C3100 - Winter 2011
V. GRAVIMETRIC AND COMBUSTION ANALYSIS
GRAVIMETRIC ANALYSIS
(Chapter 26)
Gravimetric analysis = any analysis that uses the mass of a product
(precipitate) to calculate the quantity of the original analyte
The precipitant (precipitating agent) can be both inorganic and organic
compounds:
e.g. Ag+ for Cl-, to form AgCl
SO42- for Ba2+ to form BaSO4
Dimethylglyoxime (HDMG) for Ni2+ to form Ni(DMG)2
The conditions under which precipitation occurs have to be carefully
controlled, and interfering species may need to be removed before analysis
Very accurate measurements possible, but tedious (low throughput)
Desired characteristics of a precipitate for gravimetric analysis:
• Pure (no co-precipitates) – known composition
• Very low solubility (cool solution!) – small KSP
• Easy to filter; colloids and very fine particles clog or pass through the filter
• High molar mass: this reduces the relative error in the mass weighed
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Dr. G. Van Biesen
C3100 - Winter 2011
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Precipitation and Crystal Growth
Precipitation starts when the solubility of the compound has been exceeded
(solution is supersaturated). The excess solute will precipitate until its
concentration equals its solubility (temperature dependent). The supernatant
is the solution left after the precipitate has formed.
Two phases of crystal growth:
¾ Nucleation: molecules in solution come together randomly to form small
aggregates (nuclei)
¾ Particle growth: nuclei grow in size by adding more molecules to form
crystals, or by collision and combination of nuclei
In gravimetric analysis, we want to allow particles to grow to sizable dimensions
(several μm) so they filter easily. Colloids (1 – ~500 nm diameter) have to be
avoided. Nucleation dominates in highly supersaturated solutions, and this can
be avoided by:
• Raising the temperature: precipitate from a hot solution
(increases solubility – decreases supersaturation)
• Adding precipitant slowly, with vigorous mixing to prevent local
supersaturation
• Using low concentrations of analyte and precipitant
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Dr. G. Van Biesen
C3100 - Winter 2011
• Often, precipitates are more soluble at low pH. In that case, precipitate
near the low end of the pH where precipitation is still quantitative
Another reason why small particles and colloids in gravimetric analysis have to
be avoided, is that they are much worse for co-precipitation of impurities
(because of the larger surface area).
Co-precipitation can be under the form of:
¾ adsorption: bound to the surface of the crystal
¾ absorption: inside of the crystal:
Inclusions: Ions which replace analyte ions in the crystal lattice. The
probability of this happening is greater if the ion has the same
charge and is of similar size as the analyte ion (e.g. K+ for NH4+ in
NH4MgPO4; Ba2+ for Pb2+ in PbSO4)
Occlusions: Pockets of impurity trapped within the growing crystal.
Occlusions are more likely during rapid growth of the crystal. They are
decreased by using a digestion period (see further).
Note: an example where we do want a precipitate with a large surface area
is when using an adsorption indicator in a titration (e.g. Fajans)
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Precipitation in the presence of electrolytes
Particles in solution are almost always charged due to adsorption of ions at
their surface
e.g. a colloidal AgCl particle forming in 0.1 M HNO3 (excess Ag+)
Surface of AgCl has pos. charge because of
adsorption of Ag+ (in excess relative to Cl-).
Adsorption of Ag+ preferred over H+
because of better size compatibility.
Primary layer
Region of solution directly surrounding
the particle = Ionic atmosphere – has a
net neg. charge
Particle + ionic atmosphere =
electric double layer – moves
as a unit through solution
In order for colloidal particles to coalesce and grow, they have to collide.
However, their ionic atmospheres repel one another.
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Dr. G. Van Biesen
C3100 - Winter 2011
Common solutions to this problem:
¾Increase their kinetic energy by heating (collisions are more energetic so that
electrostatic repulsion is overcome)
¾Increase electrolyte conc., which decreases the volume (thickness) of the ionic
atmosphere and allows particles to come closer together before electrostatic
repulsion becomes significant
After precipitation, most procedures call for a digestion period; the precipitate is
left in the presence of the (hot) mother liquor. This allows for slow recrystallization
and overall increase in particle size (because small particles have a higher surface
energy, they dissolve easier than larger ones, and are precipitated on the larger
particles), and expelling of impurities
Washing of precipitates:
After collection of the precipitate on the filter, it has to be washed to remove any
droplets of the mother liquor. This is done by breaking the suction first, and stirring
the precipitate in a minimum amount of water, or better: electrolyte solution. Water
washes away the ionic atmosphere of the crystals (electrical double layer expands,
increasing repulsion by the primary layer) and the product breaks up = peptization.
