Lec4-6 Solutions (OH) PDF

SOLUTIONS Lectures. 4-6  https://attend-class.com/  Objectives ► Mixtures and Solutions ► Solubility ► The Effect of Temperature on Solubility ► The Effect of Pressure on Solubility: Henry’s Law ► Units of Concentration ► Ions in Solution: Electrolytes ► Properties...

SOLUTIONS Lectures. 4-6  https://attend-class.com/  Objectives ► Mixtures and Solutions ► Solubility ► The Effect of Temperature on Solubility ► The Effect of Pressure on Solubility: Henry’s Law ► Units of Concentration ► Ions in Solution: Electrolytes ► Properties of Solutions ► Osmosis and Osmotic Pressure ► Dialysis 2  Reference Textbook “Fundamentals of General, Organic, and Biological Chemistry” by McMurry, Castellion, Ballantine, Hoeger, Peterson (Pearson, 7th Edition), Chapter 9 (Solutions) 3  Cu, O2, N2 H2O CaCO3 10 g 100 ml 10 % Colloid suspension 4  Mixtures and Solutions ► Definition: A mixture is a combination of two or more substances, both of which retain their chemical identities. ► Mixtures may be solid, liquids, gases. ► Mixtures are important for medicine and pharmacy. ► Heterogeneous mixture: A non-uniform mixture that has regions of different composition. ► Homogeneous mixture: A uniform mixture that has the same composition throughout e.g. solutions and colloids.. 5 Comparison of solutions, colloids, and heterogeneous mixtures. 6  Blood ►Blood has the characteristics of solutions, colloids and heterogeneous mixtures. ►Blood is constituted of ~ 45% suspended RBC (heterogeneous mixtures ) and ~ 55% plasma contains ions (solutions) and protein (colloids) molecules. 7 Solutions may exist in gaseous, liquid or solid states. Examples G/G: Air (N2, O2 and other gases) G/L: CO2/water (Soda Water) L/L: Gasoline ( Mix. of Hydrocarbons) C8H18 L/S: Dental amalgam (Hg/Ag) S/L: Seawater (e.g. NaCl/water) S/S: Metal alloy (Copper/Gold) ► G: Gas, L: liquid, S: Solid 8  The Solution Process Generally, the formation of solutions is based on a phenomena Like Dissolves Like. Polar or ionic substances dissolve in polar solvents, while non- polar or non-ionic substances are soluble in non-polar solvents. e.g. NaCl (ionic) or glucose (polar) are soluble in water, but insoluble in Hexane. e.g. Iodine, fats and oils (non-polar) are insoluble in water, but soluble in non-polar solvents e.g. Hexane e.g. Slightly polar organic compounds e.g. CHCl3 are slightly soluble in water. ► Which of these compounds dissolves in water: I2, C5H12, CCl4 and CH3OH ? 9  Solution process of NaCl crystal in water. Polar water molecules surround individual ions in the crystal pulling them from the crystal surface into solution. Oxygen atoms point to (+) ions and hydrogen atoms point to (-) ions (ion-dipole intermolecular force). δ- δ+ 10 ► Ionic substances e.g. NaCl dissolves in water due to solvation (Hydration) process, whereas polar substances e.g. glucose are soluble in water due to hydrogen bonding with water molecules. ► The solution process may be accompanied by heat change which might be exothermic (release heat and warming) e.g. CaCl2 in water or endothermic (heat absorption and cooling) e.g. NH4NO3 in water. ► Solid Hydrates: Some compounds attract water strongly forming Solid Hydrates (water molecules present in chemical formula) e.g. AlCl3.6H2O (antiperspirant),CuSO4.5H2O (pesticide, fungicide). ► Some compounds can attract water vapor from humid air and called hygroscopic substances and are used as drying agents e.g. Silica gel, CaCl2. 