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- 1. Properties of Solutions AP Chem Unit 11
- 2. Solutions SectionsSolution Composition The Energies of Solution Formation Factors Affecting Solubility The Vapor Pressures of Solutions Boiling-Point Elevation and Freezing-Point Depression Osmotic Pressure Colligative Properties of Electrolyte Solutions Colloids
- 4. Solution Composition Solutions are homogeneous mixtures that can be gases, liquids or solids. Solutions can be described as dilute or concentrated. molarity mass percent mole fraction molality
- 5. Concentrations Molarity = Mass % = Mole fraction of A= Molality = M = moles liters mass of solute mass of solution æ èç ö ø÷ x 100% XA = nA nA + nB moles of solute kg of solvent
- 6. Practice Problem 1 A Solution is prepared by mixing 1.00g of ethanol (C2H3OH) with 100.0g of water to give a final volume of 101 ml. Calculate the molarity, mass percent, mole fraction and molality of ethanol in this solution. M=.215, %=.990%, X=.00389, m=.217
- 7. Normality Normality is defined as the number of equivalents per liter of solution. equivalents depends on the reaction taking place in the solution example: for an acid base reaction, the equivalent is the mass of acid or base that can furnish or accept exactly 1 mole of protons (H+ ions). H2SO4 or Ca(OH)2 equivalents = molar mass/2 one equivalent of acid reacts with one equivalent of base.
- 8. Normality • For oxidation - reduction reactions, the equivalent is defined as the quantity of oxidizing or reducing agent that can accept or furnish 1 mole of electrons. one equivalent of reducing agent will react with exactly one equivalent of oxidizing agent.
- 9. Example The equivalent mass of an oxidizing or reducing agent can be calculated from the number of electrons in a half reaction. MnO4- + 5e- + 8H+ Mn2+ + 4H2O Since the MnO4- ion present in 1 mole of KMnO4 in acidic solution consumes 5 moles of electrons, the equivalent mass is the molar mass divided by 5: Eq mass of KMnO4 = molar mass 5 = 158g 5 = 31.6g
- 10. Practice Problem 2 The electrolyte in automobile lead storage batteries is a 3.75 M sulfuric acid solution that has a density of 1.230 g/ml. Calculate the mass percent, molality, and normality of the sulfuric acid. 7.50 N
- 11. ENERGIES OF SOLUTION FORMATION
- 12. Solubility What factors affect solubility? The cardinal rule of solubility is that like dissolves like. Polar solvents are used to dissolve a polar or ionic solute and nonpolar solvents are used to dissolve a nonpolar solute.
- 13. Solubility Solubility and formation of a solution takes place in three distinct steps: 1. Separating the solute into its individual components (expanding the solute). 2. Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent). 3. Allowing the solute and solvent to interact to form the solution.
- 14. Solubility 1. Expanding the solute: endothermic 2. Expanding the solvent: endothermic 3. Interaction: exothermic Enthalpy of solution = ΔHsoln = ΔH1 + ΔH2 + ΔH3
- 15. Solubility ΔHsoln could be overall negative sign (exothermic). ΔHsoln could be overall positive sign (endothermic).
- 16. Example Oil slicks to not dissolve in water: 1. ΔH1 is endothermic but small. C chains held together with LDF need separated. 2. ΔH2 is endothermic and large. Hydrogen bonds in the water are difficult to separate. 3. ΔH3 will be small since polar and nonpolar interactions are minimal. ΔHsoln is an overall endothermic process and unlikely to produce a solution.
- 17. Example 2 Most ionic substances dissolve in water: 1. ΔH1 is endothermic and large. Ionic forces must be overcome. 2. ΔH2 is endothermic and large. Hydrogen bonds in the water are difficult to separate. 3. ΔH3 will be very exothermic because most ionic substances interact very well with water. ΔHsoln is usually small but can be overall exothermic or endothermic.
- 18. Enthalpy of Hydration Enthalpy of hydration (ΔHhyd ) combines the terms ΔH2 (expanding the solvent) and ΔH3 (solvent-solute interactions). The heat of hydration represents the enthalpy change assoicated with the dispersal of a gaseous solute in water. H2O(l) + Na+ (g) + Cl- (g) Na+ (aq) + Cl- (aq) ΔHsoln = ΔH1 + ΔHhyd
- 19. Example 3 Heat of solution of NaCl and water: 1. NaCl(s) Na+ (g) + Cl- (g) ΔH1=786 kJ/mol 2. & 3. H2O(l) + Na+ (g) + Cl- (g) Na+ (aq) + Cl- (aq) ΔHhyd=783 kJ/mol ΔHsoln=3kJ, endothermic but small.
