PrecipitationPrecipitation

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.

Crystallization1. Nucleation2. Particle

Growth

Molecules form smallAggregates randomly

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

C

O

H2N NH2

+ 3H2OHeat

CO2+ 2NH4

+ + 2OH-

pH gradually increasesC

O

H OH

+ OH- HCO2- + H2O

Formate

Formic Acid

3HCO2- + Fe3+ Fe(HCO2)3

.nH2O(s)

Fe(III)formate

Large particle size

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 chargedAdsorption 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)

Net +ve Charge on Colloidal Particle because of Ag+ Adsorbed

Digestion and Purity

Digestion: Period of standing in hot mother liquor.

Promotion of recrystallisationCrystal particle size increase 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 CaR2(s) + 2H+ Analyte Precipitate

Mn2+ + 6CN- Mn(CN)64-

Impurity Masking agent Stays in solution

Postprecipitation: Collection of impurities on ppt during digestion: a supersaturated impuritye.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 CompositionProduct 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

(1 Hour)Fe2O3 + CO2(g) + xH2O(g)

2Mg(NH4)PO4.6H2O Mg2P2O7 + 2NH3 + 13H2O

1100oC

(1 Hour)

Thermogravimetric AnalysisThermogravimetric Analysis

CO2CaO2C

H2O 200oC

OH OH

CO2CaO2C

OH OH

O

Ca

O

O

300oC

500oCCaCO3

700oCCaO

Calciumcarbonate

Calciumoxide

Calcium salicylate monohydrate

Heating a substance and measuring its mass as a function of temperature.

2Mg(NH4)PO4.6H2O Mg2P2O7 + 2NH3 + 13H2O

1100oC

(1 Hour)

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 1100 oC for 1hour and weighed as Mg2P2O7. The mass of Mg2P2O7 is 0.2168 g.

Calculate the percentage of magnesium in the sample.

Solution:

The gravimetric factor is:

Grams of Mg in analyte

Grams of Mg2P2O7=

2 x (24.305)

222.553

Relative atomic massOf Mg

FM of Mg2P2O7

Grams of Mg in analyte = Grams of Mg2P2O7 formed2 x (24.3050)

222.553

553.222

)3050.24(22168.0

xg

Note: 2 mol Mg2+ in 1 mol Mg2P2O7.

Mass of Mg 2+ = 0.0471 g

% Mg =Mass of Mg2+

sample Mass(100) = 0.0474

0.352(100)

= 13.45 %

Combustion Analysis

Determination of the Carbon and Hydrogen content of organic compounds burned in excess oxygen.

H2O absorption

CO2 AbsorptionPrevention of entrance of

atmospheric O2 and CO2.

Note: Mass increasein each tube.

C, H, N, and S Analyser: Modern Technique

Thermal Conductivity, IR,or Coulometry for Measuring products.

Sample size usually 2 mg in tin or silver capsule.

Capsule melts and sample is oxidised in excess of O2.

C, H,N, S1050 oC; O2 CO2(g) + H2O(g) + N2(g) + SO2(g) + SO3(g)

(95 % SO2)

Products Hot WO3 catalyst: CarbonHeat

CrO3 Cat. CO2

Then, metallic Cu at 850 oC:

Cu + SO3850 oC

SO2 + CuO(s)

2Cu + O2850 oC

2CuO(s)

Dynamic Flash combustion: Short burst of gaseous products

Oxygen Analysis:

Pyrolysis or thermal decomposition in absence of oxygen.

