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Physical Properties of Solutions

Chapter Twelve. Physical Properties of Solutions. Some Types of Solutions. Solution: Solute dispersed in a solvent. Solution Concentration. Amount of solute. Most concentration units are expressed as:. Amount of solvent or solution. Molarity : moles of solute/ liter of solution

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Physical Properties of Solutions

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  1. Chapter Twelve Physical Properties of Solutions

  2. Some Types of Solutions Solution: Solute dispersed in a solvent.

  3. Solution Concentration Amount of solute Most concentration units are expressed as: Amount ofsolventor solution • Molarity: molesof solute/literof solution • Molality: molesof solute/kgof solvent • Percent by mass: grams of solute/gramsof solution (then multiplied by 100%) • Percent by volume: milliliters of solute/millilitersof solution (then multiplied by 100%) • Mass/volume percent: gramsof solute/milliliters of solution (then multiplied by 100%)

  4. Example 12.1 How would you prepare 750 g of an aqueous solution that is 2.5% NaOH by mass? Example 12.2 At 20°C, pure ethanol has a density of 0.789 g/mL and USP ethanol has a density of 0.813 g/mL. What is the mass percent ethanol in USP ethanol?

  5. Example 12.1 How would you prepare 750 g of an aqueous solution that is 2.5% NaOH by mass? • ? g NaOH = 750 g solution x 2.5% = 19 g NaOH • ? g H2O = 750 g solution – 19 g NaOH = 731 g H2O Example 12.2 At 20 °C, pure ethanol has a density of 0.789 g/mL and USP ethanol (= 95% ethanol by volume) has a density of 0.813 g/mL. What is the mass percent ethanol in USP ethanol? • Assume you a working with 100 ml sample of the USP ethanol, then: • 95.0 ml ethanol x 78.9 g ethanol = 75.0 g ethanol 100 ml solution • Mass percent ethanol = 75.0 g ethanol x 100 % = 92.3 % 81.3 g solution

  6. Solution Concentration (cont’d) Amount of solute Most concentration units are expressed as: Amount of solution • Parts per million (ppm): grams of solute/gramsof solution (then multiplied by 106 or 1 million) • Parts per billion (ppb): grams of solute/gramsof solution (then multiplied by 109 or 1 billion) • Parts per trillion (ppt): grams of solute/gramsof solution (then multiplied by 1012 or 1 trillion) • ppm, ppb, ppt ordinarily are used when expressing extremely low concentrations (a liter of water that is 1 ppm fluoride contains only 0.001 g F–!)

  7. PPM and PPB are Tiny Amounts ...................................................................................................................................................... ...................................................................................................................................................... ...................................................................................................................................................... ...................................................................................................................................................... ...................................................................................................................................................... ...................................................................................................................................................... ...................................................................................................................................................... ...................................................................................................................................................... ...................................................................................................................................................... ...................................................................................................................................................... ...................................................................................................................................................... ...................................................................................................................................................... ...................................................................................................................................................... ...................................................................................................................................................... ......................................................................................................................................................

  8. Example 12.3 The maximum allowable level of nitrates in drinking water in the United States is 45 mg NO3–/L. What is this level expressed in parts per million (ppm)?

  9. Example 12.3 The maximum allowable level of nitrates in drinking water in the United States is 45 mg NO3–/L. What is this level expressed in parts per million (ppm)? The density of water with only small amounts of dissolved substances is essentially 1.0 g/ml. So one liter of dilute aqueous nitrate solution has a mass of 1000 g. 45 mg NO3–1 g water45 mg NO3– 1000 g water 1000 mg 1,000,000 mg water X = = 45 ppm NO3–

  10. Solution Concentration (cont’d) Amount of solute Most concentration units are expressed as: Amount of solvent or solution Molality (m): moles of solute/kilogramsof solvent. • Molarity varies with temperature (expansion or contraction of solution). • Molalityis based on mass of solvent(not solution!) and is independent of temperature. • We use molality in describing certain properties of solutions when temperature changes are occurring.

  11. Example 12.4 What is the molality of a solution prepared by dissolving 5.05 g naphthalene [C10H8(s)] in 75.0 mL of benzene, C6H6 (d = 0.879 g/mL)? Example 12.5 How many grams of benzoic acid, C6H5COOH, must be dissolved in 50.0 mL of benzene, C6H6 (d = 0.879 g/mL), to produce 0.150 m C6H5COOH?

  12. Solution Concentration (cont’d) Amount of solute Concentration expressed as: Amount of solvent or solution • Mole fraction (xi):moles of component i per molesof all components (the solution). • The sum of the mole fractions of all components of a solution is ____. • Mole percent: mole fraction times 100%.

