. It was found that 10 cm3 of the hydrogen
peroxide solution reacted with 20 cm3 of 0.02M potassium permanganate solution
when titrated. Calculate the concentration of the hydrogen peroxide solution.
2 MnO4–(aq) + 6 H+(aq) + 5 H2O2(aq) ➝ 2 Mn2+(aq) + 4 H2O(l) + 5 O2(aq)
moles of MnO4– = (M.V)MnO4 = 0.02x20 = 0.4 mmole
moles of H2O2 =(5/2) moles MnO4– = (5/2)x 0.4 = 1mmole
(M)H2O2 = (M/V)H2O2 =1mmole/10mL
54
⇨ (M)H2O2 =0.1M
55
CHAPTER 9
56
Complexometric
Titrations
COMPLEXOMETRIC
TITRATIONS
Key Concepts
 The technique involves titrating metal ions with a complexing agent or chelating
agent (Ligand)
 Ligands or complexing agents or chelating agents can be any electron donating
entity, which has the ability to bind to the metal ion and produce a complex ion, Ex:
H2O, NH3, Cl-, Br-, I-…….
 Ethylenediaminetetraacetic acid, or EDTA, is an aminocarboxylic acid. In its fully
deprotonated form (Y4−), it has six binding sites—four negatively charged
carboxylate groups and two tertiary amino groups—that can donate six pairs of
electrons to a metal ion. The resulting metal–ligand complex, in which EDTA forms a
cage-like structure around the metal ion, is very stable. The actual number of
coordination sites depends on the size of the metal ion, however, all metal–EDTA
complexes have a 1:1 stoichiometry.
M 2
57


COMPLEXOMETRIC
TITRATIONS
Factors influencing EDTA reactions:
 The nature and activity of metal ion.
 The pH at which the titration is carried out.
 The presence of interfering ions such as CN-, Citrate, Tartrate, F- and other complex
forming agents.
 Organic solvents also increase the stability of complex.
58
Effect of pH:
EDTA TITRATIONS


The basic form of EDTA (Y4-) reacts with most metal ions to form a 1:1 complex.
Fraction (a) of the most basic form of EDTA (Y4-) is defined by the H+ concentration
and acid-base equilibrium constants
[Y 4 ]
[Y 4 ]
αY 
 αY 
EDTA 
[H 6 Y 2 ]  [H 5 Y  ]  [H 4 Y ]  [H 3Y  ]  [H 2 Y 2 ]  [HY 3 ]  [Y 4 ]
4
4
where [EDTA] is the total concentration of all free EDTA species in solution
aY4- is depended on the pH of the solution
K1K 2 K 3 K 4 K5 K6
{[H  ]6  [H  ]5 K1  [H  ] 4 K1K 2  [H  ]3 K1K 2 K 3  [H  ]2 K1K 2 K 3 K 4  [H  ]K1K 2 K 3 K 4 K5  K1K 2 K 3 K 4 K5 K6 }
59
aY 4  
EDTA TITRATIONS
The concentration of Y4- and the total concentration
of EDTA is solution [EDTA] are related as follows:
[Y 4 ]  αY4 EDTA 
⇨ The formation constant will be
60
at fixed pH conditional formation constant (K’f)
could be calculated as following
EDTA TITRATIONS
Example:
Calculate the molar Y4- concentration in a 0.0200M EDTA solution buffered to a pH of 10.00
[Y 4 ]  α Y 4 EDTA 
 0.3 x 0.0200  0.0060M
Example
Calculate the equilibrium concentration of Ni2+ in a solution with
an analytical NiY2- concentration of 0.0150M at pH 8.0
(Kf=4.2x1018) .
EDTA

22+
NiY
⇌ Ni
+
43(Y  HY  H 2 Y 2-  H3Y-  H 4 Y)
0.015-x
x
x
61
K’f=4.2x1018 x 4.2x10-3 = 1.76x1015 = 0.015/x2
x = [Ni2+] = √(0.015/1.76x1015)=2.9x10-9M
EDTA TITRATIONS
Question:
What is the concentration of free Fe3+ in a solution of 0.10 M FeY at (a) pH
2.0, (b) pH 8.0? (Kf=1.3x1025,
aY4-(pH 2)=2.6x10-14, aY4-(pH 8)=4.2x10-3).
Answer: [Fe3+] = 5.4x10-7 at pH 2.0 and [Fe3+] = 1.4x10-12 at pH 8.0


Note that the metal –EDTA complex becomes less
stable as pH decreases, K’f decreases
In order to get a “complete” titration (K’f ≥106), EDTA
requires a certain minimum pH for the titration of
each metal ion.
By adjusting the pH of an EDTA titration: metal ion
(e.g. Fe3+) can be titrated without interference from
others (e.g. Ca2+)
Minimum pH for Effective
Titration of Metal Ions
62

