1.Basics of Pharmaceutical Chemical Analysis. 1. Some important terms to describe analytical results Error types (quantitatively) The absolute error E

1.Basics of Pharmaceutical Chemical Analysis

1. Some important terms to describe analytical results

• Error types (quantitatively) • The absolute error E of a measurement is the difference

between the measured value and the true value. • E = Xi - Xt , where Xi is the measured value, Xt is the true

• Relative Error Er

t


• 1) Systematic errors: Those which can be avoided• 2) Random errors: those errors can not be avoided.


• Precision• Accuracy


• Selectivity

• Sensitivity

• Robustness

1. Concentration

• Physical units interested in knowing how many grams (mg, μg,..) present in unit volume (L, ml, μl,…)

• Chemical units interested in knowing how many moles of the material present in each unit of volume (L) of the solution

1. Concentration

• Most common physical units used in pharmaceutical analysis: • mg/ml• μg/ml• part per million (ppm)• (%w/w)• (%w/v)• (%v/v)• (%v/w)

1. Concentration

• Chemical Units include:• Molarity: number of moles in each liter of solution

A_ Analytical molarityB_Equilibrium, or species molarity e.g. [H2SO4]

• Normality: number of equivalents in one liter of solution

1. Example of molarity

• Example. Calculate the analytical and equilibrium molar concentrations of the solute species in an aqueous solution that contains 285 mg of trichloroacetic acid, Cl3CCOOH (163.4 g/mol)in 10 ml (the acid is 73% ionized in water)

• What is the molarity of K+ in aqueous solution that contains 63.3 ppm of K3Fe(CN)6 (329.2 g/mol).

• M = 5.77 X 10-4 mol/L

1. Preparation Examples

• Q1. Describe the preparation of 500 ml of 0.0740 M Cl- solution from solid BaCl2.2H2O (244.3 g/mol).

• 4.52 g of BaCl2.2H2Oin 500 ml water

• Describe the preparation of 100 ml of 6.0 M HCl from a concentrated solution that has density of 1.18 and is 37% (w/w) HCl (36.5 g/mol). Mconc. X Vconc. = Mdil. X Vdil.

1. pH value of aq. Solutions of acids and bases

• The pH of a solution is defined as –log[H+].• In pure water the conc. of hydrogen ions is governed by the

equilibrium:

1. Examples

• What is the pH of 0.054M solution of HCl?• What is the pH of 0.01M KOH?• What is the pH of 0.01 M acetic acid? Pka = 4.76

1. pH value of Buffers

• Solutions of mixtures of a weak acid and a conjugate base or weak base with conjugate acid.

• Henderson–Hasselbalch equation

• Calculate the pH of a solution prepared by dissolving 242.2 mg of a primary amine (base) in 10 ml of 0.17 M HCl and dilution to 100 ml with water. Pka = 8.08 (Mwt of amine 121.1 g/mol)

1. Buffer Capacity

• Buffer capacity: defined as the number of moles of a strong acid or a strong base that causes 1.00 L of the buffer to undergo a I.00 unit change in pH.

• Buffer capacity depends onA_ the concentration of acid and its conjugate base or base and

its conjugate acid B_the proximity between pH and pKa value

2. Titremetric analysis

2. Concept of Titremetric analysis

• Titrant• titrand or called analyte• burette• indicator• equivalence point • end point

2. Example

• Example: The organic matter in a 3.776 g sample of a mercuric ointment is decomposed with HNO3, After dilution, the Hg2+ is titrated with 21.30 ml of a 0.1144 M solution of NH4SCN, Calculate the percent Hg (200.59 g/mol) in the ointment.

Hg2+ + 2SCN- Hg(SCN)2


• In order to be suitable for titration, the reaction could be of an type as long as the following conditions are met:

1) complete reaction2) spontaneous (fast)3) stoichiometric reaction4) There must be some means of detecting the end points.


• Primary standard are stable chemical compounds that are available in high purity and which can be used to standardize the standard solutions used in titrations.

• Titrant such as NaOH and HCl cannot be considered as primary standards since their purity is quite variable.


