Precipitation Reactions and Gravimetric Analysis.pdf

8/19/2019 Precipitation Reactions and Gravimetric Analysis.pdf

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Precipitation Reactions andPrecipitation Reactions and

Gravimetric AnalysisGravimetric Analysis

• Titrations where the titrant forms a precipitate with theanalyte.

• Not always so straightforward – a number ofrequirements need to be met.

• Precipitation reactions are often slow and have thetendency to absorb and co-precipitation other species.

• Major applications: determination of halides by the

precipitation silver salts.the determination of sulphate byprecipitation as barium sulphate.


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Precipitation Equilibria: The solubility

product

What do we mean by the term “insoluble”?

AgCl → Ag+ + Cl-

We can write the equilibrium constant or the solubility

product:

Ksp = [Ag+][Cl-]

Precipitation will not take place unless the product of[Ag+] and [Cl-] exceeds the Ksp

E.g. AgCl = 1.82 x 10-10


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Solubility product constants of selected slightlysoluble salts.


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Factors affecting Ksp

1. The common ion effect

If there is an excess of one ion over the other, the

solubility of the precipitate will decrease - Le Chatelier’sprinciple.

2. The effect of complex-ion formation

The presence of complexing agents that are able tocombine with either the cation or anion of a slightlysoluble compound - will result in an increase in itssolubility.

AgCl → Ag+ + Cl-

Ag+ + NH3 → AgNH3+

AgNH3+ + NH3 → Ag(NH3)2

+


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Precipitation titrationsCalculate pCl for the titration of 100.0 ml of 0.100 M Cl- with 0.100 M

AgNO3 after the addition of 0.00, 20.00, 100.0 and 110.0 ml of AgNO3.


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The smaller the Ksp, the sharper the end point.

0

2

4

6

8

10

12

14

16

0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0

vol Ag (ml)

p

A g

AgI Ksp = 8.3 x10-17

AgBr Ksp = 5.0 x10-13

AgCl Ksp = 1.82 x10-10


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The more concentrated the reagents, the bigger the end

point break.

0

2

4

6

8

10

12

14

16

18

0.0 10.0 20.0 30.0 40.0 50.0 60.0

vol Ag (ml)

p A g

0.1 M NaI

0.5 M NaI


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What about solutions that contain a

mixture of anions?The compound that is least soluble

precipitates first.

0

2

4

6

8

10

12

14

16

0 10 20 30 40 50 60 70 80 90 100 110 120

vol Ag (mL)

p A g

If we look at the K sp value for AgI

(8.3 × 10-17) we see that thecalculated solubility at equivalence

point is

Will the chloride start to

precipitate?

If we started with 25 ml of 0.1 M

chloride, at the first equivalence

point it will be

Ksp = [9.1 x 10-9

][0.03] = 3×

10-10

.which is greater than Ksp(AgCl)

AgCl starts to precipitate just

before the equivalence point of

AgI.

AgI

AgCl

-9(AgI)

- 109.1Ksp ][I ][Ag ×===+

M0.0333mL50mL25

M0.1mL25=

+

×

1.82 x 10-10

8.3 x 10-17


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Detection of the end point: Indicators

1. Indicators reacting with the titrant

2. Adsorption indicators

1. Indicators reacting with the titrant

Mohr method• Chloride is titrated with standard silver nitrate solution

using a soluble chromate salt indicator.

CrO4

-2 + 2Ag+ → Ag2

CrO4(yellow) (red)

• The concentration of the indicator is important. The Ag2CrO4 should just start precipitating at theequivalence point.


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• From Ksp, the concentration of Ag+ at the equivalencepoint is 1 x 10-5 M, meaning that Ag2CrO4 should

precipitate just when [Ag+] = 1 x 10-5 M. If the solubility

product of Ag2CrO4 is 1.1 x 10-12, we can calculate what

the concentration of CrO4-2 should be:

Ksp = [Ag+]2[CrO4

-2]

[CrO4-2] = 1.1 x 10-2 M

• If greater : Ag2CrO4 will begin to precipitate before theequivalence point.

• If less: Ag2CrO4 will only begin to precipitate after theequivalence point has been reached.


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Volhard titration

• Indirect titration for determining anions that precipitatewith silver (Cl-, Br -, SCN-).

• A measured excess of AgNO3 is added in acidic solutionto precipitate the anion.

X- + Ag+ → AgX + excess Ag+

• Excess Ag+ is then back-titrated with standard potassiumthiocyanate solution.

Excess Ag+

+ SCN-

→ AgSCN

• The endpoint is detected by using Fe3+, which forms asoluble red complex with the first excess of titrant (SCN-):

Fe3+ + SCN- → Fe(SCN)2+


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2. Adsorption indicators (Fajan's method)

• These are dyes that adsorb to the surface of a

precipitate near the equivalence point.

• The best–known example is fluorescein, which is used

to indicate the equivalence point in the titration of Cl –

with Ag+.

Fluorescein Fluoresceinate anion (yellow green)


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Consider the titration of Cl – with Ag+ in the presence of

fluorescein.

• Before equivalence point: Cl- is in excess and is theprimary adsorbed layer around the AgCl particles.

• This repels the negatively charged fluoresceinate anions. AgCl : Cl-

• When Ag+ is added in excess, the surface of theprecipitate become positively charge.

• The fluoresceinate anions become adsorbed in thecounter–ion layer of the AgCl colloids.

AgCl : Ag+ :: In-

This gives these particles a red colour, thus indicating end

point.


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• If an indicator is adsorbed more strongly than the analyte

ion, it cannot be used.

