Precipitation titration

1. 1
Mrs. Prajakta B. Kothawade
Assistant Professor,
PES Modern College of Pharmacy, for ladies, Moshi, Pune

2. Precipitation is combination of two ionic species to form a
very insoluble product. Precipitation of this product forces
the reaction to completion
2

3. REQUIREMENTS
The requirements for precipitation reaction. useful in titrimetric analysis are:
1. The precipitate must be practically insoluble.
2. The precipitation reaction should be rapid and quantitative.
3. The titration results should not be hampered by adsorption (co-precipitation)
effects.
4. It must be possible to detect the equivalence point during the titration.
3

4. 4
Theory of precipitation
1. Solubility
2. Solubility product

5. 5
1. SOLUBILITY
Solubility, which is dependent on the solvent and temperature, is
the concentration of the dissolved solute in moles per litre, when
the solution is in equilibrium with a solid solute.
In the solid state the solute molecules occupy space in a fixed
repeating pattern to form what is called a crystal of solid. This
repeating pattern depends upon the molecular structure of the
compound. The solute molecules are held together in that pattern
by intermolecular forces of attraction.

6. 6
Now in order to dissolve a solid, these forces of attraction must
be overcome so that solute-solute attraction is replaced by
solute-solvent attraction.
The solvent should compete with crystal forces and overcome
them, which often means that the solvent environment must be
similar to that provided by the crystal structure. This is the basis
for the simple rule "like dissolves like".
During precipitation, however, the opposite condition is aspired
for, where the intermolecular forces between the molecules of
product are high and solute-solute forces replace the solute-
solvent forces

7. 7
2. SOLUBILITY PRODUCT
Consider an aqueous solution of a slightly soluble salt BA in
equilibrium with excess of solid at constant temperature. The
equilibrium can be represented by:
BA(s) - [B+] [A-] (1)
Where BA(s) represents the solid phase. In dilute solution
essentially no undissociated BA will be present in the solution.
Since the activity of solid is constant, the equilibrium constant
for (1) may be written as:
K sp = [B+] [A-] (2)

8. 8
Ksp is the solubility product, which is a constant for a given
solute, solvent and temperature.
In a more complex case
BmAn (s)  mBn+ + nAm- ` (3)
Here, Ksp = [Bn+]m [Am-]n
Solubility product is of importance as it premits the calculation of
one of the ion concentration if the other is known. A substance
precitipates out when the product of the ionic concentration
exceeds the Ksp value, i.e; in equation (1) solid BA will
precitipate out when the product of [B+] and [A-] exceeds Ksp

9. 9
FACTORS AFFECTING SOLUBILITY
1. Common Ion Effect :
The solubility of any slightly soluble salt can be decreased by
adding an excess of either of its ions.
e: g. The dissociation of a slightly soluble salt BA is
BA(s) - [B+] [A-]
And Ksp = [B+] [A-]
This is the equilibrium condition.

10. 10
If, however an excess of either B+ or A- are added in the
form of another salt (whose solubility) is greater than that of
BA then the product of ionic concentrations [B+] [A-] will
exceed the solubility product and hence BA will precipitate.
The common ion effect provides a valuable method for
controlling the concentration of the ions furnished by a
weak electrolyte.

11. 11
2. Effect of temperature on solubility
The solubility of the precipitate encountered in quantitative analysis increases
with the rise in temperature,. With some substances the influence of
temperature is small, but with others it is quite appreciable. Thus, the
solubility of AgCI at 10°C and 100°C is 1.72 and 21.1 mg/litre, while that of
BaSO4 is 2.2 and 3.9 mg/litre respectively. In many instances the common ion
reduces the solubility to so small a value that the temperature effect, which is
otherwise appreciable, becomes very small.