• Important requirements for a primary standard are the following:

1. High purity2. Atmospheric stability. 3. Absence of hydrate water 4. Modest cost. 5. Reasonable solubility in the titration medium. 6. Reasonably large molar mass


• Titration curve: are plots of a concentration-related variable as a function of reagent volume. In acid-base titration, it’s a plot between the pH of solution being titrated and the volume of the added titrant.

Potential end point


• Back titration• 1) Volatile substances, e.g. ammonia• 2) Insoluble substances, e.g. Calcium carbonate• 3) Slow reactions• 4) Substances which require heating

Neutralization Titration (Acid/base titration)

Acid/Base Titration

• Acid/Base Indicators

• Indicator pH range = pKa ± 1

• Indicator should fall within the range of inflection of the strong acid/strong base titration curve.

Acid/Base Titration

Titration Of Strong Acids And Strong Bases

• Consider the titration of 50 ml of 0.05 M HCl (analyte) with 0.1M NaOH (titrant).

• First we determine equivalent point• M1V1 = M2V2• V = 25ml• Initial Point • Before any base is added, the solution is 0.0500 M in H30+, and

pH = -log[H30+] = -log 0.0500 = 1.30

• After Addition of 10.00 mL of Reagent • The H+ ion concentration is decreased as a result of both reaction with the

base and dilution. • pH = 1.6

HCl + NaOH NaCl + H2O


• After Addition of 25.00 mL of Reagent (The Equivalence Point) • At the equivalence point, neither HCl nor NaOH is in excess, and so the

concentrations of hydronium and hydroxide ions must be equal• pH = 7 • After Addition of 25.10 mL of Reagent • The solution now contains an excess of NaOH, pH = 10.12


Titration curves for HCI with NaOH. Curve A: 50.00 mL of 0.0500 M HCI with 0.1000 M NaOH. Curve B: 50.00 mL of 0.000500 M HCI with 0.001000 M NaOH.

The Effect of Concentration

Titrating a Strong Base with a Strong Acid

• Titration curves for NaOH with HCl. Curve A: 50.00 mL of O.0500 M NaOH with 0.1000 M HCl. Curve B: 50.00 mL of 0.00500 M NaOH with 0.0100 M HCl.

Weak acid/strong base and weak base/strong acid

• Example. Titration of 50.0 ml of 0.1 M acetic acid (pKa 4.76) with 0.1 M NaOH.

• Initial pH: calculated from [H+] = (ka. c)1/2

• pH before equivalence point: A state of equilibrium between the acid and its conjugate base formed, and the pH is calculated from Henderson–Hasselbalch equation


• At equivalence point: all acetic acid is consumed and converted to sodium acetate, and pH is calculated starting from [OH-] = (Kb. c)1/2

• pH after equivalence point: After the addition of excess amount of NaOH

• pH = 14 – pOH


Titration curve of 25 ml of 0.1M solution of aspirin (pKa 3.5) titrated with 0.1M NaOH


Titration curve of 25 ml of 0.1M solution of quinine(pKa 8.05) titrated with 0.1M HCl


• For an acid, the measured pH when the acid is exactly half neutralized is numerically equal to pKa.

• For a weak base, the pH at half titration must be converted to pOH, which is then equal to pKb.

Some applications of neutralization titration

• Blank titration


• 1)Estimation of alcohols and hydroxyl values by reaction with acetic anhydride (AA)

• Example. 1.648 g sample of castor oil was refluxed with 5 ml of acetic anhydride for 6 h. 44.6 ml of KOH (0.505 M, 56.1 g/mol) was added to neutralize the excess acetic anhydride and acetic acid. 53.5 ml of KOH required to blank-titrate the 5 ml of the reagent (acetic anhydride). Calculate the hydroxyl value.



• 2) Saponification value as an example of back-titration with blank determination.

• The saponification value for a fixed oil (vegetable oil) is the number of mg of KOH equivalent to 1 g of oil.

KOH residual + HCl KCl + H2O

RCOOR' + KOHexcess RCOOK + R'OH + KOHresidualref lux


• 3) Elemental analysis of nitrogen (Kjeldahl determination)

Example for Kjeldahl method

• 0.9092-g sample of a wheat flour was analyzed by the Kjeldahl procedure. The arnmonia formed was distilled into 50.00 mL of 0.05063 M HCI, and then back titrated with 7.46-mL of 0.04917 M NaOH. Calculate the percentage of protein in the flour.