  13. Example 12.6 An aqueous solution of ethylene glycol HOCH2CH2OH used as an automobile engine coolant is 40.0% HOCH2CH2OH by mass and has a density of 1.05 g/mL. What are the (a) molarity, (b) molality, and (c) mole fraction of HOCH2CH2OH in this solution? Example 12.7 An Estimation Example Without doing detailed calculations, determine which aqueous solution has the greatest mole fraction of CH3OH: (a) 1.0 m CH3OH,(b)10.0% CH3OH by mass, or (c) xCH3OH = 0.10.

  14. Enthalpy of Solution Solution formation can be considered to take place in 3 steps: • Move solvent molecules apart (breaking solvent-solvent bonds) to make room for the solute molecules. DH1 > 0 (endothermic) • Separate the molecules/ions of solute (breaking solute-solute bonds) . DH2 > 0 (endothermic) • Allow the separated solute and solvent molecules to mix randomly (form solute-solvent bonds). DH3 < 0 (exothermic) DHsoln = DH1 + DH2 + DH3

  15. Visualizing Enthalpy of Solution For dissolving to occur, the magnitudes of DH1 + DH2 and of DH3 must be roughly comparable.

  16. Intermolecular Forcesin Solution Formation • An ideal solution exists when all intermolecular forces are of comparable strength, DHsoln = 0. • When solute–solvent intermolecular forces are somewhat stronger than other intermolecular forces, DHsoln < 0. • When solute–solvent intermolecular forces are somewhat weaker than other intermolecular forces, DHsoln > 0. • When solute–solvent intermolecular forces are much weaker than other intermolecular forces, the solute does not dissolve in the solvent. • Energy released by solute–solvent interactions is insufficient to separate solute particles or solvent particles.

  17. Intermolecular Forces in Solution For a solute to dissolve, solvent–solvent forces … … must be comparable to solute–solvent forces. … and solute–solute forces …

  18. Non-Ideal Solutions … when mixed, give less than 100 mL of solution. 50 mL of ethanol … … and 50 mL of water … In this solution, forces between ethanol and water are > / < / = other intermolecular forces.

  19. Aqueous Solutions of Ionic Compounds • The forces causing an ionic solid to dissolve in water are ion–dipole forces, the attraction of water dipoles for cations and anions. • The attractions of water dipoles for ions pulls the ions out of the crystalline lattice and into aqueous solution. • The extent to which an ionic solid dissolves in water is determined largely by the competition between: • interionic attractions that hold ions in a crystal, and • ion–dipole attractions that pull ions into solution.

  20. Ion–Dipole Forces in Dissolution Negative ends of dipoles attracted to cations. Positive ends of dipoles attracted to anions.

  21. Example 12.8 Predict whether each combination is likely to be a solution or a heterogeneous mixture: (a) methanol, CH3OH, and water, HOH (b) pentane, CH3(CH2)3CH3, and octane, CH3(CH2)6CH3 (c) sodium chloride, NaCl, and carbon tetrachloride, CCl4 (d) 1-decanol, CH3(CH2)8CH2OH, and water, HOH Remember rule of thumb: “like dissolved like”

  22. Some Solubility Terms • Liquids that mix in all proportions are called miscible. • Ones that don’t = immiscible • When there is a dynamic equilibrium between an undissolved solute and a solution, the solution is saturated. • The concentration of the solute in a saturated solution is the solubility of the solute. • A solution which contains less solute than can be held at equilibrium is unsaturated.

  23. Formation of a Saturated Solution Eventually, the rates of dissolving and of crystallization are equal; no more solute appears to dissolve. Solid begins to dissolve. As solid dissolves, some dissolved solute begins to crystallize. Longer standing does not change the amount of dissolved solute.

  24. Solubility as Function of Temperature • Most ionic compounds have aqueous solubilities that increase significantly with increasing temperature • A few have solubilities that change little with temperature • A very few have solubilities that decrease with increasing temperature (see following slide) • If solubility increases with temperature, a hot, saturated solution may be cooled carefully without precipitation of the excess solute. This creates a supersaturated solution. • Supersaturated solutions ordinarily are unstable …

  25. Some Solubility Curves What is the (approx.) solubility of KNO3 per 100 g water at 90 °C? At 20 °C?

  26. A Supersaturated Solution Solute immediately begins to crystallize … … until all of the excess solute has precipitated. A single “seed crystal” of solute is added.

