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Pharmaceutical Analytical Chemistry

Pharmaceutical Analytical Chemistry. Dr. Rania Mohammed Elgamal. Course Topics. Acid-Base Titration 6 hours Precipitation and Complex-formation Titration 4 hours Oxidation-reduction Titration 4 hours Electrochemical methods 4 hours

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Pharmaceutical Analytical Chemistry

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  1. Pharmaceutical Analytical Chemistry Dr. Rania Mohammed Elgamal

  2. Course Topics • Acid-Base Titration 6 hours • Precipitation and Complex-formation Titration 4 hours • Oxidation-reduction Titration 4 hours • Electrochemical methods 4 hours • Ultraviolet/visible spectrophotometry 6 hours • Introduction to chromatographic separation 4 hours

  3. Objectives By the end of the course you will be able to :  • Check the feasibility of titrimetric reactions. You shall be also able to choose the suitable indicator and derive the titration curve. • Calculate the percentage purity of solid (powdered) samples and the concentration of liquid samples.  • Obtain the absorption spectrum of the light-absorbing compound and define its max.  • Know the principles of chromatographic separation of pharmaceutical compounds in a mixture. • Check the feasibility of electrochemical methods of analysis.

  4. Course Evaluation • Continuous Assessment: • First Assessment Test 15% • Second Assessment Test 15% • Term Activity* 5 % • Laboratory attendance 5% • Laboratory reports and participation 10% • Final Laboratory Test 10% • Final Examination: • Final Paper testFinal Exam 40%

  5. References • Fundamentals of Analytical Chemistry, Douglas A. Skoog and Donald M. West. Fourth edition. Sanders College Publishing, Philadelphia (1984). • Analytical Chemistry, Douglas A. Skoog; Donald M. West, F. James Holter, Standey R. Crouch, 7th ed. Harcourt College Publishers (2000). • Principales of Quantitative Chemical Analysis, Robert de Levie. McGraw Hill, New York (1997). • Vogel’s Textbook of Quantitative Inorganic Analysis, 4th ed. J. Baisett, R.C. Denney, G.H. Jefferg and J. Mendham, Longman, Essex (1978) The handouts are for guidance and studying must be from textbooks

  6. Pharmaceutical Analytical Chemistry • Analytical chemistry deals with methods used for determining the composition of various materials. • The process of material identification called Qualitative Analysis . • The process of material quantitation called Quantitative Analysis .

  7. Areas of Chemical Analysis and questions they answer • Identification What is the identity of the substance in the sample? • Quantitation How much of the substance x is in the sample? • Detection Does the sample contain substance X or not? • Separation How the species of interest can be separated?

  8. Quantitative Chemical Analysis • Classification of Quantitative methods: a-According to the quantity to be analyzed • 1- Micro methods used for the determination of quantities less than 1 mg. • 2- Semi-micro methods used for determination of quantities ranging from 1-100 mg. • 3- Macro methods used for determination of quantities more than 100 mg.

  9. Quantitative Chemical Analysis b-According to technique • I- Volumetric or Titrimetric methods Analysis by volume. • II- Gravimetric methods Analysis by weight. • III- Instrumental methods (Physicochemical methods) • Electrochemical methods • Spectroscopic methods • Separation methods

  10. Volumetric Analysis • It is the quantitative chemical analysis carried out by determining the volume of a solution of accurately known concentration which is required to react quantitatively with measured volume of solution of the substance to be analyzed. • The solution of accurately known concentration is called the standard solution (titrant).

  11. Volumetric Analysis • The process of adding standard solution gradually to the sample until the reaction is just completed is termed as titration. • The point at which the reaction is completed is called end point orequivalence point. • The concentration of the substance to be analyzed is calculated from the volume of the standard solution.

  12. Detection of End Point 1- Physical change produced by the standard solution itself (Self indicator). 2-The Addition of a substance known as indicator. Indicators: are Compounds which have different colors at different conditions.

  13. Requirements for Quantitative Titrimetric Analysis • The reaction between the sample and the standard solution must be simple and can be represented by a chemical equation. • The reaction must be instantaneous (relatively fast or rapid). Sometimes catalyst is needed. • The substance to be determined should react completely with the titrant in stoichiometric manner (definite ratio). • The end point of the reaction can be detected easily. (indicator is available).