12. 12
Determination of End Points in Precipitation
Titrations
1. Formation of a coloured precipitate
2. Formation of a soluble coloured compound
3. Use of Adsorption Indicators or Fajans method
4. Guy Lussac's Method

13. 13
Types of Precipitation Titrations
1. Direct titration
Mohr’s method
by Formation of a coloured precipitate
2. Indirect titration
Volhard’s method
by formation of a soluble coloured compound

14. 14
1. Direct titration (Mohr’s method)
by Formation of a colored precipitate
principle of fractional , precipitation is used.
First, the precipitating reagent reacts with the ion to be
analyzed and precipitate is formed. At the end point when no
analyte is present, the precipitating reagent reacts with an
indicator ion added in very small quantity to form a colored
'precipitate. The appearance of such a colored precipitate
marks the end point.
The concentration of an indicator ion is chosen such that, it
starts

15. 15
It involves the titration of sodium chloride with silver nitrate
Using dilute potassium chromate as indicator.
Let us consider the titration of 0.1 M NaCI with 0.1 M AgNO3 in
the presence of few ml of dilute K2CrO4 solution.
Ksp(AgCl) = 1.2 x 10-10 = [Ag+] [Cl-]
Ksp(Ag2CrO4) ,=1.7 x 10 = [Ag+2]2 [ CrO4]

16. 16
As the AgN03 solution is added, AgCl will precipitate first. We
usually expect the salt with smaller solubility product to
precipitate first, but this is true only if both salts dissociate to yield
the same number of ions. Also, choride ions in this case are in
great excess of that of chromate ions. At the first point where the
red silver chromate is just precipitated, we shall have both salts in
equilibrium with the solution, hence,

17. 17
[Ag+] = Ksp(AgCl) = √Ksp(Ag2CrO4)
[Cl-] [CrO4
-]
[Cl-] = Ksp(AgCl) = 1.2 x 10-10 (1)
[CrO4
-] √Ksp(Ag2CrO4) √1.7 x10-12
= 9.2 X 10-5
As seen earlier, at equivalence point, no excess of Ag+ and Cl-
should be present which marks the end of the titration.
Ksp(AgCl) = [Ag+] [Cl-]
At equivalence [CI-] = [Ag+]

18. 18
Ksp(AgCl) = [Cl-] [Cl-]
√Ksp(AgCl) = [Cl-]
[Cl-] = √ 1.2 x 10-10
[Cl-] = 1.1 x 10-5
[Cl-] = 9.2 X 10-5
√[CrO4
-]
[CrO4
-] = [Cl-]2
(9.2 X 10-5 )2
= 1.4 x 10-2

19. 19
So, the concentration of potassium chromate solution should be
0.0014M.
Mohr's procedure also applies to the determination of bromides
in the similar manner.
It is necessary that Mohr's method should be applied only to
neutral or slightly alkaline solution i.e;. within pH range6.5 to9.
In the acidic solution, the following reaction occurs:

20. 20
We have already discussed the effect of pH on the solubility of a
salt. Here H2CrO4, acts as a acid, consequently the chromate ion
concentration is reduced and solubility product of silver chromate
may not be exceeded at all. In markedly alkaline solutions, silver
hydroxide (Ksp =2.3x 10-8)may be precipitated. A simple method
of making an acid solution neutral is to add excess of pure
calcium carbonate or sodium hydrogen carbonate. Any alkaline
solution may be acidified with acetic acid and then a small excess
of calcium carbonate is added.

21. 21
.The solubility product of silver chromate increases with rising
temperature. The danger of accidentally raising or lowering the
pH beyond acceptable limits is minimised by using a mixture of
potassium chromate and potassium dichromate in proportions so
as to give a neutral solution. In the presence of ammonium salts
the pH must not exceed 7.2 because appreciable concentrations
of ammonia do have an effect upon the solubility of silver salts.
Titration of iodide and thiocyanate is not successful because AgI
and AgSCN adsorb chromate ions strongly and a false indistinct
end point results.

22. 22
2. Effect of solvent on solubility
The solubility of most inorganic compound is reduced by the
addition of organic solvents such as methyl, ethyl and n-
propyl alcohol etc. e.g. the addition of about, 20 ,% by
volume of ethanol renders the solubility of lead sulfate
practically negligible, thus permitting quantitative separation.

