Chapter 17 - Properties of Solutions 17.1 Solution Composition 17.2 Thermodynamics of Solution Formation 17.3 Factors Affecting Solubility 17.4 Vapor Pressures of Solutions 17.5 Boiling-Point Elevation and Freezing-Point Depression 17.6 Osmotic Pressure 17.7 Colligative Properties of Electrolyte Solutions 17.8 Colloids Definition of Solution (Ch. 16 dealt with pure substances rather than mixtures) Solution when the components of a mixture are uniformly intermingled; the mixture is homogenous Examples: air, seawater, steel NOT: water (pure), wood (not homogeneous) Colloid a suspension of tiny particles in some medium is called a colloidal dispersion, or a colloid; the mixture is heterogeneous Examples: gelatin, fog, butter Definitions for Solutions Solute - The smaller (in mass) of the components in a solution; the material dispersed into a solvent. Solvent - The major component of the solution; the material that the solute is dissolved into. Solubility - The maximum amount that can be dissolved into a particular solvent to form a stable solution at a specified temperature. Miscibility - The ability of two substances to dissolve in one another in any proportion. Solutions do not need to be liquid 1
Types of Colloids Methods for Quantifying How Much Solute is in a Solvent A quick note on parts per notation One part per hundred: Typically represented by the percent (%) symbol and denotes one part per 100 parts, one part in 10 2, and a value of 1 10 2. This is equivalent to one drop of water diluted into 5 milliliters (one spoon-full) or one second of time in 1⅔ minutes. One part per million (ppm): Denotes one part per 1,000,000 parts, one part in 10 6, and a value of 1 10 6. This is equivalent to one drop of water diluted into 50 liters (roughly the fuel tank capacity of a compact car), or one second of time in approximately 11½ days. One part per billion (ppb): Denotes one part per 1,000,000,000 parts, one part in 10 9, and a value of 1 10 9. This is equivalent to 1 drop of water diluted into 250 chemical drums (50 m 3 ), or one second of time in approximately 31.7 years. One part per trillion (ppt): Denotes one part per 1,000,000,000,000 parts, one part in 10 12, and a value of 1 10 12. This is equivalent to 1 drop of water diluted into 20, two-meter-deep Olympic-size swimming pools (50,000 m 3 ), or one second of time in approximately 31,700 years. Remember!! Concentrations are ratios. They are not additive! Volumes are additive: V(total) = V 1 + V 2 + V 3 + (in general) Masses are additive: Moles are additive: m(total) = m 1 + m 2 + m 3 + n(total) = n 1 + n 2 + n 3 + Concentrations are not additive: c = c 1 + c 2 + c 3 + 2
Calculating Molality - I To calculate molality (mol solute/kg solvent), we need the number of moles of solute and the mass of solvent used to dissolve the solute! Normally, we are given the mass of solute and mass of solvent, therefore we calculate the moles of the solute from the mass, then use the mass of the solvent to calculate the molality. Mass (g) of solute Remember: Molality is different from MM (g/mole) molarity. Molality is based on mass, and is independent of Moles of Solute temperature or pressure (unlike molarity). Because 1 L of H 2 O weighs 1 kg, molality and divide by kg water molarity of dilute aqueous solutions are nearly identical. Molality (m) of solution Calculating Molality - II Problem: Determine the molality and molarity of a solution prepared by dissolving 75.0 g Ba(NO 3 ) 2 (s) in 374.0 g of water at 25 0 C. Plan: We convert the quantity of Ba(NO 3 ) 2 to moles using the molar mass and then divide by the mass of H 2 O in kg or the volume of H 2 O in liters (using water density = 0.99707 g/ml). Solution: molar mass of Ba(NO 3 ) 2 = 261.32 g/mol 75.0 g moles Ba(NO 3 ) 2 = = 0.287 mole 261.32 g/mol 0.287 mole Ba(NO molality= 3 ) 2 = 0.767 m 0.37400 kg H 2 O molarity - we need the volume of solution, and can assume that addition of the salt did not change the total volume. 374.0 g H 2 O = 375.1 ml = 0.3751 L 0.99707 g/ml M = 0.287 mole 0.3751 L = 0.765 M Expressing Concentrations in Parts by Mass Expressing Concentrations in Parts by Volume Problem: Calculate the parts by mass of iron in a 1.85 g Fe supplement pill that contains 0.0543 mg of Fe. Plan: Convert mg Fe to grams and then use mass of Fe/mass of pill. Solution: 0.0543 mg Fe = 5.43 x 10-8 g Fe 5.43 x 10-8 g Fe 1.85 g pill = 2.94 x 10-8 If 1 x 10-9 = 1 part per billion, 29.4 x 10 9 = 29.4 parts per billion (ppb) Problem: The label on a can of beer (340 ml) indicates 4.5% alcohol by volume. What is the volume in ml of alcohol in the can? Plan: We know the vol% and the total volume so we use the definition of parts by volume to find the volume of alcohol. Solution: Vol. Alcohol = x 3