  14. Reactions Used in Titrimetric Analysis I- Neutralization Reactions (Acid-Base Reactions) II-The Precipitation Reactions (Precipitimetry) III- Complex Formation Reactions (Complexometry) IV-Electron-transfer Reactions (Redoximetry)

  15. Standard Solutions • These are solutions of exactknown concentration • Types of standard solutions • 1-Molar standard solution (M) • It is the solution which contains the gram molecular weight of the substance in 1L of solution. • 1M solution contains 1 x gm m.wt of substance/L of solution. 2M solution contains 2 x gm m.wt of substance/L of solution. M/10 solution contains 0.1 x gm m.wt of substance/L of solution.

  16. Molar Standard Solutions (M) -Examples Molar standard solution (M) 1M solution of H2SO4 contains 98.07 gm/L of solution. 2M solution of H2SO4 contains 196.14 gm/L of solution. M/10 solution of H2SO4 contains 9.8 gm/L of solution. 1M solution of NaOH contains 40 gm/L of solution. 2M solution of NaOH contains 80 gm/L of solution. M/10 solution of NaOH contains 4 gm/L of solution. 1M solution of Na2CO3 contains 106 gm/L of solution. 2M solution of Na2CO3 contains 212gm/L of solution. M/10 solution of Na2CO3 contains 10.6 gm/L of solution.

  17. Normal Standard Solutions (N) • Solution which contains gm equivalent weight /L of solution. Equivalent Weight • Eq.Wt of acids = m.wt / no. of replaceable H+ Example Eq.Wt of HCl = m.wt / 1 Eq.Wt of H2SO4 = m.wt / 2 • Eq.Wt of bases = m.wt / no. of replaceable OH- Example Eq.Wt of NaOH = m.wt / 1 Eq.Wt of Ba(OH)2 = m.wt / 2 • Eq. Wt For Salts = m. wt/( number of metal x its charge ) Examples NaCl eq. wt = m.wt / 1 CaCl2 eq. wt = m.wt / 2 -N.B. Equal volumes of equal normalities contain equal number of molecules, that means equal normalities react 1 to 1 ratio.

  18. Neutralization Reactions Acid-Base Titrations In Aqueous Solution • Solutions • Solution is a homogenous mixture of two or more substances. The component (solid, gas or liquid ) present in small quantity is called the solute, while the one present in large quantities is called the solvent . • Solutions may be 1- Saturated solutions . 2- Unsaturated solutions . 3- Supersaturated solutions .

  19. Electrolytes and Non-electrolytes I-Electrolytes: Electrolytes are substances when dissolved in water undergo dissociation and give electricity-conducting solutions. Electrolytes may be : 1-Strong Electrolytes : Substances when dissolved in water dissociate or ionize to a high degree. Examplesof strong electrolytes . Acid: HCl , HNO3 , H2SO4 , HBr , HI . Base: NaOH , KOH , Ca(OH)2 , Ba(OH)2. Salt: NaCl , CH3COONa , NH4Cl.

  20. Electrolytes and Non-electrolytes 2-Week Electrolytes: Substances when dissolved in water dissociate or ionize to a slight degree. Examplesof week electrolytes . - Acid: CH3COOH , HCN , H2S , H3BO3 , HF . - Base: NH4OH , N2H4 . - Salt: HgCl2 , CdCl2 , HgBr2, CH3COONH4 . II-Non-electrolytes: Non-electrolytes are substances when dissolved in water do not undergo dissociation and give a non-conducting solutions. Examples: Sugar , Glycerin , Ethyl acetate.

  21. Electrolytic Dissociation Theory • Pure water is a bad conductor for electricity. • When an electrolyte is dissolved in water, it dissociates into negatively charged ions (anions) and positively charged ions (cations). • Solutions conduct the electric current due to the presence of ions. • The degree of dissociation is directly proportional to the degree of dilution.

  22. Degree of Dissociation (α) • It is the ratio of the ionized fractions to the total amount of the dissolved solute. • For each concentration there is a state of equilibrium between the un-dissociated molecules and the dissociated molecules (ions). Molecule = Cation (+ve) + Anion (-ve) CH3COOH = H+ + CH3COO- NH4Cl = NH4+ + Cl- • The degree of dissociation characterizes the chemical activity of the respective substance.

