Precipitation Titrations
Dr. Riham Ali Hazzaa
Analytical chemistry
Petrochemical Engineering
Precipitation titration:
It is a titration in which the reaction
between the analyte and titrant
involves a precipitation.
SOLUBILITY RULES
1. Most nitrates are soluble
2. Most salts with Grp 1A ions and NH4+ are
soluble.
3. Most salts with Cl-, Br-, I- are soluble EXCEPT
those with Ag+, Pb2+, Hg22+
4. Most sulfates are soluble EXCEPT those with
Ba2+, Pb2+, Hg22+, Ca2+.
5. Most hydroxides are slightly soluble EXCEPT
the strong bases.
6. Most sulfides, carbonates, chromates and
phosphates are slightly soluble.
Solubility Equilibria:
Equilibrium exists between an undissolved
solute and its saturated solution when the rate
of precipitation equals the rate of dissolution.
An aqueous solution containing aqueous NaCl
and solid NaCl contains the following
equilibrium.
NaCl(s) → Na+(aq) + Cl-(aq)
Ksp = [Na+][Cl-]
This equilibrium is described by the
equilibrium constant, Ksp(solubility product).
Solubility (S)
It is defined as the concentration of a dissolved
solute at equilibrium with its undissolved form.
The solubility can be calculated from the
solubility product Ksp.
The solubility product (Ksp)
It is the equilibrium constant for a chemical
reaction in which a solid ionic compound dissolves
to yield its ions in solution.
Solubility products of some compounds
Compound
AgCl
Ag Br
Ag I
Solubility product Ksp
1.8×10 -10
5×10 -13
8.3×10 -17
Examples:
Calculate the solubility of AgCl. Ksp= 1.8  10-10
AgCl  Ag+ + Cl[Ag+] = [Cl-] = S
Ksp = [Ag+][Cl-] = S2
S =
Ksp
=
1.8 10 10
Molar solubility S = 1  10-5M
Calculate the solubility of Ag2CrO4. Ksp = 1.2  10-12
[
]
Ag2CrO4(s)  2Ag+(aq) + CrO42-(aq)
Ksp =
Ksp =
Molar solubility S of Ag2CrO4 =
= 4S3
3
Ksp 3 1.2 1012

 6.7 105
4
4
Common-Ion Effect and Solubility
The solubility of a slightly soluble solute is decreased
in the presence of a common ion.
Consider PbI2 in the presence of KI.
PbI2(s)
PbI2 ppt out
→
Pb2+(aq) +
[Pb2+(aq)] decreases
2I-(aq)
[I-(aq)] is greater
What is the solubility of PbI2 in (a) water
Ksp= 7.110-9
(b) 0.20 M KI?
PbI2(s)  Pb2+ + 2I-.
Ksp = (S) (2S)2 = 4S3
9
Ksp
7
.
9
*
10
-3M
3
3
=
1.3

