Mansoura National University Lecture Notes- Pharmaceutical Analytical Chemistry 1 (PC101)

Mansoura National University Faculty of Pharmacy Pharm D-Clinical Pharmacy Program Department of Pharmaceutical Analytical Chemistry Level 1 Pharmaceutical Analytical Chemistry -1 (PC101) Lec. 1...

Mansoura National University Faculty of Pharmacy Pharm D-Clinical Pharmacy Program Department of Pharmaceutical Analytical Chemistry Level 1 Pharmaceutical Analytical Chemistry -1 (PC101) Lec. 1 Prof. Dr. Jenny Jeehan Nasr [email protected] Dr. Galal Magdy Dr. Heba Samir Elama [email protected] [email protected] Pharmaceutical Analytical Chemistry Department, Faculty of Pharmacy, Mansoura University  ‫املس توى ا ألول – الفصل ادلراىس ا ألول‬ ‫‪Pharmaceutical Analytical Chemistry 1 PC 101‬‬ ‫ميدترم‪ /‬أأنشطة‬ ‫معىل‬ ‫نظرى‬ ‫شفوى‬  Analytical Chemistry Qualitative Analysis (= What?) Identification of analytes (Yes or No) by physical property [color, taste, solubility,…..]. Quantitative Analysis (=How much?) Determine the concentration (amount of substance) [Determination or Estimation or Assay]. 4  Pharmaceutical Analytical Chemistry 1 Pharmaceutical Analytical Chemistry 1 General Inorganic Quantitative Physical Chemistry Volumetric Chemistry Analysis Chemical Chemical Acid-base Kinetics Equilibrium titration Precipitation Mole Concept titration 5  Learning Outcomes In this lecture, we will learn about: 1. What is Chemistry? 2. Solution Terminology 3. Solubility 4. Mole concept 5. Concentration units 6. Stoichiometry of reaction 6 What is chemistry? 7  What is chemistry? ▪ The branch of science that deals with chemicals. ▪ Chemicals include all the day-to-day things you touch, see, and smell. ▪ Chemicals are everywhere (except in the vacuum). 8  What are chemicals composed of? ▪ Chemicals are composed of different substances (substance means chemical material of which an object is composed e.g. ice is composed of substance water, or chemical material water). 9  What is chemistry? What Is Risk Benefit Balance ▪ Despite the benefits of chemistry, we have to be careful in dealing with chemicals. ▪ Many chemicals are toxic, others are potential cancer producers. Therefore, chemicals must be handled with control. 10  Learning Outcomes In this lecture, we will learn about: 1. What is Chemistry? 2. Solution Terminology 3. Solubility 4. Mole concept 5. Concentration units 6. Stoichiometry of reaction 11  Solution terminology ▪ Solvent: A substance that dissolves other materials to form a solution. ▪ Solute: A substance that dissolves in a solvent. ▪ Solution: A homogenous mixture of one or more solutes dissolved in a solvent. 12  Solution terminology ❖ Concentrated solution: has a relatively large proportion of solute to solvent. ❖ Diluted solution: has a relatively smaller proportion of solute to solvent. 13 Solution Saturation 14 Solution Saturation 15 Solution Saturation 16  Solution Saturation Unsaturated solution: A solution that can dissolve a further amount of solute at a particular temperature. Saturated solution: A solution which cannot dissolve a further amount of solute at a particular temperature. 