- 20. Solubility The dissolving process often requires a small amount of energy (stirring, heat), but ionic substances, even though their expansion is very endothermic, dissolve due to their tendency toward increased probability. Processes that require large amounts of energy tend not to occur.
- 21. Practice Problem 3 Decide whether liquid hexane (C6H14) or liquid methanol (CH3OH) is the more appropriate solvent for the substances grease (C2oH42) and potassium iodide (KI). Hexane works best for grease and methanol serves as a better solvent for potassium iodide
- 23. Structure and Solubility Since polarity is a big factor in solubility and molecular structure determines polarity, structure has a lot to do with solubility. Vitamins are divided into fat soluble and water soluble due to their structures:
- 24. Pressure Effects Pressure has little effect on the solubilities of solids or liquids, but it does significantly increase the solubility of a gas.
- 25. Henry’s LawHenry’s law states that the amount of a gas dissolved in a solution is directly proportional to the pressure of the gas above the solution. The relationship between gas pressure and the concentration of dissolved gas is given by: C=kP C = the concentration of the dissolved gas, k is a constant characteristic of a particular solution and P represents the partial pressure of the gaseous solute above the solution. Henry’s law is obeyed most accurately for dilute solutions that do not dissociate or react with the solvent.
- 26. Practice Problem 4 A certain soft drink is bottled so that a bottle at 25°C contains CO2 gas at a pressure of 5.0 atm over the liquid. Assuming that the partial pressure of CO2 in the atmosphere is 4.0 x 10-4 atm, calculate the equilibrium concentrations of CO2 in the soda both before and after the bottle is opened. The Henry’s law constant for CO2 in aqueous solution is 3.0 x 10-2 mol/Latm at 25°C. .16 mol/L and 1.2 x 10-5 mol/L
- 27. Temperature Effects Solubility doesn’t always increase with temperature. The dissolving of a solid occurs more rapidly at higher temperatures, but the amount of solid that can be dissolved may increase or decrease with increasing temperature. Predicting the temperature dependence of solubility is very difficult. The only sure way to determine the temperature dependence of a solid’s solubility is by experiment.
- 29. Temperature Effects The behavior of gases dissolving in water typically decrease with increasing temperature. Thermal pollution in lakes and rivers Boiler scale – 2Ca2+ + HCO3(aq) H2O + CO2(aq) + CaCO3(aq)
- 31. VAPOR PRESSURE OF SOLUTIONS
- 32. Vapor Pressure and Solutions Liquid solutions have physical properties significantly different from those of the pure liquid solvent. Antifreeze with water delays freezing and boiling. Salt and water lowers freezing point.
- 33. Vapor Pressure and SolutionsA nonvolatile solute lowers the vapor pressure of a solvent. 1. Vapor pressure of the pure solvent is greater than that of the solution. 2. The equilibrium vapor pressure of the pure solvent (water) is greater than that of the solution equilibrium vapor pressure. 3. Water vaporizes and adds to solution.
- 34. Vapor Pressure and SolubilityRaoult’s Law: Psoln = Xsolvent Po solvent • Psoln is the observed vapor pressure of the solution, Xsolvent is the mole fraction of solvent, and Po solvent is the vapor pressure of the pure solvent. • In a solution consisting of half nonvolatile solute molecules and half solvent molecules (typically water), the observed vapor pressure is half of that of the pure solvent, since only half as many molecules can escape. • If vapor pressures are known, moles can be determined.
- 35. Practice Problem 5 Calculate the expected vapor pressure at 25°C for a solution prepared by dissolving 158.0 g of sucrose, molar mass= 342.3 g/mol, in 643.5cm3 of water. At 25°C, the density of water is 0.9971 g/cm3 and the vapor pressure is 23.6 torr. 23.46 torr
- 36. Vapor Pressure and Solubility The lowering of vapor pressure depends on the number of ionic solute particles present in the solution. Example: 1 mole of sodium chloride dissolved in water lowers the vapor pressure approximately twice as much as expected because the solid has two ions per formula unit, which separates when it dissolves.
