This presentation covers different titration methods principle and formula used to quantify pharmaceuticals products like table or capsules.

1. TITRATION METHODS FOR
QUANTIFICATION OF
PHARMACEUTICAL
PRODUCTS

2. Introduction
 Titration is analytical technique which allows the quantitative determination of a
specific substance (analyte) dissolved in a sample.
 It is performed by the slow addition of one solution of a known concentration
(called a titrant) to a known volume of another solution of unknown concentration
until the reaction reaches neutralization, which is often indicated by a color
change.
 The solution called the titrant must satisfy the necessary requirements to be a
primary or secondary standard.
 In a broad sense, titration is a technique to determine the concentration of an
unknown solution

3. Introduction
 The volume is considered as the signal of titration process.
 The end point for titration is determined using either color indicator or
potentiometric
 In Potentiometric, we measure the electric potential across the substance
 Titration use simple apparatus
 Titration techniques is not expensive

4. TYPES OF TITRATION
Titrimetric methods are classified also into four groups based on the type of reaction
involved.
These groups are:
 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;
 Redox titrations, where the titrant is an oxidizing or reducing agent;
 Precipitation titrations, in which the analyte and titrant react to form a
precipitate.

5. ACID-BASE TITRATION
 A titration in which the reaction between the analyte and titrant is an acid–base
reaction.
 According to Arrhenius: the definition of acid is a substance which yields
hydrogen ion (H+) in an aqueous medium’; and that of a base is—‘a substance
which yields hydroxyl ions (OH–) in an aqueous medium’.
 According to Lowry and Bronsted’s theory—‘an acid is a substance capable of
yielding a proton (hydrogen ion), while a base is a substance capable of accepting
a proton’.
 According to Lewis—‘an acid is an electron pair acceptor, whereas a base is an
electron pair donor’.

6. ACID-BASE TITRATION ( Cont.)
 According to the Lux-Flood concept—‘an acid is the oxide-ion acceptor while a
base is the oxide donor’.
 According to Usanovich
 Acid : It is a chemical species that reacts with a base thereby giving up cations or
accepting anions or electrons.
 Base : It is a chemical species that reacts with an acid thereby giving up anions or
electrons or combines with cations
 In an acid-base titration, the acid will not release a proton unless the base capable
of accepting it is simultaneously present

7. ACID –BASE TITRATION( cont.)
 The acid may be strong or weak and the base may be also Strong or weak
 Strong acid and bases titrated in aqueous medium
 Weak acid and bases can not be titrated in aqueous medium because water itself
can act as an acid or base in this case the titration is performed in a non-aqueous
environment
 in acid base titration, we use indicator solution to indicate an end point
 Acid –base titration may be carried out in different manner like, direct method and
back titration

8. Acid- base titration: direct titration
 The sample is titrated directly with suitable standard solution
 The amount of reagent consumed to the equivalent point is the amount of
substance to be determined
 Mathematically: M1× V1 = M2×V2
Where:
M1: Concentration of standard solution
M2: concentration of sample
V1: Volume of standard solution consumed to reach end point
V2: Volume of Sample

9. Acid- base titration: direct titration( cont)
 Example: Assay of Aspirin Tablets BP 75 mg
Two tablet of aspirin tablet taken during quantification analysis of new brand of aspirin. Each tablet have
been claimed to contain 75 mg of API, those tablet was dissolved with 50% ethanol and titrated with
phenolphthalein indicator to the end point with 8ml of 0.1N NaOH solution
 Molecular Formula, C9H8O4 or CH3COOC6H4COOH
 acetylsalicylic acid is a weak Acid
 Ethanol is used to dissolve aspirin
 Sodium hydroxide(0.1M NaOH) is used as standard solution
 Phenolphthalein is used as indicator

10. Acid-base titration: direct titration( cont)
 Procedures for quantitative determination of aspirin
i. Standardize sodium hydroxide solution
ii. Take two tablets and weight them to get its mass
iii. Dissolve them in 50% ethanol solution
 Data and calculations
i. Mass of two tablets: 0.168g
ii. Average weight of tablet: 0.084g
iii. Volume of NaOH used to neutralize aspirin: 8ml
iv. Molecular mass of aspirin: 180.158g/mol

11.  Aspirin (acetyl salicylic acid reacts with sodium hydroxide according to the
following equation:
C9H8O4 + 2OH- → C7H5O3
- + C2H3O2
- + H2O
 1mole of aspirin consume 2mole of sodium hydroxide
 Calculations:
i. Number of moles of NaOH reacted:
8 𝑚𝑙 ×0.100𝑚𝑜𝑙
1000𝑚𝑙
= 0.0008 mole
ii. 0.0008 moles neutralize 0.0004 moles of Aspirin
iii. 1 moles of Aspirin contains 180.158g