17  Solution Saturation Supersaturated solution: A solution that contains more amount of solute than required to saturate it at a particular temperature. N.B. - Supersaturated solutions are unstable and release their extra-solid reaching the point of saturation when: I. A tiny crystal (seed) of a solid is added, and additional solute crystallizes on this "seed" crystal. II. Further cooling of the supersaturated solution. 18 19  Electrolyte and Nonelectrolyte Solutions Nonelectrolytes: Molecular substances have no tendency to dissociate in water. Their solutions can't conduct electricity. e.g. Sugar and ethyl alcohol. 20  Electrolyte and Nonelectrolyte Solutions Electrolytes 1. Strong electrolytes: Substances that are completely dissociated in an aqueous solution. Their solution can conduct electricity. e.g. NaCl and HCl Examples: Strong acid, strong bases, salts of strong acid and base. HCl(aq) + H2O → Cl-(aq) + H3O+(aq) 21  Electrolyte and Nonelectrolyte Solutions Electrolytes 2. Weak electrolytes: Substances that dissociate to a small extent in an aqueous solution. They are weakly conductive to electric current. e.g. acetic acid, water a) H2O ionizes to only 10-7 M: H2O(aq) + H2O H3O+(aq) + OH-(aq) b) CH3COOH ionizes to only 0.42% w/v: CH3COOH(aq) + H2O CH3COO-(aq) + H3O+(aq) c) NH4OH and HCN. 22 Electrolyte and Nonelectrolyte Solutions  Learning Outcomes In this lecture, we will learn about: 1. What is Chemistry? 2. Solution Terminology 3. Solubility 4. Mole concept 5. Concentration units 6. Stoichiometry of reaction 24 Solubility 25  Solubility ▪ Solubility is a measure of how a solute will dissolve in a solvent at a constant temperature. Soluble: can dissolve in a solvent. Insoluble: can not dissolve in a solvent. ▪ Ionic compounds are soluble in polar solvents & have low solubility in nonpolar solvents. E.g.: NaCl dissolves in water. 26  Liquids Dissolving in Liquids ▪ Liquids that are soluble in one another (“mix”) are MISCIBLE. “LIKE dissolves LIKE” POLAR liquids are generally soluble in other POLAR liquids. NONPOLAR liquids are generally soluble in other NONPOLAR liquids. ▪ CCl4 and C6H6 (non-polar liquids) are miscible with each other. ▪ CH3OH is miscible with water (hydrogen bonding). 27 1. Which of the following best describes the components and properties of electrolytes? Electrolytes consist of ….. (a) ions & are conductive (b) metals & are non-polar (c) nonmetals & are anions (d) covalent bonds & are polar 2. A solution that contains more than the maximum amount of solute that is capable of being dissolved at a given temperature. (a) Supersaturated (b) Unsaturated (c) Solubility (d) Saturated 28  Learning Outcomes In this lecture, we will learn about: 1. What is Chemistry? 2. Solution Terminology 3. Solubility 4. Mole concept 5. Concentration units 6. Stoichiometry of reaction 29  Mole Concept Molecules and atoms are extremely small objects - both in size and mass. The mole is a standard scientific unit for measuring quantities of very small entities such as atoms or molecules.  Mole Concept 1 Mole 6.022 × 1023 (Atoms/Molecules/…….) 