COMPLEXOMETRIC EDTA
TITRATION CURVES
 Complexation titration curve shows the change in pM, where M is the metal ion,
as a function of the volume of EDTA
Calculating the Titration Curve
Step 1: Calculate the conditional formation constant for the metal–EDTA
complex.
Step 2: Calculate the volume of EDTA needed to reach the equivalence point.
Step 3: Calculate pM values before the equivalence point by determining the
concentration of unreacted metal ions.
Step 4: Calculate pM at the equivalence point using the conditional formation
constant.
63
Step 5: Calculate pM after the equivalence point using the conditional
formation constant.
COMPLEXOMETRIC EDTA
TITRATION CURVES
Let’s calculate the titration curve for 50.0 mL of 5.00×10–3 M Cd2+ using a
titrant of 0.0100 M EDTA at pH 10.0 (assume Kf 2.9x1015 and aY4-(pH 10) is 0.3)
Step 1: Calculate the conditional formation constant for the metal–EDTA complex.
K’f = Kf x aY4- = 2.9x1015 x 0.3 = 8.7x1014
Step 2: Calculate the volume of EDTA needed to reach the equivalence point.
at the equivalence point
moles EDTA = moles Cd2+
MEDTA×VEDTA = MCd×VCd
⇨ Veq = VEDTA = MCdVCd / MEDTA
64
= (5.00×10−3 M)(50.0 mL) / (0.0100 M) = 25.0 mL
COMPLEXOMETRIC EDTA
TITRATION CURVES
Step 3: Calculate pM values before the equivalence point by determining the
concentration of unreacted metal ions.
Before the equivalence point, Cd2+ is present in excess and pCd is determined by the
concentration of unreacted Cd2+.
For example, after adding 5.0 mL of EDTA, the total concentration of Cd2+ is
CCd = (initial moles Cd2+ − moles EDTA added) / total volume
= (MCdVCd − MEDTAVEDTA)/ (VCd + VEDTA)
= ((5.00×10−3 M)(50.0 mL) − (0.0100 M)(5.0 mL)) / (50.0mL + 5.0mL)
= 3.64×10−3 M
65
pCd = -Log(3.64×10−3) = 2.44
COMPLEXOMETRIC EDTA
TITRATION CURVES
Step 4: Calculate pM at the equivalence point using the conditional formation
constant.
at equivalence point all Cd2+ and EDTA converted to CdY2-,
[CdY2−] = initial moles Cd2+ / total volume = MCdVCd / (VCd + VEDTA)
= (5.00×10−3 M)(50.0 mL) / (50.0 mL + 25.0 mL) = 3.33×10−3 M
(volume of equivalence was calculated and found to be 25.0 mL)
Now calculate Ccd
CdY2−
3.33×10−3 -x
⇌
Cd2+ +
x
Y4−
x
K’f = 8.7x1014 = [CdY−2] / CCdCEDTA = (3.33×10−3 −x)/ (x.x)
⇨ CCd = x =√(3.33×10−3 / 8.7x1014) = 1.95x10-9
66
pCd = -Log(1.95x10-9)= 8.7
COMPLEXOMETRIC EDTA
TITRATION CURVES
Step 5: Calculate pM after the equivalence point using the conditional formation constant.
At this stage CdY2− formed and excess of EDTA present, their concentrations
should be calculated. For example when 30.0 mL of EDTA added:
[CdY2−] = initial moles Cd2+ / total volume = MCdVCd / (VCd + VEDTA)
= (5.00×10−3 M)(50.0 mL) / (50.0 mL + 30.0 mL) = 3.125×10−3 M
[EDTA] = (moles of EDTA added-initial moles Cd2+) / total volume
= MEDTAVEDTA -MCdVCd / (VCd + VEDTA)
= ((0.0100 M)(30.0 mL) − (5.00×10−3 M)(50.0mL)) / (50.0 mL + 30.0 mL)
= 6.25×10−4 M
Now calculate Ccd
CdY2−
3.125×10−3 -x
⇌
Cd2+
x
+
Y4−
6.25×10−4 +x
K’f = 8.7x1014 = [CdY−2] / CCdCEDTA = (3.125×10−3 −x)/ (x)(6.25×10−4 +x)
⇨ CCd = x =3.125×10−3 / (8.7x1014 x 6.25×10−4) = 5.75x10-15
67
pCd = -Log(5.75x10-15)= 14.2
68
COMPLEXOMETRIC EDTA
TITRATION CURVES
DETERMINATION OF
EDTA TITRATION END
POINT

Metal Ion Indicator: a compound that changes color when it binds to a metal ion
- Similar to pH indicator, which changes color with pH or as the compound binds H+

For an EDTA titration, the indicator must bind the metal ion less strongly than EDTA
- Needs to release metal ion to EDTA
69
End Point indicated by a color
change from red to blue
DETERMINATION OF
EDTA TITRATION END
POINT
Common Metal Ion Indicators
70

COMPLEXOMETRIC
EDTA TITRATION
Example: The concentration of a solution of EDTA was determined by standardizing
against a solution of Ca2+ prepared using a primary standard of CaCO3. A
0.4071-g sample of CaCO3 was transferred to a 500-mL volumetric flask,
dissolved using a minimum of 6 M HCl, and diluted to volume. After
transferring a 50.00-mL portion of this solution to a 250-mL Erlenmeyer flask,
the pH was adjusted by adding 5 mL of a pH 10 NH3–NH4Cl buffer. After
adding calmagite as an indicator, the solution was titrated with the EDTA,
requiring 42.63 mL to reach the end point. Report the molar concentration of
EDTA in the titrant. (M.wt. of CaCO3 =100.09 g/mol)
Solution:
The primary standard of Ca2+ has a concentration of
[0.4071 g CaCO3 / 100.09 g CaCO3] / (0.5000 L) = 8.135×10−3 M Ca2+
The moles of Ca2+ in the titrant is
8.135×10−3 M Ca2+ × 0.05000 L Ca2+ = 4.068×10−4 mol Ca2+
which means that 4.068×10–4 moles of EDTA are used in the titration.
The molarity of EDTA in the titrant is
71
4.068×10−4 mol EDTA / 0.04263 L EDTA = 9.543×10−3 M EDTA