- 37. Practice Problem 6 Predict the vapor pressure of a solution prepared by mixing 35.0 g solid Na2SO4 (molar mass= 142.05 g/mol) with 175 g water at 25°C. The vapor pressure of pure water at 25°C is 23.76 torr. (how many moles of solute particles are present?) 22.1 torr
- 38. Nonideal Solutions When both solvent and solute are volatile, both contribute to the vapor pressure over the solution. A modified form of Raoult’s law is used: Ptotal = PA + PB = XAPo A + XBPo B • Ptotal is the total vapor pressure of solution AB. XA and XB are the mole fractions of A and B. PA and PB are the partial pressures of A and B.
- 39. Ideal vs. Nonideal A liquid-liquid solution that obeys Raoult’s law is called an ideal soution. Raoult’s law is to solutions what the ideal gas law is to gases. Nearly ideal behavior is often observed when solutes and solvents are similar. • When strong interactions occur (ΔHsoln is very exothermic), a negative deviation of Raoult’s law results • When weak interactions occur (ΔHsoln is endothermic), a positive deviation of Raoult’s law results.
- 40. Behavior Summary
- 41. Practice Problem 7 A solution is prepared by mixing 5.81 g acetone (C3H6O, molar mass = 58.1 g/mol) and 11.9 g chloroform (HCCl3, molar mass = 119.4 g/mol). At 35°C, this solution has a total vapor pressure of 260 torr. Is this an ideal solution? The vapor pressures of pure acetone and pure chloroform at 35°C are 345 and 293 torr, respectively. Acetone (CH3)2CO and chloroform CHCl3 319 is the expected total vapor pressure; this is not an ideal solution. This is a negative deviation, therefore interactions must be strong.
- 42. BOILING-POINT ELEVATION AND FREEZING-POINT DEPRESSION
- 43. Vapor Pressure and Freezing/Boiling Points Since changes of state depend on vapor pressure, the presence of a solute also affects the freezing point and boiling point of a solvent. • Freezing point depression, boiling point elevation and osmotic pressure are called colligative properties. • These properties are dependent on the number of solute particles and not the identity of the particles.
- 44. Boiling Point Elevation The normal boiling point of a liquid occurs at the temperature at which the vapor pressure is equal to 1 atmosphere. A nonvolatile solute lowers the vapor pressure of the solvent; therefore such a solution must be heated to a higher temperature than the ‘pure’ boiling point for the vapor pressure to reach 1 atmosphere. • A nonvolatile solute elevates the boiling point of the solvent.
- 45. Boiling Point Elevation Water and a nonvolatile water solution. The boiling point increases and the freezing decreases with the solution. The effect of a nonvolatile solute is to extend the liquid range of a solvent.
- 46. Boiling Point Elevation The magnitude of the boiling point elevation depends on the concentration of the solute. The change in boiling point can be calculated by: ΔT = Kbmsolute • ΔT is the boiling point elevation, Kb is a constant that is characteristic of the solvent and is called the molal boiling-point elevation constant. msolute is the molality of the solute in the solution. • An observed boiling point elevation can determine molar mass.
- 48. Practice Problem 8 A solution was prepared by sissolving 18.00 g glucose in 150.0g water. The resulting solution was found to have a boiling point of 100.34°C. Calculate the molar mass of glucose. Glucose is a molecular solid that is present as individual molecules in solution. 180 g/mol
- 49. Freezing Point Depression Vapor pressures of ice and liquid water are the same at 0°C (freezing point). When a nonvolatile solute is dissolved in water, the vapor pressure of the solution lowers and therefore the freezing point of the solution is lower than that of the pure water. ΔT = Kfmsolute • ΔT is the freezing point depression, Kf is a constant that is characteristic of the solvent and is called the molal freezing-point depressionconstant. msolute is the molality of the solute in the solution.
