types of titrations
TRANSCRIPT
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
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.
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.
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 M
2+ + 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)
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)
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)
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.
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.
Estimation of Ironi) 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.
Concentration systems
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
Concentration systems
The concentration of a solution may be expressed interms 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
Molality
• The molality is defined as the number of moles of solute per kg of the solvent.
• Molality (m) =solventkg
soluteofMoles
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 = solutionofLiter 1
111
solventofkg 1
111
Formality
Formality (F) = solutionofLiter
solute of weightsformula ofNumber
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
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
Normality
Normality (N) =litresinsolutionofVolume
sequivalentgramofNo
mLinsolutionofVolume
sequivalentgrammilliofNo
Normality (N) =
convert gram to milligram equivalents and express normality in terms of Milligram equivalents
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
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
mginweightmmol
=
147
1470
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
ValencemginweightmEq
×=
147
2 x 1470
• Example: How many a) millimoles, b) milliequivalents, of calcium carbonate (CaCO3 – m.w. 100) are represented in 3000 mg calcium carbonate solution?
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