12.  The mass of aspirin in tablets= actual number of moles x molecular mass
 Mass of aspirin = 0.0004 moles x 180.158g/mol=0.0720603g
 Each tablet contains 72.06 mg
 Percentage assay:
𝑎𝑐𝑡𝑢𝑎𝑙 𝑚𝑎𝑠𝑠
𝑡ℎ𝑒𝑜𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑚𝑎𝑠𝑠
× 100 =
72.06 𝑚𝑔 ×100
75 𝑚𝑔
= 96.1%
 Shortly percentage assay=
V×N×Mm×100×Av wt
LC×wt spl taken
 By considering the specification of USP where the API must be not less than 90
and not more than 110, we can conclude that the sample meets the specifications

13. Acid-base titration ( back titration)
 A titration in which a reagent is added to a solution containing the analyte, and
the excess reagent remaining after its reaction with the analyte is determined by a
titration.
 Back titration or residue titration is normally employed in the following two
situations, namely : Case I : when a chemical reaction proceeds rather slowly or
sluggishly, Case II : when the substance under determination fails to give a sharp
and distinctly visible end-point with an indicator by direct titration.
 For example: assay of Aspirin tablet

14. Example: assay of aspirin
 1.427 g of aspirin was dissolved in 50% ethanol and 50.00 mL of 0.500 mol/L
sodium hydroxide solution was added. Then the excess sodium hydroxide have
been titrated to a phenolphthalein end-point with 31.92 mL of 0.289 mol/L
hydrochloric acid.
 sample solution was heated in water bath to speed up hydrolysis
 only base remaining after the reaction is the excess base that has not reacted with
the aspirin
 Titrate the excess sodium solution with HCl until the pink color disappear
 You are doing a strong acid-strong base titration.

15. Data and Calculations
 Step 1. Start with the balanced equation.
C9H8O4 + 2OH- → C7H5O3
- + C2H3O2
- + H2O
 Step 2. Calculate the moles of NaOH used.
Moles NaOH=0.050 00L NaOH×0.500 mol NaOH1L NaOH=0.025 00 mol
 Step 3. Calculate the moles of HCl used in the back-titration.
Moles of HCl=0.031 92L HCl×0.289 mol HCl1L HCl=0.009 225 mol
 Step 4. Calculate the moles of excess NaOH
NaOH+HCl→H2O+NaCl
Moles of excess NaOH=0.009 225mol HCl×1 mol NaOH1mol HCl=0.009 225 mol

16. Data and Calculations
 Step 5. Calculate the moles of NaOH used in the reaction.
Original moles = moles reacted + excess moles
Moles reacted=original moles - excess moles=0.02500 - 0.009 225=0.015 775 mol
 Step 6. Calculate the moles of aspirin.
1 moles of aspirin consumed by 2 moles of NaOH
Moles of Asp=
0.015 78mol
2mol
=0.007 888 mol
 Step 7. Calculate the mass of aspirin.
Mass of Asp=0.007 888mol Asp×180.16 g Asp1mol Asp=1.421 g
 Step 8. calculate the percentage Assay of the aspirin.
Percent by mass
mass of pure Asp
mass of impure Asp
× 100=
1.421g
1.427g
×100 = 99.6%

17. Complexation Titration
 A titration in which the reaction between the analyte and titrant is a complexation
reaction.
 Complex is a compound that is formed by the combination of a metal ion with a
molecule that is capable of donating electrons,
 It is used in determination of metal salts
 It is performed by either direct or back titration
 EDTA is commonly used as titrant
 Titration with EDTA is pH dependant

18. COMPLEXATION TITRATION
 The metals that react strongly with EDTA can be titrated in acidic solution. Zinc is an
example. The metals that react more weakly with EDTA must be titrated in alkaline
solution. Calcium and magnesium are examples
 The end point is detected using dye indicator
 The general equation is:
 Insoluble metals are estimated by back titration
 Excess EDTA is titrated with salts solution containing Zn or Mg ion of known
concentration.

19. Example
 An accurately weighed portion of calcium acetate tablet powder, equivalent to
about 300 mg of calcium acetate, Dissolved and diluted to 100ml with
hydrochloric acid and sodium hydroxide.300 mg of hydroxy naphthoblue were
used as indicator. Then 2ml of sample of the total solution were titrated with the
0.01 M edetate disodium solution to a blue endpoint. The mean corrected titration
volume was 14.6mL

20. Answer
 Step 1: write a balanced Equation
Ca2+
(aq) + EDTA4-
(aq) → Ca(EDTA)2-
 Step 2: calculate moles of EDTA
mol of EDTA:Volume EDTA (L) × Molarity EDTA (mol/ L
moles ol of EDTA =
14.6mL×0.01mol
1000ml
=0.000146 mol

21. Answer
 Step 3: calculate number of moles of Calcium
from the equation one moles of EDTA consumed by one mole of Calcium
Moles of Calcium=0.000146 moles
 Step 4: calculate mass of Calcium reacted
mass of calcium
0.000146mol ×40.08g×1000mg
1mol×1g
=5.85168mg
The total mass of Calcium=
5.8516mg ×100ml
2ml
= 292.584mg
 Percentage assay=
𝑎𝑐𝑡𝑢𝑎𝑙 𝑚𝑎𝑠𝑠
𝑡ℎ𝑒𝑜𝑟𝑖𝑡𝑖𝑐𝑎𝑙
×100 =
292.584mg
300mg
×100 = 97.5%