10

2+
a) Solubility =S= [Pb ] = 4
4
b)
(Pb2+)
=
KsP
I 
 2
7.9  10 9 = 2×10-7M

(0.2) 2
The solubility has decreased upon the addition
of an excess of I-
Precipitation Titration Methods
• Argentometric precipitation titrations
Mohr method for determining chloride: (Direct
titration)
Titration reaction:
AgNO3 + NaCl → AgCl ↓ +NaNO3
Indicator reaction:
2 AgNO3 + Na2CrO4 →Ag2CrO4(s)↓+2NaNO3
Yellow
red ppt
• Volhard method (Indirect or back titration method )
• A measured excess of AgNO3 is added to precipitate the
anion CL-, Br-, I-, and the excess of Ag+ is determined by
back titration with standard potassium thiocyanate
solution:
Ag+ (aq) + Cl– (aq) → AgCl(s)↓ + excess Ag+
Excess :
AgNO3 (aq) + KSCN (aq) → AgSCN(s) ↓+KNO3(aq)
(soluble red complex)
Oxidation and reduction titration
• Oxidation-reduction titration is a type of
titration based on a redox reaction between
the analyte and titrant.
Redox titration may involve the use of a redox
indicator and/or a potentiometer.
Oxidation and Reduction
Oxidation-Reduction (Redox) reactions involve
transfer of e between reactants to form different
products.
Electrons must be balanced, so if oxidation takes place,
reduction must also.
OXIDATION-REDUCTION REACTIONS
• A redox reaction involves the transfer of
electrons between reactants
• Electrons gained by
one
species
must
2
equal electrons lost by another
• Both oxidation and reduction must occur
simultaneously.
Loss of electrons = Oxidation
Gain of electrons = Reduction
OIL-RIG
Oxidation Is Loss
Reduction Is Gain
• Oxidation removal of electrons
Mg(s) →Mg2+ + 2e
the reagent causing the loss of electrons is
called the oxidising agent
• Reduction gain of electrons
Fe3+ + 3e→ Fe(s)
the species donating the electrons is called the
reducing agent
Redox reactions
MnO4- + 8H+ + 5e → Mn2+ + 4H2O
Fe2+ → Fe3+ + 1e
5Fe2+ → 5Fe3+ + 5e
Now add the reduction and the oxidation half
equations
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
This represents the redox process
Oxidation Numbers
1. Atoms in elemental form, oxidation number is zero.
(Cl2, H2, P4, Ne are all zero)
2. Monoatomic ion, the oxidation number is the charge
on the ion.
(Na+: +1; Al3+: +3; Cl-: -1)
3. Oxidation number of O is usually -2. But in peroxides
(like H2O2 and Na2O2) it has an oxidation number of -1.
4. Oxidation number of H is +1 when bonded to
nonmetals and -1 when bonded to metals.
(+1 in H2O, NH3 and CH4; -1 in NaH, CaH2 and AlH3)
5. The oxidation number of F is -1
6. The sum of the oxidation numbers for the molecule is
the charge on the molecule (zero for a neutral
molecule).
• Calculate the oxidation number of sulphur in
sulphuric acid H2SO4
Hydrogen = +1 oxidation number
Oxygen = -2 oxidation number
(2 x H) + S + (4 x O) = 0
2 + S -8 = 0,
S=6
OXIDATION
• If atom X in compound A loses electrons and
becomes more positive (OX# increases), we
say X (with charge) or A is oxidized.
Fe2+ → Fe3+ + 1e
Also, we say that A is the reducing agent (RA)
or is the electron donor.
5Fe2+ + MnO4- + 8H+ →5Fe3+ + Mn2+ + 4H2O
REDUCTION
• If atom Y in compound B gains electrons and
becomes more negative (OX# decreases), we
say Y (with charge) or B is reduced.
MnO4- + 8H+ + 5e→ Mn2+ + 4H2O
• Also, we say that B is the oxidizing agent (OA)
or is the electron acceptor
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
• Define oxidising and reducing agents
• An oxidizing agent is an element which causes
oxidation (and is reduced as a result) by
removing electrons from another species.
• A reducing agent is an element which causes
reduction (and is oxidized as a result) by giving
electrons to another species.
Most of the oxidation-reduction reactions fall into
one of the following simple categories:
• combination reaction
2 Na(s) + Cl (g) →2 NaCl(s)
• decomposition reaction
2HgO(s) → 2Hg (l) + O2 (g)
• displacement reaction
Zn (s) + 2HCl (aq) → ZnCl2 (aq) + H2(g)
• combustion reaction
4Fe(s) + 3O2 (g) → 2Fe2O3(s)
Redox Indicators
• Standard oxidizing agents
potassium permanganate KMNO4,
potassium dichromate K2Cr2O7,
iodine I2
• Standard reducing agents
Sodium thiosulfate, Na2S2O3
Fe+2
Reactivity series
• It is possible to organize a group of similar
chemicals that undergo either oxidation or
reduction according to their relative
reactivity.
• Oxidation (and reduction) is a competition for
electrons. The oxidising species (agents)
remove electrons from other species and can
force them to become reducing agents
(releasers of electrons)
Reactivity series
Example
• The zinc metal is more reactive than copper
metal and so it can force the copper metal
ions to accept electrons and become metal
atoms.
Zn(s)→ Zn2+(aq) + 2e
Cu2+(aq) + 2e→ Cu(s)
• placing a strip of zinc metal in a copper
sulfate solution will produce metallic
copper and zinc sulfate
• Copper displaces silver from a solution of
silver nitrate
Molecular equation:
2AgNO3 + Cu(s)→ 2Ag(s) + Cu (NO3)2
ionic equation:
2Ag+ (aq) + 2NO3- (aq) + Cu(s) →
2Ag(s) + Cu2+ (aq) + 2NO3- (aq)
Net ionic equation:
2Ag+ (aq) + Cu(s) → 2Ag(s) + Cu2+ (aq)
Electricity from chemical reactions
• Galvanic cells: chemical energy converts to
electrical energy
• In this cell the Zinc anode dissolves and releases
electrons which pass around the external wires to
the Copper electrode where they are given to the
Copper ions (2+) which are then deposited as Copper
atoms on the electrode.
At the anode: Zn → Zn2+ + 2e
At the cathode: Cu2+ + 2e → Cu