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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.
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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.
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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).
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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
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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)
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Net +ve Charge on Colloidal Particle because of Ag+ Adsorbed
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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.
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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.
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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.
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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)
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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.
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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.
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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.
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Mass of Mg 2+ = 0.0471 g
% Mg =Mass of Mg2+
sample Mass(100) = 0.0474
0.352(100)
= 13.45 %
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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.
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C, H, N, and S Analyser: Modern Technique
Thermal Conductivity, IR,or Coulometry for Measuring products.
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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
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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
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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
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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
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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 %
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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
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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
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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+
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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
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[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,
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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.
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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
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[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
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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+
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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
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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
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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.
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Titration Curve for 40.00 mL of 0.05000 M KI and 0.05000 M KCl with 0.084 M AgNO3.
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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.
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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.
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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?
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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
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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-.