602200000000000000000000 3 Moles 3 × 6.022 × 1023 8 Moles 8 × 6.022 × 1023 31  Mole 6.022 × 1023 atoms /molecules/……. AVOGADRO’S NUMBER Represent quantities 32  Mole Concept It is also equal to the formula weight (atomic or molecular) of a substance expressed in grams. Note: weight = mass for atoms: it is termed atomic weight or atomic mass for molecules: it is termed molecular weight or molecular mass Notice that: Na, K, are atoms, H2O, N2, H2SO4 are molecules. ▪ Formula or molecular weight: The sum of the atomic weights of the atoms that make up a compound. e.g. MW of NaOH = 23 + 16 + 1 = 40 g  1 Mole For Atoms Mole is gram atomic weight (Atomic weight expressed in grams). Atomic weight of Na is 23 ----> 1 mole Na = 23 g = 6.022 × 1023 atoms For Molecules: Mole is gram molecular weight (Molecular weight expressed in grams). M.W. of NaOH is 40 ----> 1 mole NaOH=40g = 6.022 × 1023 molecules 34 34  Mole 1 Mole = 6.022 × 1023 atoms /molecules/particles 1 Mole = atomic, molecular, or formula weight in grams 35  Mole Example: Calculate the weight of one mole of CaSO4· 7H2O. (Ca: 40.08, S: 32.06, O: 16, H: 1) - One mole is the formula weight expressed in grams. - One mole of CaSO4· 7H2O contains (1 Ca + 1 S + 11 O + 14 H) - The formula weight = 40.08 +32.06 + (11 × 16) + (14×1) = 262.14 g - One mole of CaSO4· 7H2O = 262.14 g 36  Mole How many moles are contained in 36 g of water? (atomic weights H=1, O=16). - One mole is the molecular weight expressed in grams. - One mole of H2O = (2×1) + 16 = 18 g 1 Mole 18 g ?? Mole 36 g No. of moles = (36×1)/18 = 2 moles 37  Mole to molecule conversion Determine the number of molecules of H2O in 3 moles H2O 1 mole H2O = 6.022 X 1023 molecules H2O 3 moles H2O contains: 3 x 6.022 X 1023 molecules H2O 3 moles H2O = 18.066 X 1023 molecules H2O 38  Mole to atoms conversion How many atoms are present in 17 moles of water? ‫لم احدد نوع الذرات‬ In order to convert moles of a substance to atoms one must first convert moles to molecules Step 1 moles molecules Step 2 molecules atoms N.B. this depends upon the formula of the 1 mole H2O= 6.022 X 1023 molecules substance which in this case is H2O. According to the formula for H2O: 17 moles H2O contains: 1 molecule H2O = 2 atoms H + 1 atom O = 3 atoms total 17 x 6.022 X 1023 molecules 1 molecule H2O 3 atoms = 1.02 X 1025 molecules H2O 1.02 X 1025 molecule X atoms X=(3 x 1.02 X 1025)/1 = 3.06 X 1025 atoms H2O 39 1. How many grams are in 1.2 moles of neon? (Ne: 20) a) 0.059 g b) 7.2 x 1023 g c) 2.0 x 10-24 g d) 24 g 2. How many moles are in 6.02 x 1023 molecules of H2O? a) 1 mole b) 2 moles c) 6.02 moles d) 602000000 moles 3. The 3 moles of carbon will have -------- Avogadro no. of particles. 40 Given 5 moles of Sulfuric Acid having a formula of H2SO4 answer the following questions: 1 - How many grams of Sulfuric Acid? 2 - How many molecules of Sulfuric Acid? 3 - How many Hydrogen atoms? ‫هنا حددت نوع الذرات‬ 4 - How many Oxygen atoms? (Atomic weights H=1, S= 32, O=16) 41 42  Learning Outcomes In this lecture, we will learn about: 1. What is Chemistry? 2. Solution Terminology 3. Solubility 4. Mole concept 5. Concentration units 6. Stoichiometry of reaction 43  Solution, Solute, and Solvent Solvent: A substance that dissolves other materials to form a solution. Solute: A substance that dissolves in a solvent. Solution: A homogenous mixture of one or more solutes dissolved in a solvent. 