Gaseous products: Nickelised Carbon1075 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 MassSpectrometry

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 + 15/2O2

FW = 122.123

7CO2 + 3H2O44.010 18.015

4.635 mg of C6H5CO2H = mmolmmolmg

mg03795.0

/123.122

634.4

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)

mmolmg

mgyx

/010.44

999.21

053.442

161.846

mmolmg

mgyx

/010.44

999.21

053.442

161.846

X + y = 7.290 mg x = 7.290 - y

CO2

mmolmg

mgyy

/010.44

999.21

053.442

161.84

290.76

y = mass of C2H4O = 0.767 mg

Therefore, % Weight Oxirane = )100(294.7

767.0

mg

mg

= 10.52 %

The Precipitation Titration CurveThe 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 CurveTitration CurveA 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:

Before the Equivalence Point:

Addition of 20 mL of Ag+:

This reaction: I- + Ag+ AgI(s) goes to completion.

Ve = 0.05000 L = 50.00 mL

(0.02500 L)(0.1000 mol I-/L) (Ve)(0.05000 mol Ag+/L)=

mol I- mol Ag+

I

KAg sp [I-] due to I- not precipitated by

20.00 mL of Ag+.

Fraction of I- reacted: )00.50(

)00.20(

mL

mL

Fraction of I- remaining: )00.50(

)00.30(

mL

mL

Some AgI redissolves: AgI(s) I- + Ag+

Therefore,

mL

mLM

mL

mLI

00.45

00.25)1000.0(

00.50

00.30][

FractionRemaining

OriginalConc.

DilutionFactor

Original volumeof I-

Totalvolume

[I-] = 3.33 x 10-2 M

I

KAg sp

2

17

1033.3

103.8

x

xAg

[Ag+] = 2.49 x 10-15 M

pAg+ = -log[Ag+] = 14.60

The Equivalence PointThe Equivalence Point::

All AgI is precipitated

AgI(s) I- + Ag+Then,

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 PointAfter the Equivalence Point:

[Ag+] is in excess after the equivalence point.

Note: Ve = 50.00 mL

Suppose that 52.00 mL is added:

Therefore, 2.00 mL excess Ag+

mL

mLMAg

00.77

00.2)05000.0(][

Original Ag+

Concentration Dilution Factor

Volume of excessAg+

Total volumeof solution

[Ag+] = 1.30 x 10-3 M

pAg+ = -log[Ag+] = 2.89

Shape of the Titration Curve:

Steepest slope:dx

dyhas maximum value

Equivalence point: point of maximum slope

Inflection point: 02

2

dx

yd

Titration Curves: Effect of Diluting the reactantsTitration 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 reactantsTitrations 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 KEffect of Kspsp on the Titration Curve on the Titration Curve

AgI is least solubleSharpest change at

equiv. point

Least sharp, but steep enough for Equiv.

point location

K = 1/Ksp largest

Titration of a MixtureTitration 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 increases 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-/Cl- mixture than for pure I-.

1. Random experimental error: both +tive and –tive.

2. Coprecipitation: +ve error

High nitrate concentration to minimise coprecipitation.

Example: Some Cl- attached to AgI ppt and carries down an equivalent amount of Ag+.

NO3- competes with Cl- for binding sites.

Coprecipitation error lowers the calculated concentration of the second precipitated halide.

Separation of Cations by PrecipitationSeparation of Cations by Precipitation

Consider a solution of Pb2+ and Hg22+: Each is 0.01 M

PbI2(s) Pb⇌ 2+ + 2I-

Hg2I2(s) Hg⇌ 22+ + 2I-

Ksp = 7.9 x 10 -9

Ksp = 1.1 x 10 -28

Smaller Ksp

ConsiderablyLess 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) Hg⇌ 22+ + 2I-

Initial Concentration: 0 0.010 0 Final Concentration: solid 1.0 x 10-6 x

spKIHg 222 ]][[ (1.0 x 10-6)(x)2 = 1.1 x 10-28

X = [IX = [I--] = 1.0 x 10 ] = 1.0 x 10 –11–11 M M

Will this [I [I--] = 1.0 x 10 ] = 1.0 x 10 –11–11 M M precipitate 0.010 M Pb2+?

21122 )100.1)(010.0(]][[ xIPbQ

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-.