• Note. Since most proteins contain approximately the same percentage of nitrogen, multiplication of this percentage by a suitable factor (6.25 for meats, 6.38 for dairy products, and 5.70 for cereals) gives the percentage of protein in a sample.

• %N = Mass of N/ Sample mass x 100% • % of crude protein = %N x F


• Polyfunctional acids and bases

pKa 10.32pKa 6.38


Titration of 1 M sodium carbonate with 1M HCl


Non-aqueous Acid-Base Titrations

Non-aqueous Titrations

• Theory• suitable for the titration of very weak acids and bases .

• commonly based on perchloric acid in acetic acid.

• Increase the basicity or acidity of the analyte.

• Sharper inflection is resulted from non aqueous titration compared with aqueous



• Acetic anhydride is used to remove water from aqueous perchloric acid (the commercially available form).

• Standardized with KHP• Crystal violet

• Mercuric acetate

2RNH3+Cl- + Hg(CH3COO)2 HgCl2 + 2RNH3

+ + 2CH3COO-

Non-aqueous Titrations for very weak acids

• Titrant:• Lithium methoxide (CH3OLi), Sodium or potassium methoxide

(CH3ONa, CH3OK), tetrabutyl ammonium hydroxide


• Practical application• Example 1. 4 g tablet of methacholine cloride (195.69 g / mol), dried and

stored in a vacuum desiccator, dissolved in 50 ml of glacial acetic acid, 10 ml of mercuric acetate solution and one drop of crystal violet was added and then titrated with 56 ml of 0.1 M perchloric acid to a blue-green end point. Blank titration was conducted and 1.7 ml of perchloloric aid used. Calculate % w/w of methacholine cloride.

Complexation Titrations

The formation of complexes

• Ligand,complexing agent, chelator or chelating agent

• Metal ion and coordinate bond.

• Complex might be + or - charged or neutral. Cu(NH3)42+; a

neutral complex with glycine, Cu(NH2CH2COO)2 and an anionic complex with chloride ion, CuCI4

2-.

• Multidentate:. e.g. Ethylenediaminetetraacetic acid (EDTA)

Complexometric titrations

Curve A: tetradentate ligand. Curve B: bidentate ligand Curve C: involves a unidentate ligand.


• EDTA is a hexadentate ligand. • EDTA is fully protonated (H4Y) at pH < 3

• Half protonated (H2y2-) between pH 3-10

• Fully deprotonated (Y4-) at pH > 10• EDTA is commercially available as Na2H2y.2H20

• EDTA is a valuable titrant because it form stable 1:1 complex with almost all cations.



• Effect of pH on complex formation• since the actual complexing species is Y4-, complexes will form

more efficiently and stable in alkaline solution.


• Eriochrome Black T (Mordant Black, Solochrome).

• Back titartion • The excess EDETA is back-titrated with Mg2+ or Zn2+ of known

concentration using a suitable indicator.

General principles involved in disodium edetate titrations

• Example: 0.3 g sample of magnesium sulphate (MgSO4, 120.36 g/mol) was dissolved in water and buffered with ammonia at pH 10, mixture of Eriochrome Black T and sodium chloride (1:99) was added as indicator. The mixture was titrated with 45 ml of 0.05 M disodium edetate until the solution becomes full blue. 0.3 ml was consumed in blank titration. Calculate the %w/w of MgSO4.

• Answer: 89.7%w/w

General principles involved in disodium edetate titrations

• Water Hardnes• Ca, Mg, and other heavy metals replace sodium or potassium

ions in soaps and form "scum”.• To determine Ca2+ hardness strongly alkaline solution (pH 12)

is used to precipitate Mg2+ as Mg(OH)2.

• Titration of Ca2+ and Mg2+ in a 50.00mL sample of hard water required 23.65 mL of 0.01205 M EDTA. A second 50.00mL aliquot was made strongly basic with NaOH to precipitate Mg2+ as Mg(OH)2. The supernatant liquid was titrated with 14.53 mL of the EDTA solution. Calculate:

• (a) the total hardness of the water sample, expressed as ppm of CaCO3.