11  Solubility ► Solubility is a physical property that expresses the number of grams of solid solute per 100 ml of solvent (gm/100 ml). ► In case of liquids e.g. water, ethanol, ► miscibility instead of solubility is used. Ex. Solubility of KBr at 20oC = 65 g/100 ml Ex. Solubility of KBr at 90oC = 100 g/100 ml ► Saturated Solution: A solution that contains the maximum amount of solute i.e. ► When the concentration of the solute exceeds its solubility, the solute will not dissolve. ► Unsaturated Solution : A solution which does not contain the maximum amount of solute i.e. more solute can be added. 12  Cont. Solubility Factors affecting solubility ► Effect of particle size: Decrease of particle size of solid solute increases the surface area and increases the solubility e.g. powders are more soluble than crystals in water. ► Effect of temperature: Generally, the solubility of a solid solute in water increases by an increase of temperature (S/L). ► Effect of pressure: The solubility of gases in water increases by increase of applied pressure (G/L). 13  The Effect of Temperature on Solubility ► Solids are more soluble at high temperature to form supersaturated solutions, which contain more solute than saturated solutions. ► These solutions are unstable and crystallization process occur when a tiny seed crystal is added. saturated supersaturated saturated 20oC 70 g 20oC 20oC 65 g 70 g 5 g ppt 90oC, 70 g 70 g/100 ml 65 g/100 ml + 5 gm ppt KBr at 20oC = 65 g/100 ml KBr at 90oC = 100 g/100 ml 14 ► Solubility of some (a) solids and (b) gases, in water as a function S/L of temperature. ► Most solid substances become more soluble in water as temperature rises. ► The solubility of gases in water decreases as temperature rises. ► A decrease of the O2 dissolved in warm water leads to death of G/L fish.  Above the line (supersaturated)  On the line (saturated)  Below the line (unsaturated) 15 Health Note on solubility ► Gout disease: Increase concentration of Uric Acid in blood beyond the saturated concentration results in deposits of needle-shaped crystals in cartilage, soft tissues causing severe pain. ► Formation of kidney stones: Ingestion of food rich in calcium phosphate and calcium oxalate, results in deposits of crystals of these compounds in kidneys causing severe pain. 16 The Effect of Pressure on Solubility: of Gases in Water (Henry’s Law) Henry’s law: The solubility of gas is directly proportional to its partial pressure. If [T] is constant, C  Pgas , or C/Pgas = k , So C1/P1 = C2/P2 C: solubility 17  Problem At a partial pressure of O2 of 159 mmHg, the solubility of O2 in blood is 0.44 g/100 ml. What is the solubility of O2 at 11,000 ft, where the partial pressure of O2 is 56 mmHg? P1 = 159 mmHg, C1 = 0.44 g/100ml P2 = 56 mmHg, C2 = ??? g/100ml Notice: C1/P1 = C2/P2 * 1 atm = 760 mmHg 18  Homework ► 1. From the graph shown of solubility versus temperature for oxygen gas, determine the concentration of dissolved oxygen in water at 25oC and at 35oC. ► 2. A solution is prepared by dissolving 62.5 g of KBr in 100 ml of water at 60oC. Is this solution saturated, unsaturated or supersaturated? What will happen if the solution is cooled to 10oC? ► (Use solubility curve of KBr page 15) 19  Units of concentrations of solutions ► Percent concentration ► Parts per million (ppm) concentration ► Molarity (M), Molality (molal) and Normality (N ) concentrations ► https://attend-class.com/ 20 100 mL 21 Percent Concentrations S/S S/L L/L 22 A solution contains 1.8 gm of heparin sodium dissolved in to make a final volume of 0.015 L. what is the % w/v of this solution? Weight = 1.8 g, Vol. = 0.015 L X 1000= 15 mL 23 How many milliliters of methyl alcohol are needed to prepare 75 ml of a 5% v/v solution? X 100 24 Parts per Million (ppm) and