AP Chemistry--Chapter 11: Properties of Solutions I. Solution Composition (ways of expressing concentration) 1. Qualitatively, use dilute or concentrated to describe 2. Quantitatively a. Mass Percentage mass % of component = mass of component in solution x 100 total mass of solution b. Parts per million (or mg/kg, or mg/l; also ppb) ppm = mass of component in solution x 10 6 total mass of solution c. Mole Fraction mole fraction of component = moles of component total moles of all components d. Molarity (M) = moles of solute/ liter of solution e. Molality (m) = moles of solute/kilogram of solvent 3. Dilution problems: M 1 V 1 = M 2 V 2 Practice Problem 11.1 A solution is prepared by mixing 1.00 g ethanol (C 2 H 5 OH) with 100.0 g water to give a final volume of 101 ml. Calculate the molarity, mass percent, mole fraction, and molality of ethanol in this solution. Practice Problem 11.2 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 mole fraction of the sulfuric acid.
Practice Problem 11.3 How much concentrated (12.1 M) HCl would be needed to prepare 500.0 ml of 2.0 M HCl? Practice Problem 11.4 How many grams of solid NaOH would be needed to prepare 500.0 ml of 2.0 M NaOH? II. The Energies of Solution Formation A. The Steps 1. Breaking up the solute into individual components (expanding the solute), labeled ΔH 1, this step requires energy 2. Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent), labeled ΔH 2, this step requires energy 3. Allowing the solute and solvent to interact to form the solution, labeled ΔH 3, this step releases energy B. The overall enthalpy change in forming a solution, ΔH soln = ΔH 1 + ΔH 2 + ΔH 3 1. The formation of a solution can be either exothermic (-ΔH, energy released) or endothermic (+ΔH, energy absorbed) 2. Comparison of ΔH 1 + ΔH 2 vs. ΔH 3 helps explain why like dissolves like C. Solution Formation, Spontaneity, and Disorder 1. Energy and disorder are involved in processes that occur spontaneously a. processes in which the energy content of the system decreases tend to occur spontaneously b. processes in which the disorder of the system increases tend to occur spontaneously 2. Usually, formation of solution is favored by the increase in disorder that comes with mixing 3. In summary, a solution will form unless the solute-solute or solventsolvent interactions are too strong relative to the solute-solvent interactions Practice Problem 11.5 Decide whether liquid hexane (C 6 H 14 ) or liquid methanol (CH 3 OH) is the more appropriate solvent for the substances grease (C 20 H 42 ) and potassium iodide (KI). III. Factors Affecting Solubility A. Structure Effects
IV. 1. The more polar bonds present, the more soluble a substance becomes in water, the fewer polar bonds (or the more non-polar bonds), the less soluble in water. 2. The term hydrophobic is used to describe non-water-soluble substances, while hydrophilic is used to describe water-soluble substances B. Pressure Effects 1. solid and liquid solubilities not appreciably affected by pressure changes (why??) 2. solubility of a gas in any solvent is increased as the pressure of the gas over the solvent increases (why?) C. Temperature Effects 1. In general, the solubility of most solid solutes in water increases as the temperature of the solution increases (not all though) 2. Temperature increases the rate of solution, not always the amount, sometimes the amount actually decreases 3. For gases, the solubility of gases in water decreases with increasing temperature The Vapor Pressures of Solutions A. (see experiment pg. 514) B. The vapor pressure of a solution is lower than that over a pure solvent (why??), effect is vapor pressure lowering C. Raoult s Law: P A = P total X A where P A = vapor pressure of solution, X A = mole fraction of the solvent (X A = moles A/total moles), and P total = vapor pressure of the pure solvent (also P total = P A + P B + P C + ) 1. Raoult s law states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent present 2. ideal solutions are solutions that obey Raoult s Law; many do not because of differences in solute-solvent intermolecular forces vs. solute-solute and solvent-solvent intermolecular forces (i.e., strong solute-solvent interactions gives a vapor pressure lower than that predicted by Raoult s law) Practice Problem 11.6 Calculate the expected vapor pressure at 25ºC for a solution prepared by dissolving 158.0 g of common table sugar (sucrose, molar mass = 342.3 g/mol) in 643.5 cm 3 of water. At 25ºC the density of water is 0.9971 g/cm 3 and the vapor pressure is 23.76 torr. Practice Problem 11.7 A solution was prepared by adding 20.0 g urea to 125 g water at 25ºC, a temperature at which pure water has a vapor pressure of 23.76 torr. The observed vapor pressure of the solution was found to be 22.67 torr. Calculate the molar mass of the urea.