- 50. Practice Problem 9 What mass of ethylene glycol (C2H6O2, molar mass = 62.1 g/mol), the main componenet of antifreeze, must be added to 10.0 L water to produce a solution for use in a car’s radiator that freezes at -10.0°F (-23.3°C)? Assume the denisty of water is exactly 1 g/mL. 7.76kg
- 51. Practice Problem 10 Mr. Wieland is trying to identify a human hormone that controls metabolism by determining its molar mass. A sample weighing 0.546 g was dissolved in 15.0 g benzene, and the freezing-point depression was determine to be 0.240°C. Calculate the molar mass of the hormone. 776 g/mol
- 52. OSMOTIC PRESSURE
- 53. Osmotic Pressure A solution and pure solvent are separated by a semipermeable membrane, which allows solvent but not solute molecules to pass through. As time passes, the volume of the solution increases and volume of the solvent decreases. The flow of solvent into the solution through the membrane is called osmosis. Eventually the liquid levels stop changing and reach equilibrium, but there is a greater hydrostatic pressure on the solution than on the pure solvent. The excess pressure is called osmotic pressure.
- 54. Osmotic Pressure Osmosis can be prevented by applying pressure to the solution. The minimum pressure that stops the osmosis is equal to the osmotic pressure of the solution.
- 55. Osmotic Pressure Osmotic pressure is primarily dependent on solution concentrations. Π = MRT • Π is the osmotic pressure in atmospheres, M is the molarity of the solution, R is the gas law constant, and T is the Kelvin temperature.
- 56. Practice Problem 11 To determine the molar mass of a certain protein, 1.00 x 10-3 g of it was dissolved in enough water to make 1.00 mL of solution. The osmotic pressure of this solution was found to be 1.12 torr at 25.0°C. Calculate the molar mass of the protein. 1.66 x 104 g/mol (proteins often have large molar masses)
- 57. Osmosis Osmosis prevents transfer of all solute particles. Dialysis (a similar phenomenon) occurs at the walls of most plant and animal cells. In this case, the membrane allows transfer of both solvent and small solute molecules and ions. Solutions that have identical osmotic pressures are said to be isotonic solutions. • Solutions having a higher osmotic pressure are hypertonic and can cause crenation. • Solutions having a lower osmotic pressure are hypotonic and can cause hemolysis.
- 59. Practice Problem 12 What concentration of sodium chloride and water is needed to produce an aqueous solution isotonic with blood (Π = 7.70 atm at 25°C)? .158 M
- 60. Reverse Osmosis When a solution in contact with pure solvent across a semipermeable membrane is subjected to an external pressure larger than its osmotic pressure, reverse osmosis occurs. The pressure will cause a net flow of solvent from the solution. This process can act as a molecular filter for many solutions. • This process is being used to produce fresh water from sea water by desalination.
- 61. COLLIGATIVE PROPERTIES OF ELECTROLYTE SOLUTIONS
- 62. Colligative Properties Colligative properties of solutions depend on the total concentration of solute particles. The relationship between the moles of solute dissolved and the moles of particles in solution is usually expressed using the van’t Hoff factor: • The expected value for i can be calculated for a salt by noting the number of ions per formula unit (NaCl =2) • These calculated values assume that when a salt dissolves, it completely dissociates into its component ions. i = moles of particles insolution moles of solute dissolved
- 63. Colligative Properties Not all ions dissociate completely and act independently in solution; ion pairing occurs in solution and some pairs will act as a single particle. • Ion pairing is most important in concentrated solutions. • Less ion pairing occurs in more dilute solutions. • The deviation of i is greatest where the ions have multiple charges.
- 65. Colligative Properties To adjust freezing point depression, boiling point elevation and osmotic pressure calculations, i can be used to better account for ion pairing. ΔT = imK • K is the freezing or boiling point constant Π = iMRT
- 66. Practice Problem 13 The observed osmotic pressure for a 0.10 M solution of Fe(NH4)2(SO4)2 at 25°C is 10.8 atm. Compare the expected and experimental values for i. i=4.4; The experimental value for i is less than the expected value.
- 67. COLLIODS
- 68. Colloids
- 69. Colliods The Tyndall effect is when suspended particles scatter light. This is used to distinguish between a suspension and true solution. A suspension of tiny particles in some medium is called a colloidial dispersion or colloid. • Suspended particles are large molecules or ions from 1-1000nm. • Suspended particles remain suspended due to electrostatic repulsion
- 70. Colloids Colloids are classified according to the states of the dispersed phase and dipersing medium
- 71. Colloids Colloids can be destroyed by coagulation; usually by heating or by adding an electrolyte. • Heating increases velocities of particles and allows them to collide with enough energy that particles aggregate. • Adding an electrolyte neutralizes the ion layers and particles precipitate out.
- 72. The End