23. 23
2. Effect of pH on solubility
The solubilty of a salt will be increased by decrease in pH, if
the anion of the salt is a conjugate base of a weak acid.
e. g. consider the slightly soluble salt BA the anion of which
(A-), is the conjugate base of a, weak acid HA.
Here, there will be two equilibriums in operation.
BA(s)  [B+] [A-] (1)
HA  [H+] [A-] (2)

24. 24
The A- from (1) will shift the equilibrium (2) towards the left
while equilibrium (1) will itself be shifted to the right. Hence,
solubility of BA is increased with increase in H+ or decrease in
pH.
The equilibrium expressions are:
Ksp = [B+] [A-] (3)
. Ka = [H+] [A-] (4)
[HA]

25. 25
Molar solubility of BA is equal to [B+) which is equal to the total
concentration of A- , i. e. the A- dissolved from BA and that which
is present in HA.
S = [B+] = [A-]+ [HA] (5)
From equations (3), (4) and (5)
S = [B+] = Ksp {1 +([H+]/Ka )}
[B+]
S = √Ksp (1 + [H+]/Ka) (6)
Equation (6) relates molar solubility to Ksp, Ka and H+ ion
concentration.

26. 26
Fractional Precipitation
We shall study the situation which arises when a precipitating
reagent is added to a solution containing two anions, both of
which form slightly soluble salts with the same cation e.g. when
silver nitrate solution is added to a solution containing both
chloride and Iodide the question arises here is which salt will
precipitate first and how completely will the first sallt
beprecipitated before thesecond ion begins to react with the
reagent.

27. 27
Fractional Precipitation
The solubility products of silver chloride and silver iodide are
1.2 x 10-10 and 1.7 x 10-16 respectively i. e.
[Ag+][CI- ] = 1.2X 10-10 [Ag+] [I-] = 1.7 X 10-16
It is evident that the solubility product of silver iodide being
less
will be exceeded first and hence will be precipitated first.
i.e, [Ag+] exceeds the value
-16 -10

28. 28
Fractional Precipitation
AgCl will precipitate when the latter value is exceeded. after
this both ions will be precipitated simultaneously. The Ag+ ions
will then be in equilibrium with both the salts.
[Ag+] = Ksp(AgI) = Ksp(AgCl)
[I-] [Cl-]
[I-] = Ksp(AgI) = 1.7 x 10-16
[Cl-] Ksp(AgCl) 1.2 x10-10
= 1.4 x 10-6

29. 29
Fractional Precipitation
So, when the concentration of iodide ions is about one millionth
part of the chloride ion concentration, silver chloride will be
precipitated.
when the iodide and bromide are in combination then,
[I-] = Ksp(AgI) = 1.7 x 10-16
[Br-] Ksp(AgBr) 3.5 x10-13
=1/ 2 x 10-3

30. 30
1. Indirect titration ( Volhard’s method)
by formation of a soluble colored compound
In this case, after the end point, the excess of precipitating
reagent added reacts with an indicator to form a soluble
colored complex.
This procedure is exemplied by the method of Volhard for
the titration of silver in the presence of free nitric acid with
standard potassium or ammonium thiocyanate solution using
Fe+++ as indicator. This,method may be applied to the
determination of halides in acidic solution.

31. 31
An excess, known volume of standard silver nitrate solution is
added to the acidic halide solution. The silver halide is allowed
to precipitate completely and the excess of silver ions is
titrated with standard thiocyanate solution.
AgN03 + X-  AgX + NO3
AgN03 + SCN- AgSCN + NO3
Fe+++ + SCN- [FeSCN]2+

32. 32
First, a precipitate of reddish brown complex ion silver
thiocyanate is formed (Ksp =7.1 x 10-13) when this reaction is
complete, the slightest excess of thiocyanate produces, a
reddish-brown coloration, due to the formation of complex ion
with Fe+++ indicator added in the form of ferric nitrate or ferric
ammonium sulphate. e. g. Determination of chloride ions. An
excess of standard AgN03 is added to the chloride solution to be
determined.