The electrolyte used for washing has to be volatile, so it will be lost during drying of
the product (e.g. dilute HNO3 and HCl, NH4Cl).
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In many procedures, after the mother liquor is washed away, the precipitate
is redissolved, and then reprecipitated. During the 2nd precipitation, the
concentration of impurities is lower than during the 1st precipitation, and the
precipitate will be more pure.
Sometimes a masking agent is added prior to precipitation to prevent
precipitation of impurities. For instance, in the gravimetric analysis of Ba2+
with N-p-chlorophenylcinnamohydroxamic acid, excess KCN is added to
keep Ag+, Mn2+, Zn2+ etc. in solution.
Product Composition
Final product must have a stable and known composition
Many precipitates contain a variable quantity of water and must be dried to
remove all water or give a known stoichiometry of H2O :
• hygroscopic water, and inclusion water: can often (but not always) be
removed by drying at 110-130 °C
• water of crystallization (e.g. in NiCl2(H2O)6) requires higher temperatures
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Dr. G. Van Biesen
C3100 - Winter 2011
Ignition (strong heating) is often used to change the chemical form of
some precipitates
e.g. igniting Fe(HCO2)3.nH2O at 850 °C gives Fe2O3
Thermogravimetric analysis can be used to find out how the composition
of a precipitate changes as it is heated, and what temperature is needed
to remove all water
Fig. 26.4, Harris 8th Ed.
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Some gravimetric problems:
Problem 1
A 11.1324 g sample of iron ore was digested in concentrated HNO3. The
resulting solution was diluted with water to 250.00 mL. Three 50.00 mL portions
were taken and the iron(III) was precipitated as Fe2O3•xH2O by the addition of
NH3(aq). After filtration and washing, the precipitate from each of the three
samples was ignited at high temperature to give an average mass of 0.5394 g
of pure Fe2O3.
Calculate the % Fe in the ore sample.
Problem 2: System with 2 unknowns
A 0.4000 g sample containing only NaCl and BaCl2 yielded 0.8415 g of
dried AgCl. Calculate the mass percent of each compound in the sample.
Problem 3: Effect of washing on accuracy
Calculate the percentage loss of Ni(DMG)2 when 250.0 mL of water is used to
wash precipitates of a) 0.2578g and b) 0.08634g. Assume that the wash water
becomes saturated with Ni(DMG)2 and that the solubility of Ni(DMG)2 in water
is 3.5 x 10-4 g L−1
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Dr. G. Van Biesen
C3100 - Winter 2011
Solution to Problem 1
0.5394g
= 3.377 8 × 10 − 3 mol
159.69 g mol -1
2 mol Fe
= 3.377 8 × 10 −3 mol Fe 2O 3 ×
= 6.755 6 × 10 − 3 mol Fe
1 mol Fe 2O 3
nFe 2O 3 =
nFe
mass Fe = 6.755 6 × 10 −3 mol Fe × 55.847 g mol -1
mass Fe = 0.3772 8 g Fe (in 50 mL)
mass Fe in 250 mL = 0.3772 8 g x
250.00 mL
= 1.886 4 g
50.00 mL
⎡ 1.8864 g Fe ⎤
%Fe = ⎢
⎥ × 100% = 16.95%
⎣ 11.1324 g ore ⎦
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Solution to Problem 2
Let x = mass of NaCl
moles of NaCl =
0.4000 – x = mass of BaCl2
moles of BaCl2 =
Moles of AgCl =
0.8415 (g)
143.32 (g/mol)
x (g)
58.442 (g/mol)
0.4000 (g) – x (g)
208.23 (g/mol)
= 0.005871 mol AgCl = mol Cl-
Each mol of NaCl contributes 1 mol of Cl-, each mol of BaCl2 contributes 2 mol of Cl0.005871 mol =
x (g)
+2
58.442 (g/mol)
0.4000 (g) – x (g)
208.23 (g/mol)
Mass of NaCl x = 0.2704 g
Mass of BaCl2 = 0.4000 g – 0.2704 g = 0.1296 g
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Dr. G. Van Biesen
C3100 - Winter 2011
Solution to Problem 3
loss = solubility x volume wash water
loss = 3.5 x 10-4 g L−1 x 0.250 L x 1000mg/1g
loss = 0.086 mg (very low)
a)
%loss = 0.033%
b)
%loss = 0.10%
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Precipitation Titrations
•
Gravimetric analysis: add excess of precipitating reagent to analyte in
solution; weigh mass of precipitate, find concentration of analyte
•
Precipitation titration: add precipitating reagent (= titrant of known conc.)