Parts per Billion (ppb) Concentrations ► 25  Molarity (Molar) & Millimolarity concentrations ► 26  Millimolarity (mM): 1 M = 1000 mM Molality (molal) ► Molality (M)= Moles of solute / weight of solvent (kg) ►If the solvent is water, density = 1 (1L =1 kg) Molarity = Molality 27  What is the molarity of a solution made by dissolving 2.355 gm of sulfuric acid in water and diluting the final solution to a volume of 50 ml. (molar mass of sulfuric acid: 98.1)? Weight = 2.355 g, Vol. 50 mL/1000 = 0.05 L 51  Normality (Equivalent concentration) Equivalent weight of an acid or base is the mass which supplies or reacts with one mole of hydrogen cations (H+). Read only 29  Dilution 1 mole 1 mole ► Dilution: Process of lowering the concentration of a solution by adding additional solvent. ► The following equation is very useful in calculating final concentration of solution after dilution. M1 V1 = M2V2 30 ► M1 and V1 refers to the initial concentration and volume of the solution and M2 and V2 refers to the final concentration and volume of the solution. ► The final concentration M2 will be equal to: M2 = M1 V1/ V2 ► The concentration units [M1 and M2 ] and [V1 and V2 ] should be similar ► Dilution of concentrated solutions to diluted solutions is required in clinical and pharmacy practices. 31 What is the final concentration, if 75 ml of 3.5 M glucose solution is diluted to a volume of 0.450 L. M1 V1 = M2V2 V1 = 75 ml, M1 = 3.5 M, V2 = 0.450 L = 450 ml, M2 = ????? M 32  Neutralization reaction ►Neutralization reaction, is a chemical reaction in which an acid and a base react quantitatively with each other. ► Applications Chemical titration methods are used for analyzing acids or bases to determine the unknown concentration by using a pH meter or a pH indicator which shows the point of neutralization by color change. Simple stoichiometric calculations can be used (Lab. 3). N1 X V1 = N2 X V2 a,b (acid, base), volume in litters and weight in grams * 33 34 There are many uses of neutralization reactions that are acid- alkali reactions, antacid is a substance which neutralizes stomach acidity used to relieve heartburn. These are designed to neutralize excess gastric acid in the stomach (HCl) that may be causing discomfort in the stomach or lower esophagus. Homework 1. How many litters of gastric juice (HCl 0.1 M) will react completely with an antacid tablet that contain 500 mg of Mg(OH)2 (MW = 58.3) ? 2 HCl + Mg(OH) 2 MgCl2 + 2H2O 2. Which of the following solution is more concentrated 0.5 M KCl or 5% (m/v) KCl, 15 ppm KCl ? 35 Electrolytes & Nonelectrolytes Electrolytes ► Substances that produce ions in solutions are called Electrolytes. These solutions are good conductors of electricity e.g NaCl, HCl, MgCl2 (Strong Electrolytes) (Complete Ionization) HF (Weak Electrolytes) (Incomplete Ionization) Nonelectrolytes ► Substances that dissolve in water, but produce no ions (No Ionization) , do not conduct electricity e.g Sugars (sucrose), glucose C6H12O6 organic compounds e.g. ethanol 36 glucose NaCl 37  Electrolyte Concentration Electrolytes ( Na+, K+, Ca 2+ , Cl- ) exist at low concentrations in body fluids (e.g. plasma) and are often expressed as Milliequivalents per liter (mEq/L) (m N). 1 Eq/L = 1000 mEq/L Normal concentrations of electrolytes in human plasma Electrolyte Concentration (mEq/L) (m Normal) (Positive ions) Na+ 138 K+ 5 Ca 2+ 4 Mg 2+ 3 (Negative Ions) Cl - 110 HCO3 - 30 H2PO4 - 4 Proteins 6 38 Conversions of electrolyte’s concentration from [mEq/L] (m N ) to moles or mg. 1. [mEq/L] to [mg]: Use the formula : 2. [mEq/L] to [moles]: Use the formula : mEq/L mEq in one Litter. 