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Ch 13. Properties of Solutions; Mixtures of Substances at the Molecular Level

Ch 13. Properties of Solutions; Mixtures of Substances at the Molecular Level. Brady & Senese, 5th Ed. Mixing Processes. Mixing occurs due to interaction between molecules “ like dissolves like ” As partition is removed, molecules are able to move freely and interact.

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Ch 13. Properties of Solutions; Mixtures of Substances at the Molecular Level

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  1. Ch 13. Properties of Solutions; Mixtures of Substances at the Molecular Level Brady & Senese, 5th Ed.

  2. Mixing Processes • Mixing occurs due to interaction between molecules “like dissolves like” • As partition is removed, molecules are able to move freely and interact 12.1. Substances mix spontaneously when there is no energy barrier to mixing

  3. The Process Of Dissolution • Polar solutes dissolve in polar solvents • Non-polar solutes dissolve in non-polar solvents • Dipoles of solvent may induce dipoles in solute, effecting dissolution 12.1. Substances mix spontaneously when there is no energy barrier to mixing

  4. Miscibility of Liquids • Liquids that can dissolve in one another are miscible, while insoluble liquids are immiscible • Ethanol and water are miscible, while benzene and water are not 12.1. Substances mix spontaneously when there is no energy barrier to mixing

  5. Learning Check Which of the following are miscible in water? water ammonia carbon disulfide acetic acid 12.1. Substances mix spontaneously when there is no energy barrier to mixing

  6. Your Turn! Which of the following are likely to be miscible with water? • CH3CH2CH2CH3 • C6H6 • CH3CO2H • All are expected to be miscible 12.1. Substances mix spontaneously when there is no energy barrier to mixing

  7. Dissolution Of An Ionic Compound In Water • Positive end of the dipole of the water surrounds the anions of the ionic solid, extracting them from the lattice • Negative end of the dipole orients toward the cations, surrounding and extracting them from the lattice 12.1. Substances mix spontaneously when there is no energy barrier to mixing

  8. Dissolution Of A Polar Compound In Water Dipole of the water interacts with the oppositely charged dipoles of the solid, extracting them from the crystal 12.1. Substances mix spontaneously when there is no energy barrier to mixing

  9. solute (undissolved) solute (dissolved) Figure 13.19 Equilibrium in a saturated solution.

  10. Saturated Solutions • Solute is at equilibrium with the dissolved solute • Addition of more dissolved solute results in supersaturation and precipitation of excess solid • The presence of less solute than the solubility results in an unsaturated solution 12.3. A substance's solubility changes with temperature

  11. Solubility Varies With Temperature • Solubility may increase or decrease with increasing temperature • The extent to which temperature has an effect is specific to the solute and solvent • Most gases are less soluble in water at high temperature, while most solids are more soluble 12.3. A substance's solubility changes with temperature

  12. Case Study: Dead Zones During the industrial revolution, factories were built on rivers so that the river water could be used as a coolant for the machinery. The hot water was dumped back into the river and cool water recirculated. After some time, the rivers began to darken and many fish died. The water was not found to be contaminated by the machinery. What was the cause of the mysterious fish kills? increased temperature lowered amounts of dissolved oxygen 12.3. A substance's solubility changes with temperature

  13. Pressure Effects On Solubility Of Gases • Cg=kHPg • C = concentration of dissolved gas (M) • kH = Henry’s Constant • P = pressure applied to system (mm Hg) • kH (M/mm Hg) • N2 8.42×10 -7 • O2 1.66×10-4 • CO2 4.48×10-5 • Gases are all more soluble at higher pressures (the cause of “the bends”) 12.4. Gases become more soluble at higher pressures

  14. Units Of Concentration • Molarity (M) = moles solute / L solution • changes with Temperature • Molality (m) = moles solute/kg solvent • mole fraction (X) • X = moles component/ total moles 12.5. Molarity changes with temperature; molality, weight percentages, and mole fractions do not

  15. Your turn! Which of the following corresponds to a 3.5M solution of NaCl with a density of 0.997 g/mL? MM H2O: 18.0153; NaCl: 58.443 12.5. Molarity changes with temperature; molality, weight percentages, and mole fractions do not

  16. Raoult’s Law • Vapor pressure of a liquid varies as a function of purity • X= mole fraction of solvent • P0= vapor pressure of pure solvent • Nonvolatile Solute  Psolution=XsolventP0solvent • Volatile Solute  Psolution=XAP0A+XBPB0 • Where A and B are both volatile components. 12.6. Solutes lower the vapor pressure of a solvent

  17. Learning Check The vapor pressure of 2-methylheptane is 233.95 torr at 55°C. 3-ethylpentane has a vapor pressure of 207.68 at the same temperature. What would be the pressure of the mixture of 78.0g 2-methylheptane and 15 g 3-ethylpentane? • Psolution=XAP0A+XBP0B • mole 2-methylheptane : 78.0g/114.23 g/mol = 0.68283 mol • mole 3-ethylpentane: 15g/100.2 g/mol = 0.1497 mol • X2-methylheptane=0.8202 ; X3-ethylpentane =1-0.8202 = 0.1798 P = 230 torr 12.6. Solutes lower the vapor pressure of a solvent