• (b) the concentration in ppm of Ca+2 in the sample. • (c) the concentration in ppm of Mg+2 in the sample.

Precipitation Titrimetry


• silver nitrate, which is used for the determination of the halides, the halide-like anions (SCN-, CN-, CNO-

• Titrimetric methods based on silver nitrate are sometimes called argentometric methods.

•Ksp = [Cl-] [Ag+] / [AgCl]

•The lower the Ksp of certain product the more complete the reaction would be.

Precipitation Titrimetry, theory

• Titration curve• Calculate the pAg for titration curve of 50.00 mL of 0.05 M

NaCI with 0.10 M AgN03 (for AgCI, Ksp = 1.82 X 10-10 ). • At equivalence point.• M1V1 = M2V2• Ksp = [Cl-] [Ag+]• [Cl-] = [Ag+]


• After addition of 10 ml (preequivalence)• Ksp = [Cl-] [Ag+] = 1.82 X 10-10

• Postequivalence-Point • 10 ml of silver nitrate after the equivalence point • [Ag+] = [AgNO3] excess

Precipitation Titrimetry• The Effect of Reaction Completeness (Kf) on Titration Curves


• Three methods to determine the end points of Argentometric Titrations :

• 1) The Mohr Method, Chromate Ion• 2) The Fajans Method, Adsorption Indicators• 3) The Volhard Method, Iron(III) Ion


• Chromate Ion; The Mohr Method • Sodium chromate can serve as an indicator for the

argentometric determination of chloride, Bromide, and cyanide ions by reacting with excess silver ion to form a brick-red silver chromate (Ag2CrO4) precipitate in the equivalence-point region.

• The silver ion concentration at chemical equivalence in the titration of chloride with silver ions is given by

•


• This method is associated with positive systematic error : chromate ion concentration of 6.6 X 10-3 M imparts such an intense yellow color to the solution that formation of the red silver chromate is not readily detected.

• • A correction for this error can readily be made either by blank

titration.

• The Mohr titration must be carried out at a pH of 7 to 10 (why?)


• Adsorption Indicators: The Fajans Method • Fluorescein is a typical adsorption indicator that is useful for

the titration of chloride ion with silver nitrate.


Ag_ _ _ _ Cl

Ag_ _ _ _ Cl

Ag_ _ _ _ Cl

AgCl

Cl-

Cl-

Cl-

Cl- Cl

-

Cl-

Cl-

Cl-

Cl-

Cl-

Flourescein-

Flourescein-

Ag_ _ _ _ Cl

Ag_ _ _ _ Cl

Ag_ _ _ _ Cl

Ag+

Just after the equivalent point (end point)

Ag+ Ag

+

Flourescein- Flourescein

-Flourescein

-

AgCl

Ag+

Flourescein-

Ag+

Flourescein-

Ag+

Flourescein-


• Iron(III) Ion; The Volhard Method • It is indirect determination of halide ions. • It is back titration with a standard thiocyanate solution. • In the Volhard method, silver ions are titrated with a standard

solution of thiocyanate (titrant) ion:

• Acidic solution to prevent precipitation of iron(III) as the hydrated oxide.

Oxidation/reduction reaction and titration

Examples

• Ce4+, is called an oxidizing agent, or an oxidant. • Fe2+A reducing agent, or reductant

Oxidation/reduction reaction

• Balancing Redox Equations• Mass balance

Oxidation/reduction reaction

• Charge balance

Redox reaction• Reduction potential is a measure of how thermodynamically

favorable it is for a compound to gain electrons.

• A high positive value is a strong oxidizing agent.

• The difference in potential between two substance is the reaction potential.

• Reductants tend to react with atmospheric oxygen, therefore, they are seldom used in direct titration. Instead, they used in back titration. E.g. Iron(lI) Solutions

Potentiometer

Salt bridge

Reference electrodeTypes:Ag/AgCl or Hg/Hg2Cl2

Indicator electrodemade from inert material:platinum, gold or graphite

Analyte solution

Components of Potentiometric CellA simple diagram

Setup for the redox titration of Fe2+ with Ce4+.

The reaction potential between analyte and titrant

Specific indicators

• Iron II Complexes of Orthophenanthrolines

• Starch/Iodine Solutions: Starch complex with iodine gives a blue color.