  27. The Solubilities of Gases • Most gases become less soluble in liquids as the temperature increases. (Explanation**) • At a constant temperature, the solubility (S) of a gas is directly proportional to the pressure of the gas (Pgas) in equilibrium with the solution. S = kPgas The value of k depends on the particular gas and the solvent. • The effect of pressure on the solubility of a gas is known as Henry’s law. ** http://antoine.frostburg.edu/chem/senese/101/solutions/faq/print-temperature-gas-solubility.shtml

  28. Effect of Temperature on Solubility of Gases Eco-Connection Thermal pollution: as river/lake water is warmed (when used by industry for cooling), less oxygen dissolves, and fish no longer thrive.

  29. Pressure and Solubility of Gases(Henry’s Law) … thus more frequent collisions of gas molecules with the surface … Higher partial pressure means more molecules of gas per unit volume … … giving a higher concentration of dissolved gas.

  30. Example 12.9 A 225-g sample of pure water is shaken with air under a pressure of 0.95 atm at 20°C. How many milligrams of Ar(g) will be present in the water when solubility equilibrium is reached? Use data from Figure 12.14 and the fact that the mole fraction of Ar in air is 0.00934. According to the graph, argon has a solubility in water of 6 mg/atm•100 g H2O. 225 g H2O x 0089 atm x 6 mg/atm•100 g H2O = 0.12 mg Ar

  31. Colligative Properties of Solutions • Colligative properties of a solution depend only on the concentration of solute particles, and not on the nature of the solute. • Non-colligative properties include: color, odor, density, viscosity, toxicity, reactivity, etc. • We will examine four colligative properties of solutions: • Vapor pressure (of the solvent) • Freezing point depression • Boiling point elevation • Osmotic pressure

  32. Review: Vapor Pressure • The pressure of the vapor phase of a substance above a liquid sample of the substance after a dynamic equilibrium has been established in a closed system.

  33. Vapor Pressure of a Solution • The vapor pressure of solvent above a solution is less than the vapor pressure above the pure solvent. • Raoult’s law: the vapor pressure of the solvent above a solution (Psolv) is the product of the vapor pressure of the pure solvent (P°solv) and the mole fraction of the solvent in the solution (xsolv): Psolv = xsolv ·P°solv • The vapor in equilibrium with an ideal solution of two volatile components has a higher mole fraction of the more volatile component than is found in the liquid.

  34. Example 12.10 The vapor pressure of pure water at 20.0 °C is 17.5 mmHg. What is the vapor pressure at 20.0°C above a solution that has 0.250 mol sucrose (C12H22O11) and 75.0 g urea [CO(NH2)2] dissolved per kilogram of water? Moles of urea = 75.0 g urea x 1 mol urea/60 g urea = 1.25 mol urea Moles of solute = .250 mole sucrose + 1.25 mole urea = 1.50 mole solute Moles solvent = 1000 g H20 x 1 mole H20 /18.0 g H20 = 55.6 moles Mole fraction of solvent = 55.6 moles solvent/57.1 moles solution = 0.974 Psolv = xsolv ·P°solv = 0.974 · 17.5 mmHg Psolv = 17.0 mmHg

  35. Example 12.11 At 25°C, the vapor pressures of pure benzene (C6H6) and pure toluene (C7H8) are 95.1 and 28.4 mmHg, respectively. A solution is prepared that has equal mole fractions of C7H8 and C6H6. Determine the vapor pressures of C7H8 and C6H6 and the total vapor pressure above this solution. Consider the solution to be ideal. Since there are equal mole fractions, the vapor pressure of each substance will be ½ of the vapor pressure of the pure substance. The total vapor pressure will be the sum of the partial pressures of the two vapors (Dalton’s law of partial pressures). Example 12.12 What is the composition, expressed as mole fractions, of the vapor in equilibrium with the benzene–toluene solution of Example 12.11? At constant temperature and number of moles, the pressures of gases are proportional to the molar amounts. So we can use the partial pressure values like moles to produce mole fractions of the two gases.

  36. Fractional Distillation The vapor here … … is richer in the more volatile component than the original liquid here … … so the liquid that condenses here will also be richer in the more volatile component.

  37. FractionalDistillation of Crude OilFractional distillation on a very large scale

  38. Example 12.13 A Conceptual Example Figure 12.16 (below) shows two different aqueous solutions placed in the same enclosure. After a time, the solution level has risen in container A and dropped in container B. Explain how and why this happens.

