Precipitation Reactions and Precipitation Reactions and

Gravimetric AnalysisGravimetric Analysis

• Titrations where the titrant forms a precipitate with the analyte.

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

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

• Major applications: determination of halides by the precipitation silver salts.

the determination of sulphate by precipitation as barium sulphate.

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

Solubility product constants of selected slightly

soluble salts.

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 to combine with either the cation or anion of a slightly soluble compound - will result in an increase in its solubility.

AgCl → Ag+ + Cl-

Ag+ + NH3 → AgNH3+

AgNH3+ + NH3 → Ag(NH3)2

+

Precipitation titrations

Calculate 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.

A titration curve can be obtained by plotting pAg (-log[Ag+]) or pCl

against the volume of AgNO3, in a similar manner as you did for

acid-base titrations.

At 110.0 ml:

mmol Ag+ = 110.0 ml x 0.100 M = 11.00 mmolAg+ left = 11.00 – 10.00 = 1.00 mmol[Ag+] = 1.00 mmol/210 ml = 4.76 x 10-3 M[Cl-] = Ksp/[Ag

+] = 2.1 x 10-8 MpCl = -log(2.1 x 10-8) = 7.67

At 100.0 ml:

All Cl- has reacted with Ag+

[Cl-] = 10sp 1.0x10K −=

= 1.0 x 10-5 MpCl = -log(1.0 x 10-5) = 5.00

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)

pAg

AgI Ksp = 8.3 x10-17

AgBr Ksp = 5.0 x10-13

AgCl Ksp = 1.82 x10-10

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)

pAg

0.1 M NaI

0.5 M NaI

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)

pAg

If we look at the Ksp value for AgI

(8.3 × 10-17) we see that the

calculated 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)

- 10 9.1 Ksp ][I ][Ag ×===+

M0.0333 mL 50 mL 25

M0.1 mL 25=

+

×

1.82 x 10-10

8.3 x 10-17

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+ → Ag2CrO4

(yellow) (red)

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

• From Ksp, the concentration of Ag+ at the equivalence

point 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 the equivalence point.

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

Volhard titration

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

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

X- + Ag+ → AgX + excess Ag+

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

Excess Ag+ + SCN- → AgSCN

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

Fe3+ + SCN- → Fe(SCN)2+

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)

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

fluorescein.

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

• This repels the negatively charged fluoresceinate anions.

AgCl : Cl-

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

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

AgCl : Ag+ :: In-

This gives these particles a red colour, thus indicating end

point.

• 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.

• 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.

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 of quantitative analysis.

• However, gravimetric methods have certain limitations!

• The ideal product of a gravimetric analysis should be insoluble, easily filterable, very pure, and should have a known composition.

Steps in gravimetric analysis:

1. Preparation of the solution

2. Precipitation

3. Digestion

4. Filtration

5. Washing

6. Drying or igniting

7. Weighing

8. Calculation

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.

Particle size

• The particle size of solids formed by precipitation varies.

• Colloidal suspensions have tiny particles (10-7 to 10-4 cm

in 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.

Rate of precipitation

• Supersaturated solutions contain more solute than should be present at equilibrium.

• Relative supersaturation =

where Q is the actual concentration of solute and S is the 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 −

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 large

crystals.

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.

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.

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

Occlusion and Inclusion

• Inclusions - impurity ions that occupy site in the crystal lattice.

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

• Occlusions - pockets of impurity that become trapped within a crystal.

• Occluded or included impurities are difficult to remove -digestion or reprecipitating may be helpful.

Surface Adsorption

• Adsorbed impurities - those bound to the surface of a crystal.

• The most common form of contamination.

• Can often be removed by washing or digestion.

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

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.

Drying or igniting the precipitate

• Precipitates must generally be heated to remove water and 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.

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

eprecipitat wt. formula

analyte wt. formulaGF =

1

1 x

PO Agwt. formula

P wt.at GF

43=

1

1 x

418.58

30.97GF =

GF = 0.07399

1

1 x

PO Agwt. formula

P wt.at GF

43=

1

1 x

418.58

30.97GF =

GF = 0.07399

1

1 x

PO Agwt. formula

P wt.at GF

43=

1

1 x

418.58

30.97GF =

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? What about 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? What about the percent Mn?

We need to convert from Mn3O4 to Mn2O3

= 1.035

= 8.58%

b

a x

eprecipitat wt. formula

analyte wt. formulaGF =

2

3 x

228.8

157.9GF =

100 x 1.52

GF x g 0.126O%Mn 32 = 1

3 x

OMn

MnGF

43=

= 5.97%

2

3 x

228.8

157.9GF =