39 41 3. A Ringer’s solution for IV fluid replacement contains 155 mEq of Cl- per 1 L solution. If a patient receives 1250 ml of solution, how many moles of chloride were given? 42  Properties of Solutions ► The properties of solutions such as vapor pressure, boiling point and freezing point are slightly different from the pure water. ► The vapor pressure of solutions are lower, the boiling points are higher and the freezing points are lower than those for pure water. ► These properties depend on the concentration of the dissolved solutes and are known as colligative properties. 43 1. Vapor pressure lowering When a solute is added to the solvent, some of the solute molecules occupy the space near the surface of the liquid, as shown in the figure below, So the number of solvent molecules near the surface decreases, and the vapor pressure of the solvent decreases. Pure solvent Solvent + solute 44 2. Boiling point elevation Since, addition of a nonvolatile solute lowers the vapor pressure of the solution, then it follows that the temperature must be raised to restore the vapor pressure to the value corresponding to the pure solvent. In particular, the temperature at which the vapor pressure is 1 atm will be higher than the normal boiling point by an amount known as the boiling point elevation. Ex. The boiling point of 1 kg (1L) of water is raised by 0.51°C on addition of 1 mol of non-ionized substances e.g. glucose, whereas the boiling point of 1 Kg (1L) of water is raised by 1.02°C (2 x 0.51°C) on the addition of 1 mol of NaCl. 45 Calculation of boiling point elevation (∆Tboiling) of a solution ∆Tboiling = 0.51 x (M of solute) x (number of particles) Ex: What is the b.p. of a solution of 0.75 mol KBr in 1 kg water? ∆Tboiling = 0.51 x 0.75 x 2 = 0.765 b.p. of pure water = 100 °C b.p. of solution = 100 + 0.765 = 100.765 °C 46 3. Freezing Point depression The freezing point of 1 kg (1L) of water is lowered by - 1.86°C on addition of 1 mole of antifreeze non-ionized substances e.g. ethylene glycol [0.00°C to - 1.86°C] and from [0.00°C to -3.72°C] by adding 1 mole of NaCl. Na2SO4 Ca(NO2)2 FeCl3 Fe+3 3 Cl-1 47 Calculation of freezing point depression of (∆Tfreezing) of a solution ∆T freezing = -1.86 x (concentration of solute) x (number of particles) Ex. What is the freezing point of 1 kg water, when: a. 1.5 mole glucose (C6H12O6) is added b. 1.5 mole NaCl is added (T freezing of pure water = 0.00 °C) a. ∆T freezing = -1.86 x 1.5 x 1 = - 2.79 T freezing = 0.00 - 2.79 = -2.79 °C b. ∆T freezing = -1.86 x 1.5 x 2 = - 5.58 T freezing = 0.00 – 5.58 = -5.58 °C 48  Read only Applications ►A mixture of salt and ice is used to provide the low temperatures needed to make ice cream. ►Salt is added to melt ice by reducing the freezing point of water ►Antifreeze substance (ethylene glycol) is added to water in the car-cooling system to prevent freezing of water inside. 49  Osmosis and Osmotic Pressure ► Osmosis: Osmosis is a process of movement of water through a semi-permeable membrane from a solution of low -solute concentration to a solution of higher-solute concentration. The flow of water will continue till a state of equilibrium is established. Osmosis is also considered as one of the colligative properties of solutions. ► In osmosis, the semi-permeable membrane allow only water molecules and small ions (Na+) to pass through. Osmotic pressure ► Osmotic pressure: Pressure opposes the flow of water from solution of low-solute to solution of higher -solute concentration. ► At equilibrium both forces, 1M 3M water osmosis and osmotic pressure, are equal. 