Expressing Concentrations in Mole Fraction Problem: A sample of alcohol contains 118 g of ethanol (C 2 H 5 OH), and 375.0g of water. What are the mole fractions of each? Plan: We know the mass and formula of each component so we convert both to moles and apply the definition of mole fraction. Solution: Moles ethanol = Moles water = 118 g ethanol 46 g ethanol/mol 395 g H 2 O 18 g H 2 O/mol = 2.565 mol ethanol = 21.94 mol H 2 O X ethanol = 2.565 21.94 +2.565 = 50.10 X water = 21.94 21.94 +2.565 = 0.895 Converting Concentration Units - I Problem: Commercial concentrated hydrochloric acid is 11.8 M HCl and has a density of 1.190 g/ml. Calculate the (a) mass% HCl, (b) molality and (c) mole fraction of HCl. Plan: We know Molarity and density. (a) For mass% HCl we need the mass of HCl and water (the solvent). Assume 1L of solution: from the density we know the mass of the solution and from the molecular mass of HCl we can calculate its mass. Solution: (a) assume 1L of HCl solution 11.8 moles HCl 36.46 g HCl 11.8 moles HCl x = 430.2 g HCl mole HCl 1 L solution x 1000 ml x 1.190 g soln = 1190. g solution 1 L solution ml soln mass % HCl = 430.2 g HCl 1190. g solution x 100 = 36.2 % HCl Converting Concentration Units - II Problem: Commercial concentrated hydrochloric acid is 11.8 M HCl and has a density of 1.190 g/ml. Calculate the (a) mass% HCl, (b) molality and (c) mole fraction of HCl. Plan: (b) From (a) we know moles of HCl, mass of solution, and mass of HCl. Calculate mass of water and use definition of molality. Solution: (b) mass of H 2 O = mass of solution - mass of HCl = 1190 g solution - = 759.7 g H 2 O 11.8 moles HCl 0.7597 kg H 2 O 430.2 g HCl = 15.5 m HCl Converting Concentration Units - III Problem: Commercial concentrated Hydrochloric acid is 11.8 M HCl and has a density of 1.190 g/ml. Calculate the (a) mass% HCl, (b) molality and (c) mole fraction of HCl. Plan: (c) From (a) we know the moles HCl and from (b) we know the mass of water. Calculate moles of water then mole fractions. Add the mole fractions to check! Solution: (c) 759.7 g H 2 O = 42.17 mole H 18.016 g H 2 O/mol H 2 O 2 O Total moles = 42.172 + 11.8 = 53.972 = 54.0 X HCl = 11.8 = 0.219 54.0 X H2O = = 0.781 42.17 54.0 4
Chapter 17 - Properties of Solutions 17.1 Solution Composition 17.2 Thermodynamics of Solution Formation 17.3 Factors Affecting Solubility 17.4 Vapor Pressures of Solutions 17.5 Boiling-Point Elevation and Freezing-Point Depression 17.6 Osmotic Pressure 17.7 Colligative Properties of Electrolyte Solutions 17.8 Colloids Ch.17 Motto: Like Dissolves Like! Polar/Ionic solutes - dissolve best in polar solvents. Polar molecules can exhibit strong dipole-dipole intermolecular interactions (ionic solutes: ion-dipole interactions) with polar solvents, hence increasing their solubility. Also called hydrophilic solutes. Non-polar solutes - dissolve best in non-polar solvents. To dissolve a non-polar molecule in a polar solvent requires breaking strong dipole-dipole interactions of the solvent! Also called hydrophobic solutes. Key Point: Forces created between solute and solvent must be comparable in strength to the forces destroyed within both. What happens when solutes dissolve? The formation of solutions from pure substances is always favored by ENTROPY! When two pure substances are mixed, the disorder of the system is increased As with any chemical reaction, solvation is a balance between enthalpy and entropy Gibbs Free Energy G = H - T S 5
