V. Gravimetric Analysis

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V. GRAVIMETRIC AND COMBUSTION ANALYSIS

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

(Chapter 26)GRAVIMETRIC ANALYSIS

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Dr. G. Van Biesen


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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:

adsorption: bound to the surface of the crystal

absorption: inside of the crystal:

Inclusions: Ions which replace analyte ions in the crystal lattice. Theprobability 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 areais 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 ofadsorption of Ag+ (in excess relative to Cl-).

Adsorption of Ag+ preferred over H+

because of better size compatibility.

Region of solution directly surroundingthe particle = Ionic atmosphere has a

net neg. charge

Particle + ionic atmosphere =

electric double layer movesas a unit through solution

In order for colloidal particles to coalesce and grow, they have to collide.

However, their ionic atmospheres repel one another.

Primary layer

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Common solutions to this problem:

Increase their kinetic energy by heating (collisions are more energetic so that

electrostatic repulsion is overcome)

Increase electrolyte conc., which decreases the volume (thickness) of the ionicatmosphere and allows particles to come closer together before electrostatic

repulsion becomes significant

After precipitation, most procedures call for a digestion period; the precipitate isleft 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. Waterwashes 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 precipitateis 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 preventprecipitation of impurities. For instance, in the gravimetric analysis of Ba2+

with N-p-chlorophenylcinnamohydroxamic acid, excess KCN is added tokeep 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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16.95%100%oreg11.1324

Feg1.8864%Fe

g1.886mL50.00

mL250.00xg0.3772mL250inFemass

mL)50(inFeg0.3772Femass

molg55.847Femol106.755Femass

Femol106.755OFemol1

Femol2OFemol103.377n

mol103.377molg159.69

0.5394gn

48

8

1-3

6

3

6

32

32

3

8Fe

3

81-OFe 32

=

=

==

=

=

==

==

Solution to Problem 1

12

+ 2

= 0.005871 mol AgCl = mol Cl-

Let x = mass of NaCl moles of NaCl =

0.4000 x = mass of BaCl2 moles of BaCl2 =

x (g)

58.442 (g/mol)

0.4000 (g) x (g)

208.23 (g/mol)

Moles of AgCl =0.8415 (g)

143.32 (g/mol)

Each mol of NaCl contributes 1 mol of Cl-, each mol of BaCl2 contributes 2 mol of Cl-

0.005871 mol =x (g)

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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loss = solubility x volume wash waterloss = 3.5 x 10-4 g L1 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 insolution; 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 C2O4

2- with Pb2+

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Calculate Ve: #moles Br- = #moles Ag+

50.00 mL 0.05000 M = Ve 0.1000 M

Ve = 25.00 mL

Before EP:

10.00 mL Ag+ added:

[Br-] = CA

Precipitation titr ations: 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-.

KspAgBr = 5.0 x 10-13

Titration reaction: Ag+ + Br AgBr(s)

Ve - V

Ve

Vi

Vi + V

25.00 mL 10.00 mL

25.00 mL= 0.05000 M

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+] = = = 2 x 10-11 M pAg+ = 10.70Ksp

[Br-]

5.0 x 10-13

0.02500

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 Br-according to the Ksp:

AgBr Ag+ + Br-

x dissolves x x

[Ag+][Br-] = Ksp

x2 = Ksp

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13-

sp x1005Kx .==

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+] = =Vi + V

(V Ve) [Ag+]0

50.00 mL + 25.10 mL

(25.10 mL 25.00 mL) (0.1000 M)

= 1.33

x 10-4 M pAg+ = 3.88

[Br-] = = 3.76 x 10-9 M pBr- = 8.42

Ksp

[Ag+]

This is essentially

a simple dilution!

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Factors that influence the shape of the titration curve:

Concentration:

Titration of I- (0.1, 0.01, 0.001 M) with Ag+

(0.05 M, 0.005 M and 0.0005 M resp.)

Before Ve: [Ag+] =

Ksp

[I-]

If [I-] , then [Ag+] and pAg+ After Ve: excess [Ag+], so with [Ag+] ,

pAg+

The larger the change in p function,

the sharper the endpoint

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.

Fig. 26-8, Harris 8th Ed.

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Solubility (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

Ve

Vi

Vi + V[X-] = CA

Thus [Ag+] is directly proportional to Ksp:

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 Ksps 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 PO4

3 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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If stoichiometries are the same: precipitate with lowest Ksp is formed first

Addition of AgNO3

to a solution

containing both KI and KClKsp(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


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Endpoint detection for precipitation reactions:Either electrodes or indicators

Fajans Method: Adsorption indicator

Precipitation of anions with cations:

e.g. Clwith Ag+: Ag+ + Cl AgCl(s)

Before EP: Excess Clin solution. Particles of AgCl have a negative chargedue to extra Cl-s adsorbed on exposed Ag+.

After EP: ExcessAg+ in solution. Particles of AgCl have a positive charge dueto extra Ag+s adsorbed on exposed Cl-.

The indicator dichlorofluorescein is negatively charged above pH ~ 5, andadsorps quickly to the positively charged crystals produced immediately after

the EP. This changes the color of the indicator (yellow pink).Precipitation of cations with anions:

e.g. Hg22+ with Cl: Hg2

2+ + 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


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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 areadded (acidifying the sample prevents hydrolysis of Fe3+). Halide ion is

quantitatively precipitated with a measured excess of standard AgNO3 solution:

X- + Ag+ AgX(s)

The unreacted Ag+ is then titrated with standard KSCN to AgSCN(s):

Ag+ + SCN- AgSCN(s)

When all Ag+ has reacted, SCN- reacts with Fe3+ to give a red complex:

Fe3+ + SCN- [FeSCN]2+ (red)

Above 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 onlyunreacted 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


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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 isadded to He stream

Capsule dropped into preheated ceramic crucible; capsule melts and

sample is instantaneously oxidized

C, H, N, S1050 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 CO2

WO3

SO3 SO2 + CuO (s)

Cu

O2 CuO (s)Cu

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Dynamic Flash Combustion: short burst of gaseous products, so thatreaction 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 convertedinto CO, measured with a GC/TCD.

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Dr. G. Van Biesen


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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 eventuallysuggested 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(Mo

12O

40)].12H

2O

spectrophotometry,

redox titration,

acid-base titration Gravimetric determination ofP2O524MoO3

Heat to 400 C

It is assumed that all P

is converted to H3PO4and that the contents ofthe vat is homogenous

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The gravimetric procedure yielded a mass of P2O524MoO3 (MM = 3596.46) of0.3718 g.

A blank sample of a mixture of nitric and sulfuric acid of the same volume (100 mL)

gave 0.0331 g of P2O524MoO3 mass of P2O524MoO3 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.

C3100 - Winter 2011

Dr. G. Van Biesen

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V. Gravimetric Analysis