  23. Molecular and Ionic Equations Molecular equations: represent the reaction species (reactant and products ) as molecules . NaOH + HCl → NaCl + H2O This equation shows that one mole of NaOH neutralize one mole of HCl to form exactly one mole of NaCl and one mole of H2O . In ionic equations: strong electrolytes are represented as ions while weak electrolytes represented as molecules . In the above equation NaOH and HCl are strong electrolytes and present as ions in the solution , so that , the equation can be written as follows: Na+ + OH- + H+ + Cl- → Na+ + Cl- + H2O OH- + H+ → H2O In the reaction of NaOH (strong electrolyte ) and CH3COOH (week electrolyte ) ,the equation is written as follows: Na+ + OH- + CH3COOH → Na+ + CH3COO- + H2O OH- + CH3COOH → CH3COO- + H2O

  24. Chemical Equilibrium • In reversible reactions products are formed from the reactants and the reactants are being produced from the products. A + B = C + D Reactants ↔ products • Under that condition the composition of the reaction mixture becomes constant and the system is said to be in a state of equilibrium which is the state at which the rate of forward reaction equal to the rate of backward reaction .

  25. Law of Mass Action The rate of chemical reaction is directly proportional to the product of the molar concentration of the reacting substances. For the reaction A + B = C + D Vfα [A] [B] or Vf = K1 [A] [B] Vbα [C] [D] or Vb = K2 [C] [D] - At equilibrium Vf = Vb K1 [A] [B] = K2 [C] [D] - K1 / K2 = k equilibrium (equilibrium constant) Keq = [C] [D] / [A] [B] - In case of : aA + bB = cC + dD Keq = [C]c [D]d / [A]a [B]b

  26. Displacement of Equilibrium Le Chatelier Principle: According to Le Chatelier principle, if a stress is applied to a system in an equilibrium state , the equilibrium will be shifted in such direction to minimize that stress. Applications of Le Chatelier Principle: In Precipitation : A + B = AB (precipitate) Precipitating agent B is used to precipitate the compound AB by combining with A to form more AB and the equilibrium is shifted to the right . In Solubility: In endothermic solution , the solubility of the solute increases by heating (equilibrium shifted to right ) . solute + solvent + heat = solution In exothermic solution , the solubility of the solute decreases by heating (equilibrium shifted to left ) . solute + solvent = solution + heat

  27. Theories of Acids and Bases 1-Arrhenius theory: -An acid forms H+ in water (upon ionization) HCl → H+ + Cl- HNO3 → H+ + NO3- H2SO4 → 2H+ + SO4-- - A base forms OH- in water (upon ionization) Na OH → Na+ + OH- Ca(OH)2 → Ca+ + + 2OH- N.B. 1- Not all acid-base reactions involve water 2- Many bases (NH3, and carbonate) do not contain any OH-

  28. Theories of Acids and Bases 2-Bronsted - Lowry theory: -Acid is a proton donor H+ Acid → H+ + Conjugate base HCl → H+ + Cl- - Base is a proton acceptor H+ Base + H+ → Conjugate acid NH3+ + H+ → NH4+ Theconjugate base of an acid is the acid minus the proton it has donated The conjugate acid of a base is the base plus the accepted proton

  29. Theories of Acids and Bases 3-Lewis theory: -Base is a substance containing an atom that has unshared pair of electrons e.g. N , O , P , S (base is an electron donor e.g. NH3, amines like triethylamine). -Acid is a substance that can accept that pair of electrons e.g. AlCl3 , BCl3 , BF3 example of Lewis acid Lewis base reaction: H3N: + BF3 → H3N: → BF3 Lewis Base + Lewis acid

  30. Dissociation of Water H2O = H+ + OH- -Water molecules ionize in very slight degree. -According to the law of mass action K = [H+ ] [OH- ] / [H2O ] K [H2O ] = [H+ ] [OH- ] Kw = [H+ ] [OH- ] Kw = ionic product of water -It was found that under normal experimental conditions and at 250c Kw = [H+ ] [OH- ] = 10-14 -Since the dissociation of water gives rise to equal number of H+ and OH- Kw = [H+ ]2 =10-14 [H+ ] =√ 10-14 = 10-7

  31. Hydrogen Ion Exponent pH is the measure of acidity or alkalinity solution . -pH is the negative logarithm of the hydrogen ion concentration pH =-log [H+] pH range 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Acidic NeutralBasic -pH is a number obtained by giving a positive value to the negative power of 10 in the expression . [H+] = 10-n pH = n [H+] = 10-5 pH = 5 Kw = 10-14 pKw = 14 -In general : for acids pH = -log [H+] for bases pOH = -log [OH-] pKw = pH + pOH pH = pKw - pOH = 14 - pOH