A quantitative description of solvation: The Solution Cycle Step 1: Solute separates into components - overcoming attractions in order to expand and make room for solvent -- Endothermic Step 2: Solvent expands - overcoming intermolecular attractions to spread out and make room for the solute -- Endothermic A quantitative description of solvation: The Solution Cycle Figure 17.1 solvent (aggregated) + heat solvent (separated) DH solvent > 0 Step 3: Solute and Solvent particles mix - particles attract each other and give off energy as they interact -- Exothermic solute (separated) + solvent (separated) solution + heat DH mix < 0 FIGURE 17.2 (a) Steps in an Exothermic Reaction (b) Steps in an Endothermic Reaction Enthalpy Balance The Thermochemical Cycle: DH solution = DH solute + DH solvent + DH mix If DH Endothermic steps <DH Exothermic steps solution becomes warmer If DH Endothermic steps >DH Exothermic steps solution becomes colder 6
Solution Cycles and the Enthalpy Components of the Heat of Solution Exothermic Solution Process Endothermic Solution Process see Fig. 17.2 7
Like dissolves like: solubility of methanol in water Hydration shells around an aqueous ion more charge density (charge/size ratio of ion) results in greater hydration Demo: non-additive volumes 250 ml ethanol + 250 ml water Does 250 +250 always equal 500? Effect of Structure on Solubility Nonpolar, fat-soluble (builds up in fatty tissues pros and cons!) Polar, water-soluble (must be consumed regularly pros and cons!) Other examples: Vitamins D, E, and K (fat soluble); Vitamin B (water-soluble) DDT pesticide (fat soluble) Dioxins group of molecules, herbicides and pollutants (fat soluble) BaSO 4 gastroenterology, enhances X-rays (insoluble) Predicting Relative Solubilities of Substances Problem: Predict which solvent will dissolve more of the given solute. (a) Sodium Chloride in methanol (CH 3 OH) or in propanol (CH 3 CH 2 CH 2 OH). (b) Ethylene glycol (HOCH 2 CH 2 OH) in water or in hexane (CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 ). (c) Diethyl ether (CH 3 CH 2 OCH 2 CH 3 ) in ethanol (CH 3 CH 2 OH) or in water. Plan: Examine each solute and solvent to determine which intermolecular forces will be active. A solute tends to be more soluble in a solvent that has the same type of intermolecular forces active. 8
Predicting Relative Solubilities of Substances - II Solution: (a) - NaCl is an ionic compound that dissolves through ion-dipole forces. Both methanol and propanol contain a polar O-H group, but propanol s longer hydrocarbon chain would interact only weakly with the ions and be less effective in stabilizing the ions. (b) - ethylene glycol has two O-H groups and is stabilized by extensive H-bonding in water. (c) - diethyl ether shows both dipolar and dispersion intermolecular forces and could form H bonds to both water and ethanol. The ether would be more soluble in ethanol because solvation in water must disrupt many more strong H-bonding interactions. Effect of Temperature on Solubility Solubility increases with increasing temperature IF the solution process is endothermic: Solute + Solvent + Heat Solution Solubility decreases with increasing temperature IF the solution process is exothermic: Solute + Solvent Solution + Heat Recall Le Châtelier s Principle Figure 17.4: Increase in gas solubility with increased pressure Predicting the temperature dependence of solubility is very difficult. The rate of dissolution might increase with temperature, but the total amount of solute that will dissolve with increasing/decreasing temperature is unpredictable. (You have to do an experiment!) At equilibrium Increase in P (and r) Increase in [gas dissolved ] 9
Henry s Law The amount of gas dissolved in a solution is directly proportional to the pressure of the gas above the solution. P = k H X P partial pressure of the gaseous solute above the solution X mole fraction of the dissolved gas k H a constant characteristic of a particular solution Requirement: This law holds only if there is no chemical reaction between the solute and the solvent and if the solution is fairly dilute. Henry s Law - Lake Nyos in Cameroon CO 2 buildup in colder, denser water layers near lake s bottom (high pressure, large mole fraction); overturn in 1986 sent water supersaturated with CO 2 to the surface and released an enormous amount of CO 2 into the area, suffocating thousands of humans and animals. For all gases: H solute 0, so H soln < 0 Gas solubility always decreases with increasing T! Colligative Properties Properties of solutions that depend ONLY on the number of molecules in a given volume of solvent and NOT the properties (e.g. size or mass) of the molecules. Includes: vapor-pressure lowering boiling-point elevation freezing-point depression osmotic pressure Colligative Properties of Solutions 17.4) Vapor Pressure Lowering - Raoult s Law 17.5) Boiling-Point Elevation and Freezing-Point Depression 17.6) Osmotic Pressure 17.7) Colligative Properties of Electrolyte Solutions We ll start by focusing on nonvolatile, nonelectrolytic solutes (e.g. sucrose in H 2 O) Later: volatile solutes (end of 17.4), electrolytic solutes (17.7) 10
