Pharmaceutical Analytical Chemistry I

• Pharmaceutical Analytical Chemistry Department will offer Pharmaceutical Analytical Chemistry (I) in 2024-2025.

Course Grading

• Practical work: 25 Marks, including a 20-mark exam and a 5-mark evaluation.
• Periodicals: 15 Marks from 3 Quizzes.
  • Q1 in week 4 is worth 4 marks.
  • Q2 in week 7 is worth 6 marks.
  • Q3 in week 10 is worth 5 marks.
• Final Exam: 50 Marks.
• Oral Exam: 10 Marks.

Analytical Chemistry Types

• Qualitative Analysis determines what is present in a sample.
• Quantitative Analysis determines the quantity and purity of a sample, thus the concentration, and amount of impurities.
• Determination: It involves the measurement of a single constituent in a simple way.
• Quantitative analysis requires preparative work, such as separating interfering substances, and good interpretation.

Quantitative Analysis Classification

Based on sample concentration:

• Macro-: 100 mg (0.1 g) or more.
• Semi-micro: 10 - 100 mg (0.01 – 0.1 g).
• Micro-: Quantities not exceeding 10 mg.

Based on technique:

• Volumetric (titrimetric) analysis: Measuring the volume of a standard that is equivalent to a certain volume of sample.
  • Includes acid-base, precipitation, complex-formation, & redox titrations.
• Gravimetric analysis involves isolating and weighing the final product with known pure, stable, and definite form.
• Physicochemical (instrumental) analysis involves determining concentration by measuring some physical properties using an instrument.

Volumetric Analysis

• Volumetric analysis, or titration, is a lab method where a substance of known concentration and volume reacts with another of unknown concentration.
• Titration determines the amount of substance A by adding a measured volume of solution with a known concentration of B until the reaction is complete.

Titrimetric Reaction Requirements

• The reaction must be rapid (instantaneous) NaOH + HCl → NaCl + H2O.
• The reaction must be complete.
• A balanced chemical equation (stoichiometric) must represent the reaction.
• The reaction should be a single reaction without side reactions.
• A suitable standard solution must be available as a titrant.
• The endpoint of the reaction should be easily detected.
  • A suitable indicator should change the solution color at the end point.
  • Changes in physical or chemical properties of the solution at the endpoint will occur.

Measurements

• Accurate measurements: tools like syringe, volumetric flask, burette, and bulb pipette.
• Rough Measurements: Tools like beakers, graduated cylinders, and graduated pipettes.

Titration Methods

• Direct titration involves stepwise addition of the titrant until the reaction completes, indicated by a color change.
• Back titration involves adding a known excess of standard solution to the sample.
• After reaction completion, the remaining unreacted excess is titrated with another standard.
• This method is suitable for:
  • Water-insoluble samples (e.g., ZnO, CaO, CaCO3).
  • Volatile samples (e.g., HCOOH).
  • Slow chemical reactions (e.g., NH4Cl, lactic acid).
  • Cases of absence of a suitable indicator.
  • Titrations needing filtration and heating.

The Mole Concept

• A mole of substance represents a distinct quantity of particles (atoms, molecules, or ions).
• The amount with exactly 6.022 × 1023 elementary entities (atoms, molecules, or ions).
• A mole contains 6.022 × 1023 particles (Avogadro's number).
• 1 mole of atoms is 6.022 x 1023 atoms, 1 mole of ions is 6.022 x 1023 ions, and 1 mole of molecules is 6.022 x 1023 molecules.
• The mass of 1 mole of an element equals its molar mass in grams.
  • For elements: Molar mass equals atomic mass, e.g., 1 mole of Na atom = 22.99 g/mol.
  • For molecules: Molar mass equals molecular mass (molecular weight), e.g., 1 mole of H2O molecule = 18.016 g/mol.

Calculating Moles

• Moles = Mass (g) / Molar Mass (g/mol).
• Mass (g) is the given mass of the substance.
• Molar Mass (g/mol) is the sum of atomic masses.
• Example: Number of moles of Water (H₂O) in 54 g.
  • Molar mass of H2O = 2 X (1.008) + 16.00 = 18.016 g/mol.
  • Moles = 54/18.016 = 3.

Molecular Weight

• Molecular weight (M.Wt) is the sum of all atomic weights in a compound.
• H=1, Cl=35.5, P=31, S=32, C=12, O=16, Na=23.
• M.Wt of H₂SO₄ = (1x2) + (32) + (16x4) = 98 g/mol.
• M.Wt of HCl = (1) + (35.5) = 36.5 g/mol.
• M.Wt of H3PO4 = (1x3) + (31) + (16x4) = 98 g/mol.