  32. pH of acids and Bases pH of strong acids and strong bases -Strong acid and strong base are completely ionized so, concentration of acid or base represents the concentration of [H+] or [OH-] . For acids pH =-log [H+] For bases pOH = -log [OH-] pH = 14 - pOH For examples: pH of 0.1 M HCl (strong acid) pH =-log [H+] = -log 10-1 = 1 pH of 0.1 M NaOH (strong base) pOH = -log [OH-]= -log 10-1 = 1 pH = 14 - pOH =14- 1= 13

  33. pH of acids and Bases pH of weak acids -A Small quantity of weak acid is dissociated with the formation of [H+] . e.g. CH3COOH CH3COOH = CH3COO- + H+ Ka = [CH3COO- ] [H+] / [CH3COOH] Where: [H+] = [CH3COO- ] [CH3COOH]= Ca (concentration of acid) Ka = [H+] 2 / Ca [H+] 2 = Ka Ca [H+] = √Ka Ca pH = ½ pKa + ½ pCa pH = ½ (pKa + pCa)

  34. pH of acids and Bases Examples Calculate the pH of 0.1 M solution of acetic acid (Ka =1.75x10-5) pH = ½ pKa + ½ pCa = ½ (-log 1.75 x 10-5)+ ½ (-log 0.1) = (0.5 x 4.757)+ (0.5 x 1) = 2.88 Calculate the pH of 0.25 M solution of formic acid (Ka =1.76x10-4) pH = ½ pKa + ½ pCa = ½ (-log 1.76 x 10-4)+ ½ (-log 0.25) = (0.5 x 3.754)+ (0.5 x 0.602) = 1.877+ 0.301 = 2.18

  35. pH of acids and Bases pH of salts -Salts of strong acids and strong bases e.g. NaCl is neutral pH = 7 -Salts of strong acids and weak bases e.g. NH4Cl pH = ½ (pKw - pKb + pCs ) -Salts of weak acids and strong bases e.g. CH3COONa pH = ½ (pKw + pKa - pCs ) -Salts of weak acids and weak bases e.g. CH3COONH4 pH = ½ (pKw + pKa - pKb )

  36. Buffer Solutions Buffer solutions are solutions which resist the change in the pH of solution upon addition of small amount of strong acid or strong base - Types of buffer solutions 1-Acidic buffer solutions Consists of weak acid and its salt of strong electrolyte. e.g. acetic acid and sodium acetate (CH3COOH/CH3COONa) -Upon addition of a strong acid: sodium acetate react with it giving weakly ionized acetic acid H+ + CH3COONa → CH3COOH + Na+ -Upon addition of a strong base: acetic acid react with it and unionized water is formed OH- + CH3COOH → CH3COO- + H2O

  37. Buffer Solutions 2- Basic Buffer solutions Consists of weak base and its salt of strong electrolyte. e.g. ammonium hydroxide and ammonium chloride (NH4OH/NH4Cl) -Upon addition of a strong acid: H+ + NH4OH → NH4+ + H2O -Upon addition of a strong base: OH- + NH4Cl → NH4OH + Cl-

  38. Henderson Equation for calculation of pH of buffer solutions pH of acidic buffer: pH = pKa + log Cs/Ca pH of basic buffer: pOH = pKb + log Cs / Cb pH = pKw - pOH pH = pKw - pKb - log Cs/Cb OR pH = pKw - pKb + log Cb/Cs

  39. Examples 1- Calculate the pH of a buffer solution consisting of 1 M CH3COOH and 1 M CH3COONa where Ka=1.75x10-5 pH = pKa + log Cs/Ca = -log 1.75 x 10-5 + log 1/1 = 4.76 + 0 = 4.76 2- Calculate the pH of a buffer solution consisting of 0.5 M NH4OH and 0.3 M NH4Cl where Kb = 1.8 x 10 5 pH = pKw – pKb + log Cb/Cs = 14 – log 1.8 x 1 0 5 + log 0.5/0.3 = 14 – 5.255 + 0.222 = 8.967

  40. Examples Calculate the pH of a buffer solution consists of 1 M CH3COOH and 1 M CH3COONa after addition of 0.1 mol of HCl to one L of solution where Ka=1.75x10-5 after addition of 0.1 mol HCl , it will react with an equivalent amount of CH3COONa forming the same amount of CH3COOH HCl + CH3COONa → CH3COOH + NaCl 0.1 mol 0.1 mol 0.1 mol 0.1 mol Ca = 1+ 0.1 = 1.1 Cs = 1 - 0.1 = 0.9 pH = pKa + log Cs/Ca = -log 1.75 x10-5 + log 0.9/1.1 = 4.757 + ( 0.087) = 4.67