Raoult s Law: P solvent = x solvent * P solvent Pure solvent increases P vap to get to equilibrium Solution absorbs vapor to lower P vap to get to equilibrium Figure 17.8, Zumdahl x is the solvent mole fraction P solvent is the vapor pressure of pure solvent See Figure 17.9 Equilibrium can only be achieved after all the pure solvent has been transferred to the solution. Example: Vapor Pressure Lowering Problem: Calculate the vapor pressure lowering (DP) when 175 g of sucrose is dissolved into 350.0 ml of water at 75 0 C. The vapor pressure of pure water at 75 o C is 289.1 mm Hg, and its density is 0.97489 g/ml. Plan: Calculate the change in pressure from Raoult s law using the vapor pressure of pure water at 75 o C. We calculate the mole fraction of sugar in solution using the molecular formula of sucrose and density of water at 75 o C. Solution: molar mass of sucrose (C 12 H 22 O 11 ) = 342.30 g/mol 175 g sucrose = 0.511 mol sucrose 342.30 g sucrose/mol 350.0 ml H 2 O 0.97489 g H 2 O = 341.2 g H 2 O ml H 2 O 341.2 g H 2 O 18.02 g H 2 O/mol = 18.94 mol H 2O x water = Vapor Pressure Lowering (cont) P = x solvent * P solvent moles water moles of water + moles of sucrose 18.94 mol H = 2 O = 0.9737 18.94 mol H 2 O + 0.511 mol sucrose P = x water * P solvent = 0.9737 x 289.1 mm Hg = 281.5 mm Hg DP = P solvent - P = 289.1 mm Hg - 281.5 mm Hg = 7.60 mm Hg 11
Raoult s Law is a linear equation: P solvent = x solvent P solvent y = x * m As nonvolatile solute is added moles solute increases, so x solvent decreases. x solvent = moles solvent moles solute + moles solvent Therefore, as x solvent decreases, P solvent decreases. Ideal/Non-ideal Solutions Ideal behavior is approached when solute and solvent are involved in similar intermolecular interactions ( H soln = 0) (i.e., follow Raoult s law! solute only dilutes the solvent, no interaction) When weaker solvent-solute interactions occur, heat is removed upon dissolving ( H soln > 0) the observed vapor pressure is higher than ideal ( deviation from Raoult s law) When stronger solute-solvent interactions occur (e.g. if H- bonding occurs), heat is released upon dissolving ( H soln < 0) the observed vapor pressure is lower than ideal ( deviation from Raoult s law) Non-ideal solutions Figure 17.11 Summary of Ideal/Non-ideal Solutions Ideal H soln > 0 H soln < 0 12
What if the solute is volatile? P total = P solute + P solvent P solvent = x solvent P o solvent P solute = x solute P o solute Dalton s Law of Partial Pressures Chapter 5 P total = (x solute P o solute) + (x solvent P o solvent) Problem: Consider a solution having equal molar quantities of acetone and chloroform, x acetone = x CHCl3 = 0.500. At 35 C, the vapor pressures of the pure substances are: P acetone = 345 torr P CHCl3 = 293 torr What are the mole fractions, x, of each component in the vapor phase? Determine the partial pressure of each component and the vapor pressure of the solution. P acetone = x acetone P 0 acetone = 0.500 345 torr = 172.5 torr P CHCl3 = x CHCl3 P 0 CHCl 3 = 0.500 293 torr = 146.5 torr Total Pressure = 172.5 + 146.5 = 319.0 torr P From Dalton s law of partial pressures we know that x A A = P Total P acetone x acetone = = 172.5 torr P = 0.541 Total 319.0 torr x CHCl3 = P CHCl3 = 146.5 torr = 0.459 319.0 torr P Total Vapor is enriched in acetone! Let s condense the vapor and calculate the new mole fraction of each component in the vapor phase. P acetone = x acetone P 0 acetone = 0.541 345 torr = 186.6 torr P CHCl3 = x CHCl3 P 0 CHCl 3 = 0.459 293 torr = 134.5 torr Total Pressure = 186.6 + 134.5 = 321.1 torr P acetone 186.6 torr x acetone = P = = 0.581 Total 321.1 torr x CHCl3 = P CHCl3 = 134.5 torr = 0.419 321.1 torr P Total Vapor is more enriched in acetone! 13