incrementally to analyte solution, monitor either [analyte] or [titrant] to
locate the end point of the titration
•
Examples:
determination of Cl-, Br-, I- and SCN-, with Ag+ as titrant
determination of SO42-, with Ba2+ as titrant
determination of PO43- and C2O42- with Pb2+
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Dr. G. Van Biesen
C3100 - Winter 2011
Precipitation titrations: calculating concentrations
Calculate pAg+ and pBr- for the addition of 10.00 mL, 24.90 mL, 25.00 mL, and
25.10 mL of 0.1000 M Ag+ to 50.00 mL of 0.05000 M Br-.
Ksp AgBr = 5.0 x 10-13
Titration reaction:
Ag+ + Br⎯
¾Calculate Ve:
AgBr(s)
#moles Br- = #moles Ag+
50.00 mL • 0.05000 M = Ve • 0.1000 M
Ve = 25.00 mL
¾Before EP:
•10.00 mL Ag+ added:
Ve - V
[Br-] = CA
Ve
Vi
Vi + V
25.00 mL – 10.00 mL
= 0.05000 M
25.00 mL
50.00 mL
50.00 mL + 10 ml
= 0.02500 M
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pBr- = 1.60
The concentration of Ag+ in equilibrium with this much Br- is:
[Ag+] =
Ksp
[Br-]
5.0 x 10-13
=
0.02500
= 2 x 10-11 M pAg+ = 10.70
•24.90 mL Ag+ added:
Using the same approach, we find pBr- = 3.87 and pAg+ = 8.43
Note that [Ag+] is still negligible compared to [Br-], even though the
reaction is 99.6% complete
•25.00 mL Ag+ added: at the EP: exactly enough Ag+ is added to react
with Br-; the AgBr redissolves to give equal concentrations of Ag+ and Braccording to the Ksp:
AgBr
x dissolves
Ag+ + Brx
x
[Ag+][Br-] = Ksp
x2 = Ksp
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Dr. G. Van Biesen
C3100 - Winter 2011
x = K sp = 5.0x10-13
x = 7.1 x 10-7 M = [Ag+] = [Br-]
pAg+ and pBr− = 6.20
Note that these values are independent of the original concentrations or
volumes. At the EP, the problem of finding [Ag+] and [Br-] is the same as
for dissolving pure AgBr in water.
•25.10 mL Ag+ added: post-EP
[Ag+] is determined by Ag+ added after EP, since virtually all Ag+ added before
EP has precipitated. [Ag+] is determined using the excess volume of Ag+ (with
original conc. [Ag+]0) , and taking into account the dilution :
[Ag+] =
(V – Ve) • [Ag+]0
Vi + V
= 1.33 x 10-4 M
[Br-] =
Ksp
[Ag+]
=
(25.10 mL – 25.00 mL) (0.1000 M)
50.00 mL + 25.10 mL
This is essentially
a simple dilution!
pAg+ = 3.88
= 3.76 x 10-9 M
pBr- = 8.42
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Factors that influence the shape of the titration curve:
¾Concentration:
Titration of I- (0.1, 0.01, 0.001 M) with Ag+
(0.05 M, 0.005 M and 0.0005 M resp.)