• Dichlorofluorescein/ fluorescein is adsorbed less strongly

than Cl –, Br –, I – or SCN – and can be used in the titration

of any of these ions.


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• We want the maximum surface area for adsorption (i.e. a

colloidal precipitate).

• The indicator ion must form a precipitate with the ions

adsorbed in the primary adsorption layer.

• The photocomposition of AgX can be a major source of

error in titrations involving silver - proper standardisation

is important.


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Gravimetric Analysis• The analyte is selectively converted to an insoluble form,

which can then be dried and accurately weighed.

• Can be one of the most accurate and precise methods ofquantitative analysis.

• However, gravimetric methods have certain limitations!• The ideal product of a gravimetric analysis should be

insoluble, easily filterable, very pure, and should have aknown composition.


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Steps in gravimetric analysis:


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Supersaturation and Nucleation

• Crystallisation occurs in two phases: nucleation and

particle growth.

• Nucleation - molecules in solution come together and

form small aggregates.

• The addition of further molecules to these nuclei results

in the formation of crystals.

• Supersaturated sugar water is used make rock candy,with the sugar crystals nucleating and growing into

crystals.


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Particle size

• The particle size of solids formed by precipitation varies.

• Colloidal suspensions have tiny particles (10-7 to 10-4 cmin diameter) – cannot be easily filtered.

• Crystalline suspensions - particles settle quickly and are

readily filtered.

• Large particles are also less prone to surface adsorption

and are more easily washed free from impurities.


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Rate of precipitation

• Supersaturated solutions contain more solute thanshould be present at equilibrium.

• Relative supersaturation =

where Q is the actual concentration of solute and S isthe concentration at equilibrium.

• Highly supersaturated solutions – fast nucleation,

resulting in a suspension of many small particles.• Less supersaturated solutions produce fewer, larger

crystals.

S

SQ −


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Precipitation conditions

We want to keep Q low and S high during precipitation.

1. Precipitate from dilute solutions - keeps Q low.

2. Add dilute precipitating reagents slowly while stirring -

keeps Q low and promotes the formation of largecrystals.

3. Precipitate from hot solutions - increases S.

4. Precipitate at a low pH. Many precipitates are more

soluble in acid medium and this slows the rate of

precipitation.


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Digestion

• Precipitate is allowed to stand in solution.

• Done at elevated temperatures.

• Large crystals grow at the expense of the small ones –

decreases surface area.


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Impurities in precipitates

• Precipitates may contain varying amounts of

impurities.

• Contamination of the precipitate occurs through co-

precipitation.

Two main mechanisms:

1. occlusion or inclusion

2. adsorption on the surface


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Occlusion and Inclusion

• Inclusions - impurity ions that occupy site in the crystallattice.

• Generally occurs when the impurity ion has a similar sizeand charge.

• Occlusions - pockets of impurity that become trappedwithin a crystal.

• Occluded or included impurities are difficult to remove -

digestion or reprecipitating may be helpful.


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Surface Adsorption

• Adsorbed impurities - those bound to the surface of acrystal.

• The most common form of contamination.

• Can often be removed by washing or digestion.

• Some impurities can be treated with a masking agent toprevent them from reacting with the precipitant.


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Washing the precipitate

• Some impurities can be removed by washing the

precipitate after filtering.

• Cannot always wash with pure water!

• This causes peptization (formation of colloids).

• Prevented by adding an electrolyte to the washing

solution.

• Example: nitric acid is used as a wash solution for AgCl.


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Drying or igniting the precipitate

• Precipitates must generally be heated to remove waterand adsorbed electrolytes.

• Usually be done by heating at 110 oC for 1 to 2 hours.

• Ignition - required if a precipitate must be converted to a

more suitable form for weighing.• Many metals that are precipitated using organic reagents

can be ignited to their oxides.


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Gravimetric calculations

• The precipitate we weigh is usually in a different to the

anayte whose weight we wish to report.

• A gravimetric factor (GF) is used to help us with this

conversion.

• Example: If we wish to calculate the quantity of

phosphorus in a Ag3PO4 precipitate, the GF would be

calculated as follows:

b

a x

eprecipitatwt.formula

analytewt.formulaGF =

1

1 xPO Agwt.formula

Pwt.atGF43

=

1

1 x418.58

30.97GF =

GF = 0.07399

1

1 xPO Agwt.formula

Pwt.atGF43

=

1

1 x418.58

30.97GF =

GF = 0.07399

1

1 xPO Agwt.formula

Pwt.atGF43

=

1

1 x418.58

30.97GF =


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An ore is analysed for the manganese content by

converting the manganese to Mn3O4 and weighing it.If a 1.52 g sample yields Mn3O4 weighing 0.126 g,what would be the percent Mn2O3 in the sample? Whatabout the percent Mn?

We need to convert from Mn3O4 to Mn2O3

= 1.035

= 8.58%

An ore is analysed for the manganese content by

converting the manganese to Mn3O4 and weighing it.If a 1.52 g sample yields Mn3O4 weighing 0.126 g,what would be the percent Mn2O3 in the sample? Whatabout the percent Mn?

We need to convert from Mn3O4 to Mn2O3

= 1.035

= 8.58%

b

a x

eprecipitatwt.formula

analytewt.formulaGF =

2

3 x

228.8

157.9GF =

100 x1.52

GFxg0.126O%Mn 32 = 1

3 xOMn

MnGF

43=

= 5.97%

2

3 x

228.8

157.9GF =

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Precipitation Reactions and Gravimetric Analysis.pdf