  18. Learning Check The vapor pressure of 2-methyl hexane is 37.986 torr at 15°C. What would be the pressure of the mixture of 78.0g 2-methylhexane and 15 g naphthalene which is nearly non-volatile at this temperature? • Psolution=XsolventP0solvent • mol 2-methylhexane: 78.0g/100.2 g/mol = 0.778443 mol • mol naphthalene: 15 g/128.17 g/mol = 0.11703 • X2-methylhexane = 0.869309 • Psolution = 0.869309×37.986 torr • P=33.02 torr 12.6. Solutes lower the vapor pressure of a solvent

  19. Your Turn! n-hexane and n-heptane are miscible in a large degree and both volatile. If the vapor pressure of pure hexane is 151.28 mm Hg and heptane is 45.67 at 25º, which equation can be used to determine the mole fraction of hexane in the mixture if the mixture’s vapor pressure is 145.5 mm Hg? • X(151.28 mmHg)=145.5 mmHg • X(151.28 mmHg) + (X)(45.67 mm Hg) = 145.5 mmHg • X(151.28 mmHg)+(1-X)(45.67 mm Hg)=145.5 mm Hg • None of these 12.6. Solutes lower the vapor pressure of a solvent

  20. Solute Effects On Phase Changes: • Regardless of the identity of the dissolved particles, the presence of an impurity will result in a change in the boiling point and freezing point. • The effect is solely dependent on the nature of the solvent, a factor labeled K, and the concentration of particles present (m) • ΔT=ImK* • boiling point elevation ΔT=Tmix-Tpure • freezing Point Depression ΔT=Tpure-Tmix * This equation can be used to find freezing point depression with any solute but only boiling point elevation with nonvolatile solutes. 12.7. Solutions have lower freezing points and higher boiling points than pure solvents

  21. Some BP/FP Constants 12.7. Solutions have lower freezing points and higher boiling points than pure solvents

  22. Learning Check According to the Sierra™ Antifreeze literature, the freezing point of a 40/60 solution of sierra antifreeze and water is -4 °F. What is the molality of the solution? -4°F = 1.8 (°C) + 32 -20. °C 11=m 12.7. Solutions have lower freezing points and higher boiling points than pure solvents

  23. from before: -4°F = 1.8 (°C) + 32 =-20. °C 11=m Learning Check: In the previous sample of a Sierra™ antifreeze mixture, 100 mL is known to contain 42 g of the antifreeze and 60. g of water, what is the molar mass of the compound found in this antifreeze if it has a freezing point of -4°F? 0.66 mol solute 64 g/mol solute 12.7. Solutions have lower freezing points and higher boiling points than pure solvents

  24. Learning Check: In the previous sample of a Sierra™ antifreeze mixture, the freezing point is -4°F?What will be its boiling point? T=105.6 °C 12.7. Solutions have lower freezing points and higher boiling points than pure solvents

  25. Your Turn! Beer is known to be around a 5% ethanol (C2H5OH) solution with a density of 1.05 g/mL. What is its expected freezing point?( Kf=1.86°C/m) • -1ºC • -2ºC • 0ºC • 1ºC • Not enough information given MM: H2O=18.0153; C2H5OH=46.069 12.7. Solutions have lower freezing points and higher boiling points than pure solvents

  26. Osmosis • When two solutions are separated by a semi-permeable membrane, solvent molecules flow from areas of low concentration to areas of high concentration • As this occurs, the height of liquid rises in the higher concentration solution, building up “Osmotic pressure” (π) 12.8. Osmosis is flow of solvent through a semipermeable membrane due to unequal concentrations

  27. Applied pressure needed to prevent volume increase semipermeable membrane Osmotic Pressure pure solvent solution net movement of solvent solute molecules solvent molecules

  28. Osmosis π=MRT the concentration, is in molarity, M T=Temperature, in Kelvin R=Ideal Gas Constant, 0.082057 L·atm/mol·K The basis for kidney function, rising sap, and dialysis 12.8. Osmosis is flow of solvent through a semipermeable membrane due to unequal concentrations 29

  29. Learning Check: Osmosis A solution of 5% glucose (C6H1206) in water is placed into the osmometer shown at right. It has a density of 1.0 g/mL. The surroundings are filled with distilled water. What is the expected osmotic pressure at 25°C? 12.8. Osmosis is flow of solvent through a semipermeable membrane due to unequal concentrations

  30. Biochemists have discovered more than 400 mutant varieties of hemoglobin, the blood protein that carries oxygen throughout the body. A physician studing a variety associated with a fatal disease first finds its molar mass (M). She dissolves 21.5 mg of the protein in water at 5.00C to make 1.50 mL of solution and measures an osmotic pressure of 3.61 torr. What is the molar mass of this variety of hemoglobin? PLAN: We know p as well as R and T. Convert p to atm and T to degrees K. Use the p equation to find M and then the amount and volume of the sample to get to M. atm p 760 torr RT L 2.08 x10-4 mol L 103 mL g 1 103 mg 3.12 x10-7 mol Determining Molar Mass from Osmotic Pressure 3.61 torr SOLUTION: M = = = 2.08 x10-4 M (0.0821 L*atm/mol*K)(278.1 K) (1.50 mL) = 3.12x10-7 mol 21.5 mg = 6.89 x104 g/mol

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