The larger the change in p function,
the sharper the endpoint
Before Ve: [Ag+] =
Ksp
[I-]
If [I-] ↓, then [Ag+] ↑ and pAg+ ↓
After Ve: excess [Ag+], so with [Ag+] ↓,
pAg+ ↑
Fig. 26-8, Harris 8th Ed.
Note that only when the stoichiometry of the reaction is 1:1, the EP is the
steepest (middle) part of the curve. Otherwise, the EP is not an inflection point.
Usually, conditions are such that the inflection point is a good estimate of the
EP, regardless of the stoichiometry.
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Dr. G. Van Biesen
C3100 - Winter 2011
¾Solubility (effect of Ksp):
25.00 mL of 0.1000 M I¯, Br¯,
and Cl¯
All titrated with 0.0500 M Ag+
(Ve = 50.00 mL for all)
X- + Ag+
AgX(s) (X = Cl, Br, I)
Before Ve: [Ag+] =
Ksp
[X-]
[X-] is the same for all anions:
Ve - V
Vi
[X-] = CA
Vi + V
Ve
Thus [Ag+] is directly proportional to Ksp:
Fig. 26-9, Harris 8th Ed.
The smaller Ksp, the smaller [Ag+], and the larger pAg+; therefore the lower
the concentration of analyte that can be analyzed by titration.
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After Ve:
Notice from the figure that the part of the titration curve after Ve is
independent of the anion. pAg+ is here only determined by the excess
Ag+ (see earlier)
Analysis of mixtures:
For the analysis of mixtures, it is necessary that each ion is (nearly)
quantitatively precipitated before precipitation of the next one starts. This
requires that the Ksp’s differ by at least 4 orders of magnitude.
Ksp values for ions whose precipitates have different stoichiometries cannot
be used to predict the order of precipitation
e.g. a mixture of CO32¯, Cl¯, and PO43¯ titrated with Ag+.
ion
precipitate
Ksp expression
CO32¯
Ag2CO3
Ksp = [Ag+]2[CO32¯]
Cl¯
AgCl
Ksp = [Ag+][Cl¯]
PO43¯
Ag3PO4
Ksp = [Ag+]3[PO43¯]
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Dr. G. Van Biesen
C3100 - Winter 2011
If stoichiometries are the same: precipitate with lowest Ksp is formed first
Addition of AgNO3 to a solution
containing both KI and KCl
Ksp(AgCl)= 1.8 x 10-10; Ksp(AgI)= 8.3 x 10-17
When all I- has precipitated, there is
an increase in [Ag+] and now AgCl will
precipitate (because in the figure not
all I- has completely precipitated when
AgCl starts to precipitate, VEP is the
end of the steep portion of the curve)
Because of some co-precipitation of
AgCl with AgI, VEP for AgI is slightly
overestimated.
VEP if no Cl- present
Fig. 26-10, Harris 8th Ed.
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Endpoint detection for precipitation reactions:
Either electrodes or indicators
Fajans Method: Adsorption indicator
¾Precipitation of anions with cations:
e.g. Cl⎯ with Ag+: Ag+ + Cl⎯
AgCl(s)
Before EP: Excess Cl¯ in solution. Particles of AgCl have a negative charge
due to extra Cl-’s adsorbed on exposed Ag+.
After EP: Excess Ag+ in solution. Particles of AgCl have a positive charge due
to extra Ag+’s adsorbed on exposed Cl-.
The indicator dichlorofluorescein is negatively charged above pH ~ 5, and
adsorps quickly to the positively charged crystals produced immediately after
the EP. This changes the color of the indicator (yellow → pink).
¾Precipitation of cations with anions:
e.g. Hg22+ with Cl⎯ : Hg22+ + 2 Cl ⎯
Hg2Cl2(s)
This time, the crystals have a positive charge before the EP, and a negative
charge after the EP. A cationic indicator (e.g. bromophenol blue) is now used.