Practice Problem 11.8 Predict the vapor pressure of a solution prepared by mixing 35.0 g solid Na 2 SO 4 (molar mass = 142 g/mol) with 175 g water at 25ºC. The vapor pressure of pure water at 25ºC is 23.76 torr. Practice Problem 11.9 A solution is prepared by mixing 5.81 g acetone (C 3 H 6 O, molar mass = 58.1 g/mol) and 11.9 g chloroform (HCCI 3, molar mass = 119.4 g/mol). At 35ºC 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 35ºC are 345 and 293 torr, respectively. V. Boiling Point Elevation and Freezing Point Depression A. Physical properties of solutions that depend on the quantity but not the kind of solute particles are called colligative properties; these include boiling point elevation, freezing point depression, and osmotic pressure B. Boiling point elevation 1. boiling point of a solution is higher than that of a pure solvent (why??) 2. ΔT b = ik b molality where K b = molal boiling-point elevation constant, and ΔT b = increase in boiling point relative to the pure solvent Practice Problem 11.10 A solution was prepared by dissolving 18.00 g glucose in 150.0 g water. The resulting solution was found to have a boiling point of 100.34ºC. Calculate the molar mass of glucose. Glucose is a molecular solid that is present as individual molecules in solution. C. Freezing point depression 1. freezing point of a solution is lower than that of a pure solvent (why??) 2. ΔT f = ik f molality where K f = molal freezing point depression constant, and ΔT f = decrease in freezing point relative to the pure solvent
Practice Problem 11.11 What mass of ethylene glycol (C 2 H 6 O 2, molar mass = 62.1 g/mol), the main component of antifreeze, must be added to 10.0 L water to produce a solution for use in a car s radiator that freezes at 10.0ºF ( 23.3ºC)? Assume the density of water is exactly 1 g/ml. Practice Problem 11.12 A chemist is trying to identify a human hormone, which 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 determined to be 0.240ºC. Calculate the molar mass of the hormone. VI. Colligative Properties of Electrolyte Solutions A. Since colligative properties depend on number of particles in solution and ionic substances dissolve into ions, special consideration is made in these cases B. The van t Hoff factor (i) is added to the equations for bp elevation, fp depression, and osmotic pressure (*Raoult s law adjusted with mole fraction) C. i is moles of particles in solution/moles of solute dissolved 1. For a substance like NaCl, expected i is 2, but actually 1.87 because of ion pairing 2. Can look up value or be given info to calculate it Practice Problem 11.13 A saturated aqueous solution of NaCl at 0 C is 4.28 m. How much lower will the freezing point of this solution be compared to the freezing point of pure water? How much lower will the freezing point of a 4.28 m aqueous solution of CaCl 2 be than that of pure water?
Practice Problem 11.14 What is the expected i value for a solution of Fe(NH 4 ) 2 (SO 4 ) 2? The experimental value for i is 4.41. What causes this difference? VII. Colloids A. Particles that are large enough on the molecular scale but are still small enough to remain suspended indefinitely in a solvent system are called colloids B. Tyndall Effect is the scattering of visible light by the particles in a colloid C. Separation of colloids 1. coagulation-enlarging particles either by heating or adding an electrolyte 2. dialysis-use of semipermeable membrane to separate ions from colloidal particles