Boiling Point Elevation Recall: The boiling point (T b ) of a liquid is the temperature at which its vapor pressure equals atmospheric pressure. Nonvolatile solutes lower the vapor pressure of a liquid, making it more difficult for the solution to boil greater temperature required to reach boiling point Boiling Point Elevation: T b = K b m solute K b = molal boiling point constant for given solvent m solute = solute concentration in molality K b = molal boiling point constant for given solvent K f = molal freezing point constant for given solvent Freezing Point Depression: T f = K f m solute Presence of solute lowers the rate at which molecules in the liquid return to the solid; have to lower the temp to decrease the rate of molecules leaving the solid for the liquid until the rates are equal (equilibrium) See Figure 17.3 14
See Figure 17.12 Effect of solute: extends the liquid range of the solvent Example: Boiling Point Elevation and Freezing Point Depression in an aqueous solution Problem: We add 475 g of sucrose (sugar) to 600 g of water. What will be the Freezing and Boiling points of the solution? Plan: Find the molality of the sucrose solution and apply the equations for FP depression and BP elevation, using the constants from table 17.5. Solution: 475 g sucrose = 1.39 mole sucrose 342.30 g sucrose/mol 1.39 mole sucrose molality = = 2.31 m 0.600 kg H 2 O T b = K b * m = T f = K f * m = 0.51 o C m 1.86 o C m (2.31m)= 1.2 o C BP = 100.00 o C + 1.2 o C = 101.18 o C (2.31 m) = 4.3 o C FP = 0.00 o C - 4.3 o C= - 4.3 o C Example: Boiling Point Elevation and Freezing Point Depression in a Non-aqueous Solution Problem: Calculate the Boiling Point and Freezing Point of a solution having 257 g of naphthalene (C 10 H 8 ) dissolved into 500.0 g of chloroform (CHCl 3 ). Plan: Just like the previous example. Solution: MM naphthalene = 128.16 g/mol; MM chloroform = 119.37 g/mol moles naph = 257 g nap = 2.01 mol C 10 H 8 128.16g/mol molality = moles nap = 2.01 mol = 4.01 m kg(chcl 3 ) 0.500 kg T b = K * b m = 3.63 o C (4.01m) = 14.6 o C normal BP = 61.7 o C m new BP = 76.3 o C T f = K * 4.70 f m = o C (4.01m) = 18.9 o C normal FP = - 63.5 o C m new FP = - 82.4 o C Osmosis: The flow of solvent through a semipermeable membrane into a solution The semipermeable membrane allows solvent molecules to pass, but not solute molecules See Figures 17.15 & 17.16 15
Osmotic Pressure: π = MRT Similar to ideal gas law! R = gas constant M = molar concentration of solute Osmotic Pressure Dialysis (does allow the transfer of some solute particles) Fig 17.14 Determining Molar Mass from Osmotic Pressure Problem: A physician studying hemoglobin dissolves 21.5 mg of the protein in water at 5.0 o C to make 1.5 ml of solution in order to measure its osmotic pressure. At equilibrium, the solution has an osmotic pressure of 3.61 torr. What is the molar mass (MM) of the hemoglobin? Plan: We know p, R, and T. We convert p from torr to atm, and T from o C to K, and then use the osmotic pressure equation to solve for molarity (M). Then we calculate the number of moles of hemoglobin from the known volume and use the known mass to find MM. Solution: p = 3.61 torr * 1 atm = 0.00475 atm 760 torr Temp = 5.0 0 C + 273.15 = 278.2 K Molar Mass from Osmotic Pressure Concentration from osmotic pressure: M = π = 0.00475 atm = 2.08 x10-4 M RT 0.0821 L atm (278.2 K) mol K Finding # moles of solute: n = M * V = 2.08 x10-4 mol L soln * 0.00150 L soln = 3.12 x10-7 mol Calculating molar mass of hemoglobin (after changing mg to g): MM = 0.0215 g 3.12 x10-7 mol = 6.89 x10 4 g/mol 16
Colligative Properties of Ionic Solutions For ionic solutions we must take into account the number of ions present! i = van t Hoff factor = ionic strength, or the # of ions present For Electrolyte Solutions: vapor pressure lowering: boiling point elevation: P = i X solute P 0 solvent T b = i K b m freezing point depression: T f = i K f m Non-ideal Behavior of Electrolyte Solutions (0.05m aqueous) Most likely the result of ion pairing in the solution (ions periodically interacting and behaving as one particle). osmotic pressure: π = i MRT See Table 17.6 17