  39. Vapor Pressure Lowering by a Nonvolatile Solute … the vapor pressure from the pure solvent Raoult’s Law: the vapor pressure from a solution (nonvolatile solute) is lower than … Result: the boiling point of the solution increases by DTb.

  40. Freezing Point Depression and Boiling Point Elevation DTf= –Kf× m × i DTb= Kb× m ×i

  41. DTf= –Kf× m × i DTb= Kb× m ×i Example 12.14 What is the freezing point of an aqueous sucrose solution that has 25.0 g C12H22O11 per 100.0 g H2O? 25.0 g x 1 mol sucrose/342 g = 0.0731 mol C12H22O11 ` Tf = −1.86 ºC/m x m x 1 m =0.0731 mol/0.100 kg = 0.731 m Tf = − 1.86 ∙ 0.731 m ∙ 1 Tf = −1.36ºC So, the freezing point is −1.36ºC Example 12.15 Sorbitol is a sweet substance found in fruits and berries and sometimes used as a sugar substitute. An aqueous solution containing 1.00 g sorbitol in 100.0 g water is found to have a freezing point of –0.102 °C. Elemental analysis indicates that sorbitol consists of 39.56% C, 7.75% H, and 52.70% O by mass. What are the (a) molar mass and (b) molecular formula of sorbitol? • Use freezing point depression equation to calculate molality of solution • With m, moles of solute known. Molar mass = g/mol, and you’re given 1.00 g of the solute, so 1.00 g sorbitol/moles from the calculation gives molar mass.

  42. Osmotic Pressure • A semipermeable membrane has microscopic pores, through which small solvent molecules can pass but larger solute molecules cannot. • During osmosis, there is a net flow of solvent molecules through a semipermeable membrane, from a region of lower concentration to a region of higher concentration. • The pressure required to stop osmosis is called the osmotic pressure (p) of the solution. p = (nRT/V) = (n/V)RT = M RT This equation should look familiar …

  43. Osmosis and Osmotic Pressure The solution increases in volume until … … the height of solution exerts the osmotic pressure (π) of the solution. Net flow of water from the outside (pure H2O) to the solution.

  44. Example 12.16 An aqueous solution is prepared by dissolving 1.50 g of hemocyanin, a protein obtained from crabs, in 0.250 L of water. The solution has an osmotic pressure of 0.00342 atm at 277 K. (a) What is the molar mass of hemocyanin? (b) What should the freezing point of the solution be?

  45. Practical Applications of Osmosis Ordinarily a patient must be given intravenous fluids that are isotonic—have the same osmotic pressure as blood. External solution is hypertonic; produces osmotic pressure > πint. Net flow of water out of the cell. Red blood cell in isotonic solution remains the same size. External solution is hypotonic; produces osmotic pressure < πint. Net flow of water into the cell.

  46. Practical Applications of Osmosis (cont’d) Pressure greater than π is applied here … • Reverse osmosis (RO): reversing the normal net flow of solvent molecules through a semipermeable membrane. • Pressure that exceeds the osmotic pressure is applied to the solution. • RO is used for water purification. … water flows from the more concentrated solution, through the membrane.

  47. Solutions of Electrolytes • Whereas electrolytes dissociate, the number of solute particles ordinarily is greater than the number of formula units dissolved. One mole of NaCl dissolved in water produces more than one mole of solute particles. • The van’t Hoff factor (i) is used to modify the colligative-property equations for electrolytes: DTf=i× (–Kf) × m DTb= i×Kb× m p= i× M RT • For nonelectrolyte solutes, i = 1. • For electrolytes, we expect i to be equal to the number of ions into which a substance dissociates into in solution.

  48. At very low concentrations, the “theoretical” values of i are reached. At higher concentrations, the values of i are significantly lower than the theoretical values; ion pairs form in solution.

  49. Example 12.17 A Conceptual Example Without doing detailed calculations, place the following solutions in order of decreasing osmotic pressure: • 0.01 M C12H22O11(aq) at 25 °C • 0.01 M CH3CH2COOH(aq) at 37 °C • 0.01 m KNO3(aq)at 25 °C • a solution of 1.00 g polystyrene (molar mass: 3.5 × 105 g/mol) in 100 mL of benzene at 25 °C

  50. Colloids • In a solution, dispersed particles are molecules, atoms, or ions (roughly 0.1 nm in size). Solute particles do not “settle out” of solution. • In a suspension (e.g., sand in water) the dispersed particles are relatively large, and will settle from suspension. • In a colloid, the dispersed particles are on the order of 1–1000 nm in size. • Although they are larger than molecules/atoms/ions, colloidal particles are small enough to remain dispersed indefinitely.

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