44  Concentration Units There are numerous ways of expressing the concentration of a solution, from which we will study only two: I. Molarity (M) II. Normality (N) 45  Concentration Units I. Molarity (molar concentration) (M) ❑ The number of moles of solute per 1 liter of solution. ❑ The most commonly used unit in chemistry. 𝑵𝒐 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝑴𝒐𝒍𝒂𝒓𝒊𝒕𝒚 = 𝑴 = 𝑵𝒐 𝒐𝒇 𝒍𝒊𝒕𝒆𝒓𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 ❑ Ex.: 0.5 M NaOH means 0.5 mole NaOH /1L solution ❑ Unit: (mol/L) or (mol.L-1) or M 46 46  Molarity Calculations To calculate the molarity, we must know the number of moles of solute and the number of liters of solution. 𝐍𝐨 𝐨𝐟 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 𝐌𝐨𝐥𝐚𝐫𝐢𝐭𝐲 = 𝐍𝐨 𝐨𝐟 𝐥𝐢𝐭𝐞𝐫𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 No of moles = M X No of liters (V), where M = Molarity; V = Volume No of liters = Number of moles / M 47 How many grams of silver nitrate AgNO3 are needed to prepare 500 mL 0.3 M AgNO3 solution? (Formula mass of AgNO3 = 170 g) No of liters (V) = mL of solution/1000 V= 500 mL Molarity= 0.3 M Molecular wt. = 170 g = 500/1000 = 0.5 liter Weight (g)?? Molarity = No of moles of solute/No of liters of solution 0.3 = No of moles/0.5 No of moles =0.3 X 0.5= 0.15 mol No of moles = Weight of solute (g) / Molecular wt. of solute 0.15= Weight (g) / 170 Weight = 0.15 X 170 = 25.5 g 48  Dilution and Molarity ▪ The molarity of a solution and its volume are inversely proportional. ▪ Therefore, adding water makes the solution less concentrated. ▪ This inverse relationship takes the form of: 𝑴𝒊 × 𝑽𝒊 = 𝑴𝒇 × 𝑽𝒇 ▪ So, as water is added, increasing the final volume, Vf, the final molarity, M f, decreases. 49 What is the volume of 6.0 M HCl that can be made from 5.0 mL of 12.0 M HCl? Mi= 12.0 M Vi = 5.0 mL Mf = 6.0 M Using the Vf ??? equation: 𝑴 𝒊 × 𝑽𝒊 = 𝑴 𝒇 × 𝑽 𝒇 12.0 × 5.0 = 6.0 × Vf 𝟏𝟐 × 𝟓 Vf = = 10 mL 𝟔 50  Concentration Units II. Normality (Normal concentration) (N) ❑ Number of equivalent weights of solute per 1 liter of solution. 𝑵𝒐 𝒐𝒇 𝒆𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝒘𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 Normality = 𝑵 = 𝑵𝒐 𝒐𝒇 𝒍𝒊𝒕𝒆𝒓𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 Ex.: 2 N NaOH means 2 equivalent weight of NaOH / 1L solution. Wt of solute Number of equivalent weights = Equiv. wt of solute ❑ An equivalent weight: is the weight of substance that is equivalent in its reactive power to one mole of hydrogen. 51  Concentration Units II. Normality (Normal concentration) (N) ❑ For acids and bases, the number of reacting units is based on the number of protons (i.e., hydrogen ions) an acid will furnish, or a base will react with. 𝑴.𝒘𝒕.of 𝒂𝒏 𝒂𝒄𝒊𝒅 ❑ For acids: Equiv. wt.= + 𝑵𝒐 𝒐𝒇 𝒓𝒆𝒑𝒍𝒂𝒄𝒆𝒂𝒃𝒍𝒆 𝑯 𝒊𝒏 𝒐𝒏𝒆 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒂𝒄𝒊𝒅 Ex. Equiv. wt. of H2SO4 = M. wt. of H2SO4 /2 = 98/2= 49 g H2SO4 𝑴.𝒘𝒕.of 𝒃𝒂𝒔𝒆 ❑ For bases: Equiv. wt.