33. 33
First, the Ag+ ions react with Cl- ions to give AgCl
Ag+ + Cl-  AgCl
Ksp = 1.2 X 10-10
when this reaction is complete, the excess of Ag+ ions are
titrated with standard thiocyanate using Fe+++ as indicator.
Ag+ +SCN- AgSCN
Ksp =.7.1 X 10-13
Now, the two sparingly soluble salts are in equilibrium with the
solution, hence:
[CI-] = K(spAgCl) = 1.2 X 10-10 = 169
[SCN-] Ksp(AgSCN) 7.1 x 10-13

34. 34
When, all the excess of Ag+ has reacted, the thiocyanate may
react with AgCl precipitate since silver thiocyanate is the less soluble salt
until the ratio [CI-]I[SCN-] in solution is 169.
AgCl + SCN-AgSCN + CI
This reaction will take place before reaction occurs with Fe+++ and hence a
titration error is introduced.
To overcome this,
•silver chloride precipitate is removed by filtration and the is back titrated.
•silver chloride particles are coated by an immiscible liquid e. g.
nitrobenzene
potassium nitrate is added as a coagulant, Desorption of silver ions occurs
and readsorption is prevented by the presence of potassium nitrate

35. 35
Use of Adsorption Indicators or Fajans method
Indicators are adsorbed on the surface of the precipitate at the
equivalence point and this ad-sorption"is accompanied by a
color change. These indicators are either acid dyes e.g.
fluorescein, eosin etc. or basic dyes e.g. -rhodamine series.

36. 36
The conditions which govern the choice of an adsorption
indicator are:
1. The indicator ion should have a charge opposite to that of
the ion of the precipitating reagent.
2. The solution should be concentrated enough to give a sharp
colour change.
3. The indicator should be secondarily adsorbed only after the
equivalence point

37. 37
When a sodium chloride solution is titrated with silver nitrate,
the silver chloride precipitate will adsorb chloride ions which
are initially in excess.

38. 38
Immediately after the equivalence point,

39. 39
At end point in presence of adsorption indicator

40. 40
Guy Lussac's Method
Turbidity accompanies precipitation reaction is made use of in
this method.
After the' equivalence, the precipitation reaction ceases and
addition of an extra drop will not result in turbidity.
Nephelo-turbidimetric methods can also be used.

41. Precipitation
titration
P’ceutical applications
41

42. Applications
 Determination of Chloride in Mineral Water
 Many mineral waters contain an easily titratable amount
of chloride ions. The most accurate method for chloride
determination is inflection point titration, but for
quickmeasurement, if the sample matrixis reproducible,
end point titrationcan be used with a combinedsilver
electrode.
42

43. Applications
 Principle:
 The titrant reagent for chloride determination is silver
nitrate(AgNO3); using end point titration,
 the titrant concentration must be at least 0.05 eq/l and
generally between 0.1 and 0.05 eq/l. The water sample
must be acidic at around pH 4.5. That can be obtained
by means of acetic acid but for mineral waters with
high pH, nitric acid 1M can be also used.The reaction
corresponds to: AgCl (precipitate). Results are normally
expressed as mg/l of chloride (AW = 35.45 g/mol) or
sometimes as mg/l of sodium chloride (NaCl with MW =
58.45 g/mol).
43

44. Determination of Chloride in
Water
 As tap water or surface water contains chloride ions at
low concentration levels, chloride determination should
be performed by titration with silver nitrate (AgNO3) as
titrant. Standard NF ISO 9297 uses a colorimetric
determination (with silver chromate) of the equivalent
point but it is possible with the same titrant to use a
potentiometric determination of theequivalent point.
This application uses a potentiometric titration with a
combined silver/reference electrode.
44

45. 45
Refferences
 Vogel’s Text Book of Quantitative Chemical Analysis, 6/Ed., Pearson
Education, page no: 489-492.
 Practical Pharmaceutical Chemistry Part-I by Beckett A H & Stanlake J B,
4/Ed., CBS Publisher & Distributors, page no: 197-225.
 Pharmaceutical Analysis Vol. I & K. R. Mahadik, S.G. Wadodkar, H. N, I.
More, Nirali Prakashan, page no: 8.1-8.15.