Precipitation Titrations

Dr. Riham Ali HazzaaAnalytical chemistry

Petrochemical Engineering

Precipitation titration:

It is a titration in which the reaction between the analyte and titrant involves a precipitation.

SOLUBILITY RULES1. Most nitrates are soluble2. Most salts with Grp 1A ions and NH4

+ are soluble.

3. Most salts with Cl-, Br-, I- are soluble EXCEPT those with Ag+, Pb2+, Hg2

2+

4. Most sulfates are soluble EXCEPT those with Ba2+, Pb2+, Hg2

2+, Ca2+.5. Most hydroxides are slightly soluble EXCEPT

the strong bases.6. Most sulfides, carbonates, chromates and

phosphates are slightly soluble.

Solubility Equilibria: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 Solubility product Ksp AgCl 1.8×10 -10

Ag Br 5×10 -13

Ag I 8.3×10 -17

Examples:Calculate the solubility of AgCl. Ksp= 1.8 10-10

Ksp 10108.1

AgCl Ag+ + Cl-

[Ag+] = [Cl-] = SKsp = [Ag+][Cl-] = S2

S = =

Molar solubility S = 1 10-5M

Calculate the solubility of Ag2CrO4. Ksp = 1.2 10-12

Ag2CrO4(s) 2Ag+(aq) + CrO42-(aq)

Ksp =

[

]

Ksp = = 4S3

Molar solubility S of Ag2CrO4 = 53

12

3 107.64

102.1

4

Ksp

Common-Ion Effect and Solubility

Consider PbI2 in the presence of KI.

PbI2(s) → Pb2+(aq) + 2I-(aq)

The solubility of a slightly soluble solute is decreased in the presence of a common ion.

[I-(aq)] is greater[Pb2+(aq)] decreasesPbI2 ppt out

What is the solubility of PbI2 in (a) water (b) 0.20 M KI? Ksp= 7.110-9

PbI2(s) Pb2+ + 2I-.Ksp = (S) (2S)2 = 4S3

a) Solubility =S= [Pb2+] = 3

9

3

4

10*9.7

4

Ksp

= 1.3 10-3M

b) (Pb2+) = 2

9

2 )2.0(

109.7

I

KsP = 2×10-7M

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-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 and Reduction

OXIDATION-REDUCTION REACTIONS

• A redox reaction involves the transfer of electrons between reactants

• Electrons gained by one species must equal electrons lost by another

• Both oxidation and reduction must occur simultaneously.

2

Loss of electrons = Oxidation

Gain of electrons = Reduction

OIL-RIGOxidation 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 reactionsMnO4

- + 8H+ + 5e → Mn2+ + 4H2O

Fe2+ → Fe3+ + 1e5Fe2+ → 5Fe3+ + 5e

Now add the reduction and the oxidation half equationsMnO4

- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+

This represents the redox process

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 -16. The sum of the oxidation numbers for the molecule is

the charge on the molecule (zero for a neutral molecule).

Oxidation Numbers

• Calculate the oxidation number of sulphur in sulphuric acid H2SO4

Hydrogen = +1 oxidation numberOxygen = -2 oxidation number

(2 x H) + S + (4 x O) = 02 + 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 acceptor5Fe2+ + 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 reaction2 Na(s) + Cl (g) →2 NaCl(s)

• decomposition reaction2HgO(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+ + 2eAt the cathode: Cu2+ + 2e → Cu