Permanganate

• The oxidizing strengths of permanganate and cerium(lV) solutions are comparable.

• Cerium(lV) is stable indefinitely, whereas KMnO4 is not:

Permanganate

• Standarization of KMnO4.• Sodium oxalate is a widely used primary standard.

Dichromate• Potassium Dichromate

• Potassium dichromate solutions are indefinitely stable

• The disadvantages of potassium dichromate compared with cerium (IV) and permanganate ion are its lower Eo and the slowness of its reaction with certain reducing agents.

Iodine

• Iodine

• Iodine solutions lack stability for the following reasons:• 1) Volatility of the solute. • 2) Iodine slowly attacks most organic materials, Consequently,

cork or rubber stoppers are never used

• Must be restandardized regularly using sodium thiosulfate

Applications: Hypochlorite determination

• Starch undergoes decomposition in solutions with high I2 concentrations.

• It may complex with iodine longer than the time of detecting the end point.

Applications: Standardizing Thiosulfate Solutions by KIO3

• I mol IO3- = 3 mol I2 = 6 mol S203

2-

Example

• Example. A solution of sodium thiosulfate was standardized by dissolving 0.1210 g KIO3, (214.00 g/mol) in water, adding a large excess of KI, and acidifying with HCl. The liberated iodine required 41.64 mL of the thiosulfate solution to decolorize the blue starch/iodine complex. Calculate the molarity of the Na2S203.

Applications: Iodometric analysis of ascorbic acid

Applications: Iodine-absorbing substances in penicillines

• A major stability problem in penicillines is the hydrolysis of the lactam ring

• Excess iodine is added to penicillin sample and then back titrated with Na2S2O3.

Applications: Bromination reactions

Applications: Karl Fischer (Determining of Water)

• The Karl Fischer reagent consists of a mixture of anhydrous methanol, an anhydrous base (pyridine or imidazol), iodine and sulfur dioxide (SO2).

• the stoichiometry is one mole of I2 per mole of H2O present (1:1):

• Reactions steps:SO2+ 2H2O H2SO4 + 2H

++ 2e-

I2 + 2e-

2I-

Determination of drugs based on their functional groups

Determination of aldehyde and ketons

• The determination of aldehydes depends upon the reaction with hydroxylamine hydrochloride:

• The liberated hydrochloric acid can be titrated with standard alkali.

Determination of aldehyde and ketons

• Example. Determination of the %w/w of cinnamic aldehyde in cinnamon oil.

• 100 g of cinnamon oil was reacted with adequate excess of hydroxylamine hydrochoride (15 ml). The reaction mixture was titrated with 160ml of 0.5M KOH until the red color of methyl orange changes to full yellow. Calculate the %w/w of cinnamic aldehyde (132.16 g/mol).

Determination of alcohol and phenol

• 2) Titration of acidic OH groups. • Titration of acidic OH groups. Phenols are stronger acids than

alcohol due to resonance stabilization of the conjugate base anion.

• Generally, it’s possible to determine the concentration of phenols by non-aqueous acid-base titration.

• Non-aqueous acid-base titration is good to determine phenols in the presence of other acylable groups such as alcohol or amines.

Determination of alcohol

• Titration of system.

• Such compounds exist in two tautomaric forms.

• Similar titration to phenols, by non-aqueous acid-base titration of the enol form.

Determination of Thiols

• 1) Thiols are much stronger acids than alcohol, as the conjugate’s anion charge is distributed over large sulfur atom.

• Thiols are acidic enough to be titrated by aqueous titration.• Non-aqueous titration is generally more accepted for thiols.• 2) AgNO3 reacts with thiols to ppt silver-thiolates:

• The silver method can be conducted by adding excess AgNO3 and back-titrate the excess using SCN- (thiocynate) with Fe3+ as indicator.

Determination of Thiols

• 3) The third method for analyzing thiols depend on its oxidation to disulfides.

• In this method the unreacted I2 is back titrated with thiosulfate using starch as indicator.

Determination of aniline

• Diazotization of aromatic primary amine.

Documents

1.Basics of Pharmaceutical Chemical Analysis. 1. Some important terms to describe analytical results Error types (quantitatively) The absolute error E