= − 𝑵𝒐 𝒐𝒇 𝒓𝒆𝒑𝒍𝒂𝒄𝒆𝒂𝒃𝒍𝒆 𝑶𝑯 𝒊𝒏 𝒐𝒏𝒆 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒂𝒔𝒆 Ex. Equiv. wt. of NaOH = M. wt. of NaOH /1 = 40/1= 40 g NaOH 52  Calculate the normality of NaOH solution formed by dissolving 0.2 g NaOH to make 250 mL solution. (Molecular weight of NaOH=40 g) V= 250 mL 𝐍𝐨 𝐨𝐟 𝐞𝐪𝐮𝐢𝐯𝐚𝐥𝐞𝐧𝐭 𝐰𝐞𝐢𝐠𝐡𝐭 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 Normality = 𝐍 = Molecular wt. = 40 g 𝐍𝐨 𝐨𝐟 𝐥𝐢𝐭𝐞𝐫𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 Weight= 0.2 g Normality= ?? M.wt.of NaOH 40 Equivalent weight of NaOH = − = = 40 No of OH in one molecule of NaOH 1 Wt of solute 0.2 Number of equivalent weights of NaOH = = = 0.005 Equiv. wt of solute 40 𝟎.𝟎𝟎𝟓 Normality = = 0.02 N 𝟎.𝟐𝟓 53 Determine the molarity of a solution made by dissolving 20.0 g of NaOH in sufficient water to yield a 482 cm3 solution. (atomic weights H=1 g, O=16 g, Na= 23 g) How many grams of potassium dichromate (K2Cr2O7) are required to prepare a 125 mL solution whose concentration is 1.83 M? 54  Learning Outcomes In this lecture, we will learn about: 1. What is Chemistry? 2. Solution Terminology 3. Solubility 4. Mole concept 5. Concentration units 6. Stoichiometry of reaction 55 The stoichiometry of reaction in solution The calculation of quantities in chemical reactions “Stoichio” = elements “metry” = to measure 56  The stoichiometry of reaction in solution ▪ It is the quantitative relationships/ratios of reactants and products involved in a chemical reaction. ▪ It deals with the ratios in which chemicals react. ▪ The coefficient in the balanced chemical equation provides the mole ratios needed to solve stoichiometric problems. 57  The stoichiometry of reaction in solution ▪ The reaction tells us that mixing two moles of sodium with one mole of chlorine gas will yield two moles of sodium chloride. ▪ The mole ratio describes the molar ratio at which the reaction will proceed. ▪ The reactants are in a 2:1 molar ratio ▪ What is the ratio of sodium to salt? 58  The stoichiometry of reaction in solution Na2CrO4 + Pb(NO3)2 → PbCrO4 + 2NaNO3 1 mole + 1 mole → 1 mole + 2 moles 162 g + 331 g → 323 g + 170 g (85 g X 2) 0.1 mole + 0.1 mole → 0.1 mole + 0.2 moles 16.2 g + 33.1 g → 32.3 g + 17.0 g Molar ratio 1 Na2CrO4 : 1 Pb(NO3)2 1 Na2CrO4 : 2 NaNO3 59  The stoichiometry of reaction in solution Na2CO3 + CaCl2 → CaCO3 + 2NaCl 1 mol 1 mol 1 mol 2 mol - How many mL of 0.25 M CaCl2 are needed to react completely with 50 ml of 0.15 M Na2CO3? Solution V of CaCl2=?? mL Molarity of CaCl2= 0.25 No. moles of Na2CO3 = M x VL V of Na2CO3= 50 mL = 0.15 M Na2CO3 x (50 /1000). Molarity of Na2CO3= 0.15 = 7.5 x 10-3 mol Na2CO3. Therefore, the no. of moles of CaCl2 needed will be also 7.5 x 10-3 mol. VL= No. moles of CaCl2 / M VL = 7.5 x 10-3 / 0.25 = 0.03 L = 30 mL 60  Mole relationship in balanced equations Determine the number of moles of N2O4 needed to react completely with 3.62 moles of N2H4 for the reaction 2 N2H4 + N2O4 3 N2 + 4 H2O From the balanced equation: 2 moles N2H4 is equivalent to 1 mole N2O4 2 moles N2H4 1 mole N2O4 3.62 moles X moles X= (3.62x1)/2 = 1.81 moles N2O4 61 ✓ What is Chemistry? ✓ Solution Terminology ✓ Solubility ✓ Mole concept ✓ Concentration units ✓ Stoichiometry of reaction 62