50 ► Osmolarity (osmol): This term is used to express the concentration of solute in solution. It equals to Solute Concentration (M) x No of particles of solute. Osmolarity = (M) x (No. of particles) e.g. Physiological solutions of RBC and plasma have the same NaCl concentration, which is 0.15 M or 0.3 osmol (2 x 0.15M = 0.3 osmol). ► Based of NaCl concentrations in RBC and plasma: Isotonic: Plasma and RBC have the same osmolarity (0.3 osmol). Hypotonic: Plasma has less osmolarity than RBC (< 0.3 osmol). Hypertonic: Plasma has higher osmolarity than RBC (>0.3 osmol). 51  0.3 osmol ► The cells in (a) are placed in an isotonic solution, 0.30 osmol (0.15 M NaCl). The blood cells are normal in appearance. ► The cells in (b) are placed in a hypotonic solution ( 0.3 osmol NaCl). The cells shrink because of water loss, this process is called crenation. 52 53  Dialysis ► Dialysis is similar to osmosis, except that the pores in a dialysis membrane are larger than those in an osmotic membrane so that both water molecules and small solute particles (moves opposite to water movement) can pass through, but large particles such as, proteins and hemoglobin cannot pass. ► Hemodialysis is used to clean the blood of patients with kidneys malfunction (Kidney failure) from toxic waste products (uric acid, urea, ammonia..). Blood is diverted from the body and pumped through a long cellophane dialysis tube suspended in an isotonic solution formulated to contain many of the same components as blood plasma. 54 ► Waste materials such as urea, uric acid will pass through the dialysis membrane from the blood stream to the solution-out side. Big molecules (proteins, heamoglobin) will go back to human body. ammonia Uric acid urea urea, uric acid ammonia = 0.3 osmol 55 Normality = (M) x (No. charge) Osmolarity = (M) x (No. of particles) ∆Tboiling = 0.51 x (M) x (No. of particles) ∆Tfreezing = -1.86 x (M) x (No. of particles) Compund Compund Charge Charge No. No. of of particles particles NaCl NaCl 1 2 Na Na2CO 2CO 33 2 3 AlCl AlCl3 3 3 4 CaCO3 3 CaCO 2 2 Mg3(PO Mg 3(PO ) 4)42 2 6 5 56  Homework ► Problem: Solubility of Gases From the graph shown of solubility versus temperature for oxygen gas, determine the concentration of dissolved oxygen in water at 25oC and at 35oC. By what percentage does the concentration of oxygen gas change? solubility at 25oC = 8.3 mg/L solubility at 30oC = 7.0 mg/L ► A solution is prepared by dissolving 62.5 g of KBr in 100 ml of water at 60oC. Is this solution saturated, unsaturated or supersaturated? What will happen if the solution is cooled to 10oC? ► The solution is unsaturated at 60oC. ► At 10oC the solution become supersaturated. And if we add tiny crystal of KBr 2.5 gm of KBr will precipitate. 57 Homework How many milliliters of gastric juice (0.1 M HCl) will react completely with an antacid tablet that contain 500 mg of Mg(OH)2 (MW = 58.3) ? 2 HCl + Mg(OH) 2 MgCl2 + 2H2O N1 X V1 = (Wt X n / MW)2 1: gastric juice, 2: base Mg(OH)2, volume in litters and weight in * grams N = M n (HCl) N = 0.1M X 1 = 0.1N 0.1 N X V1 = (0.500 g X 2 / 58.3) V1 = (0.500 g X 2 / 0.1 M X 58.3) V1 = (1 g/ 5.83) = 0.171 L V1 = 0.171 L X 1000 = 170 ml 58 Which of the following solution is more concentrated 0.5 M KCl, 5% (m/v) KCl or 15 ppm KCl (MW=74.5) ? Convert w/v % to M: 5 % means 5 gm in 100 mL Convert ppm to M: 15 ppm means 15 mg in 1L 15 ppm (0.003 M) < 0.5 M KCl < 5% (m/v) KCl (0.67 M) 59 a) KBr = 0.25 X 2 = 0.5 osmol Na2SO4 = 0.20 X 3 = 0.6 osmol b) 3% NaOH = ??? M (MW = 40) = 3 / (40 X 0.100) =0.75 M Osmolarity of 3% NaOH = M X 2 = 0.75 X 2 = 1.5 osmol Osmolarity of 0.3 M NaOH = M X 2 = 0.3 X 2 = 0.6 osmol 60