Equivalent Weight

• Equivalent Weight (Eq.Wt) is the weight of a substance equivalent to 1 mole of hydrogen's reactive power.
• The definition and calculation of equivalent weights depends on the reaction type, Equivalent weight = M.wt/n.
  • Eq.Wt of acids = Molecular weight of acid / number of replaceable H+.
  • Eq.Wt. of base = Molecular weight of base / number of replaceable OH-.
  • Eq.Wt of salt = Molecular weight of salt / number of metal ion (cations)(valence)

Equivalent Weight Calculations

• H=1, Cl=35.5, P=31, S=32, C=12, O=16, Na=23, Ca=40.
• Eq. Wt of HCl = (1) + (35.5) / 1 = 36.5 g/eq.
• Eq.Wt of H2SO4 = (1x2) + (32) + (16x4) / 2 = 49 g/eq.
• Eq.Wt of NaOH = (23) + (16) + (1) / 1 = 40 g/eq.

Expressing Solution Concentration

• There exist several ways of expressing concentration of a solution:
  • Molarity (M)
  • Normality (N)
  • Formality (F)
  • Molality
  • Percent
  • Mole fraction

Molarity

• Molarity (M) is the number of moles of solute in 1 liter of solution.
• Moles = Mass of solute (g) / Molar Mass or M.wt (g/mol).
• 0.5 M NaOH contains 0.5 mole of NaOH per 1 L of solution.
• If there are 4 moles of NaOH dissolved in 1 liter of water, the molarity is 4 M.
• For example, 80 grams of NaOH in 1 liter of water is a solution of 2 M concentration.
• 3.6 grams of NaOH in 2 liters of water has a molarity of 0.045 M.

Molarity Problem Examples

• For a 1 M NaOH solution (40 grams of NaOH per liter):
  • Mass of NaOH = 0.5 moles × 40.00 g/mol = 20 grams for 0.5 M NaOH in 1 L.
  • Mass of NaOH = 0.1 moles × 40.00 g/mol = 4 grams for 0.1 M NaOH in 1 L.

Normality

• Normality (N) is the number of equivalent weights of the solute per liter of solution.
• Equivalent weight = M.wt / n.
• (M) = (N) for monoprotic acids or bases.
• Relationship between normality and molarity is N = n × M, where n is the number of equivalents.

Calculations for Equivalent Weight and Normality

• Molar weight of phosphoric acid equals 97.994 g/mol.
• Equivalent Weights and Normalities determined reactions:
  • EW = 97.994/3= 32.665 and N=n x M = 3 x 6.0 = 18 N.
  • EW = 97.994/2= 48.997 and N=n x M = 2 x 6.0 = 12 N.
  • EW = 97.994/1= 97.994 and N=n x M = 1 x 6.0 = 6.0 N.

Standard Solutions

• Standard solutions have known concentration and composition.
• There are two types of standard solutions: primary standard and secondary standard.

Primary Standard

• A solution with exactly known concentration and composition, remains constant over time.
• Prepared directly by weighing an exact weight of the primary standard substance, dissolving in a volumetric flask with a proper solvent.
• Requirements:
  • Easily obtained in high purity and known composition.
  • Very stable.
  • Non-hygroscopic and non-volatile.
  • It can be dried at 105-110°C without decomposition.
  • Relatively high molecular weight to minimize error.
  • It must react quantitatively with other substances according to balanced chemical equations.
  • Potassium acid phthalate KHC8H4O4, Oxalic acid, and Benzoic acid are examples of primary standard acids.
  • Sodium carbonate and Borax (sodium borate) are primary standard bases.

Secondary Standard

• A solution with an approximately known concentration and can not be calculated directly from the solute weight and solution volume.
• The exact concentration of secondary standard is determined through standardization.
• Standardization process is used to determine the exact concentration of secondary standard and correct any error:
  • Titrating the 2ry standard solution against a 1ry standard solution.
  • Titrating the 2ry standard solution against a previously standardized 2ry standard.
• Examples: HCl and NaOH.

Standardization with 0.1N Na2CO3:

• Transfer 10 ml of 0.1N Na2CO3 into a conical flask, add 2-3 drops of methyl orange indicator.
• Titrate with 0.1N HCl until the indicator turns from yellow to orange.
• Repeat the titration and calculate the normality of HCl using: VA x NA = VB x NB.
• VA and VB are the volumes of acid and base at the endpoint.

Dissociation Theory

• Degree of dissociation (α) describes equilibrium between molecules.
• Molecules disassociate into cations and anions.
• Degree of dissociation (α) = Number of solute molecules dissociated / Number of solute molecules before dissociation.
• Complete dissociation (α ≈ 1) makes a strong electrolyte. A value far from one (α ≈ 0) is a weak electrolyte.