  41. Buffer Capacity It is the number of gram equivalents of strong acid or strong base required to cause a change of 1 pH unit in one liter of the solution B =ΔB / Δ pH - B is a buffer capacity - ΔB is a strong acid or base added - Δ pH is the change in pH Buffer capacity is directly proportional to concentration of buffer components Solution has equal concentration of acid or (base) and its salt appears to have the maximum buffer capacity - Buffer solution with high B is of high efficiency

  42. Neutralization Indicators Neutralization indicators are weak acids or weak bases which change their color according to the pH of the solution • The acid form (HA) of the indicator has one color, the conjugate base (A–) has a different color. • In an acidic solution, [H+] is high. Because H+ is a common ion, it suppresses the ionization of the indicator acid, and we see the color of HA. • In a basic solution, [OH–] is high, and it reacts with HA, forming the color of A–. • The change of color is not sudden but takes place within small interval of pH(2 pH units or less) • It is preferred to select an indicator which exhibits color change at pH close to that of salt formed at the end point

  43. Color change of some indicators

  44. Types of Neutralization Indicators 1 – Color indicators Organic dyes that exhibit different colors at different pH values - Methyl Orange (M.O.) pH range 3.3-4.4 red to orange or yellow, - Phenolphthalein(Ph.Ph.) pH range 8.3-10 colorless to pink - Methyl Red (M.R.) pH range 4.4-6.3 red to yellow 2 -Turbidity indicators Precipitation or turbidity appears at the end point e.g. Isonitrosoacetyl-p-aminobenzene 3 –Fluorescence indicators Certain compounds emit visible radiations when exposed to ultraviolet light. This radiation stop or intensify when certain pH is reached and used to detect end point. These indicators are useful when color or turbid solutions are titrated e.g. Umbelliferone

  45. Theories of Color indicators 1 – Ostwald Theory: - Neutralization indicators are either weak acids or weak bases - The color of ionized form differs from that of non-ionized form - In acidic medium basic indicators ionized and changed in color e.g. M.O. - In basic medium acidic indicators ionized and changed in color e.g. Ph.Ph. 2 – Chromophore theory - Indicators are Organic dyes which contain an unsaturated group called chromophore group e.g. C=C , N=N , C=N , NO, NO2 which is responsible for the color change. - Accumulation of unsaturated groups leads to color development - Presence of auxochromes (-OH, -NH2 ) influence the color.

  46. Effective Range of Color Indicators It is the pH units over which the indicator changes its color. The color change within the effective range is gradual. Effective range for a good indicator shouldn’t exceed 2 pH units. Example: M.O. (3.3-4.4), M.R. (4.4-6.3) Ph.Ph. (8.3-10) • Mixed indicators Sharper color produced by using mixture of two indicators have the similar pH range but contrasting color. Example: mixture of thymol blue with cresol red has: Violet color at pH 8.4 Blue color at pH 8.3 Rose color at pH 8.2

  47. Color Indicators • Screened indicators When the color change isn’t easily detectable particularly in artificial light, addition of another indicator obtain Sharper and more pronounced color change Example screened mixture of M.O. and indogocarmine has: At pH 4 yellowish green (alkaline) and violet (acidic). • Universal or multi-range indicators The pH range can be extended By suitable mixing certain indicators. Example: mixture ofBromothymol blue with Ph.Ph. has: Red color at pH 2 Orange color at pH 4 Yellow color at pH 6 Green color at pH 8 Blue color at pH 10

  48. Neutralization Titration Curves Titration curve is the plot of pH versus the volume of titrant Titration curves are constructed to - study the feasibility of titration - choosing indicator 1- Titration curve of strong acid Vs strong base: - e.g. HCl against NaOH . 1- At beginning , pH of acid. pH = - log [H+] 2- During titration, pH of strong acid. pH = pCa 3- At the end point, pH of salt of strong acid and strong base (neutral). pH = pOH = ½ pKw 4- After end point , pH of strong base. pH = pKw - pCb

  49. Strong Acid Vs Strong Base Both M.O. and Ph.Ph. are suitable

  50. Weak Acid Vs Strong Base - e.g. CH3COOH against NaOH. 1- At beginning , pH of weak acid. pH = ½ pKa + ½ pCa 2- During titration, PH of acidic buffer. pH = pKa + log Cs/Ca 3- At end point , PH of salt of weak acid and strong base. pH = ½ pKw + ½ pKa - ½ pCs 4- After end point , PH of strong base . pH = 14 -pCb So that M.O. Isn’t suitable The suitable indicator Ph.Ph. pH range 8.3-10

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