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Dr. G. Van Biesen
C3100 - Winter 2011
Volhard Method: halide analysis - formation of a soluble, coloured complex at EP
Backtitration:
•To a measured volume of sample, a solution of Fe(NO3)3 and some HNO3 are
added (acidifying the sample prevents hydrolysis of Fe3+). Halide ion is
quantitatively precipitated with a measured excess of standard AgNO3 solution:
AgX(s)
X- + Ag+
•The unreacted Ag+ is then titrated with standard KSCN to AgSCN(s):
AgSCN(s)
Ag+ + SCN•When all Ag+ has reacted, SCN- reacts with Fe3+ to give a red complex:
[FeSCN]2+ (red)
Fe3+ + SCNAbove procedure works well for Br- and I-, but a problem arises with Cl-: AgCl (Ksp
= 1.8 x 10-10) is more soluble than AgSCN (Ksp = 1.0 x 10-12).
AgCl(s) + SCN-
AgSCN(s) + Cl-
This reaction occurs especially near the EP, so that more SCN- has to be
added, and the amount of Cl- is underestimated (endpoint fades).
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Therefore, the AgCl(s) formed in the first step is either filtered (and only
unreacted Ag+ is titrated with SCN-), or for instance nitrobenzene or
polyvinylpyrrolidone is added to the sample, which coats AgCl and
prevents it from dissolving.
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Dr. G. Van Biesen
C3100 - Winter 2011
COMBUSTION ANALYSIS: Determination of C, H, N, S (and O)
• Small amount of sample accurately weighed in Ag or Sn capsule (sealed)
• Analyzer swept with pure He, just before run, a measured excess of O2 is
added to He stream
• Capsule dropped into preheated ceramic crucible; capsule melts and
sample is instantaneously oxidized
C, H, N, S
1050 °C / O2
CO2(g) + H2O(g) + N2(g) + SO2(g)
(+ some SO3(g))
• The initial reaction products of the combustion pass through several catalysts
to complete their transformation:
C, CO + O2
SO3
½ O2
Cu
Cu
WO3
CO2
SO2 + CuO (s)
CuO (s)
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Dynamic Flash Combustion: short burst of gaseous products, so that
reaction products can be analyzed via a GC with a thermal conductivity
detector (GC/TCD)
For the analysis of oxygen, the sample is thermally decomposed in the
absence of oxygen (pyrolysis). Oxygen released by the sample is converted
into CO, measured with a GC/TCD.
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Dr. G. Van Biesen
C3100 - Winter 2011
Stoichiometry to the rescue (Journal of Chemical Education 1988, 65 (9), 800)
In the beginning of the 20th century, industrial organic chemistry was blooming. Dye
manufacturers used large vats with boiling mixtures of sulfuric and nitric acid.
One day, a worker in one of those factories did not return home, and his
disappearance was an absolute mystery. The manager of the factory eventually
suggested that the worker had fallen into the acid, and had been dissolved, hair,
flesh, bones, boots and all. In order for the wife to claim insurance money, she
needed evidence of his death. The acid was analyzed for phosphorus, which is
present in the human body at ~ 0.63%.
Large vat of nitric and sulfuric acid, perhaps containing one dissolved human body
Take 100 mL samples and treat
with molybdate reagent
(NH4)3[P(Mo12O40)].12H2O ↓
spectrophotometry,
redox titration,
acid-base titration
It is assumed that all P
is converted to H3PO4
and that the contents of
the vat is homogenous
Heat to 400 °C
Gravimetric determination of
P2O5•24MoO3
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The gravimetric procedure yielded a mass of P2O5•24MoO3 (MM = 3596.46) of
0.3718 g.
A blank sample of a mixture of nitric and sulfuric acid of the same volume (100 mL)
gave 0.0331 g of P2O5•24MoO3
→ mass of P2O5•24MoO3 due to sample = 0.3718 – 0.0331 = 0.3387 g
#moles of P = 2 x [0.3387 (g) / 3596.46 (g/mol)] = 1.884 x 10-4 mol P
= 1.884 x 10-4 (mol) x 32.0 (g/mol) = 6.03 x 10-3 g P
With a volume of 8000 L of acid, the total amount of P in the vat would be:
8000 (L) x 6.03 x 10-3 (g) / 0.100 (L) = 482 g P
Assuming a mass of 70 kg, and with an average content of 0.63% P for the human
body, the expected amount of P would be:
70 x 103 (g) x 0.0063 = 441 g P
which is close to the value found (482 g).
Based upon this analysis, the insurance company paid the money owed to the
widow.
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Dr. G. Van Biesen