Transcript Slide 1

Precipitation
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
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)
Net +ve Charge on Colloidal Particle because of Ag+
Adsorbed
Digestion and Purity
Digestion: Period of standing in hot mother liquor.
Promotion of recrystallisation
Crystal 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
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 1100 oC for 1
hour 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.
Sample size usually 2 mg 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
21.999m gCO2
 x   y 
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
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-/Clmixture than for pure I-.
1. Random experimental error: both +tive and –tive.
2. Coprecipitation: +ve error
Example: Some Cl- attached to AgI 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-.