Acid base titrations, complexometric titrations

1. ANALYTICAL CHEMISTRY
(Types of Titrations)
Dr.S.SURESH
Assistant Professor
Email:avitsureshindia@gmail.com

2. Type of Titrations
Classified into four types based on type of
reaction involved;
1.Acid-base titrations
2.Complexometric titrations
3.Redox titrations
4.Precipitation titrations

3. Types of Titrations
• Acid-base titrations, in which an acidic or
basic titrant reacts with an analyte that is a
base or an acid.
• Complexometric titrations involving a
metal-ligand complexation reaction.
• Precipitation titrations, in which the analyte
and titrant react to form a precipitate.
• Redox titrations, where the titrant is an
oxidizing or reducing agent.

4. COMPLEXOMETRIC TITRATIONS
• Complexometric titrations are based on the
formation of a complex between the analyte and
the titrant. The chelating agent EDTA is commonly
used to titrate metal ions in solution.
• Example : EDTA Titrations
EDTA (Ethylenediaminetetraacetic acid) One of the most
common chelating agents used for complexometric titrations
in analytical chemistry.

5. Estimation of hardness by EDTA method
Disodium salt of Ethylene Diamine Tetra Acetic acid is a well-known complexing
agent. Calcium (Ca2+
) and Magnesium (Mg2+
) ions are present in hard water. When
EBT is (Eriochrome Black T) added to hard water EBT forms an unstable wine-red
colour complex with Ca and Mg ions at pH 9-10.
pH = 9-10
M2+
+ EBT [M – EBT] complex
NH4Cl-NH4OH less stable (wine red)
When EDTA is added into the hard water, the metal ions form a stable metal
complex with EDTA by leaving the indicator. When all the metal ions are taken by
EDTA from the indicator metal ion complex, the wine red colour changes into steel
blue, which indicates the end point.
[M –EBT ] + EDTA pH = 9-10 [M – EDTA] complex + EBT
(wine red) more stable (Steel blue)
(Colourless)

6. Precipitation titration
• Titrations with precipitating agents are useful for
determining certain analytes e.g. Cl-
can be
determined when titrated with AgNO3
Precipitation Titration - Mohr’s method
• Direct titration
• Basis of endpoint: formation of a coloured secondary
precipitate
• Indicator: Potassium chromate (K2CrO4)

7. Estimation of chloride (by Mohr’s method)
• In this method Cl‒
ion solution is directly titrated against
AgNO3 using potassium chromate (K2CrO4 ) as the indicator.
AgNO3 + Cl‒
AgCl ↓ + NO3
‒
(in water) (White precipitate)
• At the end point, when all the chloride ions are removed.
The yellow colour of chromate changes into reddish brown
due to the following reaction.
2AgNO3 + K2CrO4 Ag2CrO4 ↓ + 2KNO3
(yellow) (Reddish brown)

8. REDOX TITRATIONS
• REDOX TITRATIONS are based on an
oxidation-reduction reaction between the
analyte and titrant. Redox titrations are
carried out by using a potentiometer or a
redox indicator to determine the endpoint.

9. Estimation of Iron
Principle:
• Ferrous Iron is oxidized to Ferric iron by Potassium
Dichromate in acid solution. The completion of
oxidation reaction is marked by the appearance of
Blue violet colour of Diphenylamine, which is used as
an indicator.

10. Estimation of Iron
i) Make up given solution up to the mark with distilled water and
shake the flask for uniform concentration.
ii) Rinse the pipette with the ferrous solution and pipette out
20ml into a clean conical flask add 20ml of the acid mixture
(sulphuric acid and phosphoric acid), and four to five drop of
diphenylamine indicator.
iii) Fill the burette with potassium dichromate solution after
rinsing it, with the same solution.
iv) Titrate the solution in the conical flask against the standard
potassium dichromate from the burette till the colour changes to
blue violet.
v) Repeat the titrations for concordant titre values.

11. Concentration systems

12. How do we Express Concentrations of Solutions
Molarity (M) = moles/litre or mmoles/mL
Normality(N) = equivalence/litre or meq/mL
Formality(F) = is identical to molarity
Molality(m) = moles/1000g solvent

13. Concentration systems
The concentration of a solution may be expressed in
terms of molarity, Molality, formality and Normality:
a.) Molarity (moles/L, or M):
 Most common unit of concentration.
 The molarity of a solution is the number of moles
of solute per liter of the solution.
 It is represented by the symbol M.
 Molarity (M) =
solutionofLiter
soluteofMoles

14. Molality
• The molality is defined as the number of moles of
solute per kg of the solvent.
• Molality (m) =
solventkg
soluteofMoles

15. Molarity and Molality
Molarity
• Let us take 1 molar CaCl2
1 mol CaCl2 = 111g
1 molar CaCl2 =
Molality
• Let us take 1 molal CaCl2
1 molal CaCl2 =solutionofLiter1
111
solventofkg1
111

16. Formality
Formality (F) = solutionofLiter
soluteofweightsformulaofNumber
Formality, F, is the number of formula weight units of
solute per liter of solution . The purpose of formality
is to distinguish the number of moles of a compound
from the number of moles of ions in solutions of ionic
compounds or weak electrolytes

17. Normality
• The normality of a solution represents the
number of equivalents of solute contained in one
litre of the solution.
• A standard solution containing one gram
equivalent weight in one litre of the solution is
called 1N solution.
• If one litre of the solution contains two gram
equivalent weights of the substance then it is
called 2N solution.
Normality (N) =
litresinsolutionofVolume
sequivalentgramofNo

18. Normality
Normality (N) =
litresinsolutionofVolume
sequivalentgramofNo
mLinsolutionofVolume
sequivalentgrammilliofNo
Normality (N) =
convert gram to milligram equivalents and express
normality in terms of Milligram equivalents

19. millimoles and milliequivalents
• No of moles =
• No of millimoles =
• No of equivalents =
• No of milli equivalents =
weightMolecular
in gramWeight s
weightMolecular
in mgWeight
weightMolecular
valancegramsinweight x
weightMolecular
valanceinWeight mg x

20. 20
Millimoles (problem)
• Example: How many millimoles, of calcium chloride (CaCl2.
2H2O – m.w. 147) are represented in 1470 mg of calcium
chloride solution?
Solution:
Given weight of Calcium chloride is 1470 mg
mmol =
= 10 mmol
weightmolecular
mginweight
mmol =
147
1470

21. 21
Milliequivalents
• Example: How many milliequivalents, of calcium chloride
(CaCl2. 2H2O – m.w. 147) are represented in 1470 mg
calcium chloride solution?
Solution:
Given weight of Calcium chloride is 1470 mg
=
= 20 meq
weightmolecular
Valencemginweight
mEq
×
=
147
2x1470

22. • Example: How many a) millimoles, b)
milliequivalents, of calcium carbonate (CaCO3 –
m.w. 100) are represented in 3000 mg calcium
carbonate solution?

23. Problems based on molality
• What is the molality of a solution in which
3.0 M of NaCl is dissolved in 1.5Kg of water.
Molality (m) =
= 3.0
1.5
= 2m
solventkg
soluteofMoles