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Transcript of Osmotic Pressure
Osmotic PressureCopyright 2011 Pearson Canada Inc.General Chemistry: Chapter 13Slide 1 of 46V = nRT = RTnV = MRTFor dilute solutions of nonelectrolytes:This equation works quite well for calculating osmotic pressures of dilute solutions of nonelectrolytes. The osmotic pressure is represented by the symbol , R is the gas constant (0.08206 L atm mol-1 K-1) and T is the Kelvin temperature. The term n represents the amount of solute (in moles), and V is the volume (in liters) of solution. Notice that this equation is similar to the equation for the ideal gas law. In this case, however, it is convenient to rearrange terms to yield equation (13.4). The ratio, n/V, then, is the molarity of the solution, represented by the symbol M.
Chemistry 140 Fall 20021Osmotic Pressure2. A 0.426 g sample of an organic compound is dissolved in 225 mL of solvent at 24.5 0C to produce a solution with an osmotic pressure of 3.36 mm Hg. What is the molar mass of the organic compound? 3. What additional information would be required to determine the molecular formula as well as the molar mass? Henrys Law4. At 0 0C a 1.00L aqueous solution contains 48.98 mL of dissolved oxygen when the O2(g) pressure above the solution is 1.00 atm. What would be the molarity of oxygen in the solution if the oxygen gas pressure above the solution were instead 4.15 atm? Freezing-Point Depression and Boiling Point Elevation of Nonelectrolyte SolutionsA Colligative property.Depends on the number of particles present.Vapor pressure is lowered when a solute is present.This results in boiling point elevation.Freezing point is also effected and is lowered.
Copyright 2011 Pearson Canada Inc.General Chemistry: Chapter 13Slide 4 of 46Vapor-pressure lowering by a nonvolatile solute
FIGURE 13-19Copyright 2011 Pearson Canada Inc.General Chemistry: Chapter 13Slide 5 of 46Tf = -Kf mTb = +Kb mChemistry 140 Fall 20025The normal freezing point and normal boiling point of the pure solvent are fp0 and bp0 respectively. The corresponding points for the solution are fp and bp. The freezing-point depression, Tf, and the boiling-point elevation, Tb, are indicated. Because the solute is assumed to be insoluble in the solid solvent, the sublimation curve of the solvent is unaffected by the presence of solute in the liquid solution phase. That is, the sublimation curve is the same for the two phase diagrams.
Copyright 2011 Pearson Canada Inc.General Chemistry: Chapter 13Slide 6 of 46Tf = -Kf mTb = +Kb m
Valid for dilute solutions only, in practice the equations (13.5) and (13.6) often fail for solutions as concentrated as 1 m (molal).Chemistry 140 Fall 20026Practical ApplicationsCopyright 2011 Pearson Canada Inc.Slide 7 of 46General Chemistry: Chapter 13
Chemistry 140 Fall 20027A typical aircraft deicer is propylene glycol, CH3CH(OH)CH2OH, diluted with water and applied as a hot, high-pressure spray.
Water sprayed on citrus fruit releases its heat of fusion as it freezes into a layer of ice that acts as a thermal insulator. For a time, the temperature remains at 0C. The juice of the fruit, having a freezing point below 0C, is protected from freezing. Solutions of ElectrolytesSvante ArrheniusNobel Prize 1903.Ions form when electrolytes dissolve in solution.Explained anomalous colligative properties. Copyright 2011 Pearson Canada Inc.General Chemistry: Chapter 13Slide 8 of 46Tf = -Kf m = -1.86C m-1 0.0100 m = -0.0186C Compare 0.0100 m aqueous urea to 0.0100 m NaCl (aq)Freezing point depression for NaCl is -0.0361C.
At the time Arrhenius was awarded the Nobel Prize in chemistry (1903), his results were described thus: Chemists would not recognize them as chemistry; nor physicists as physics. They have in fact built a bridge between the two. The field of physical chemistry had its origins in Arrheniuss work.
Chemistry 140 Fall 20028Vant HoffCopyright 2011 Pearson Canada Inc.General Chemistry: Chapter 13Slide 9 of 46Tf = -i Kf m i = == 1.98measured TfTb = i Kb m expected Tf0.0361C0.0186C = i M RT Vant Hoff defined the factor i as the ratio of the measured value of a colligative property to the expected value if the solute is an electrolyte. For 0.0100 m NaCl i = 1.98. Later in this section, we explain why the experimentally determined i for 0.0100 m NaCl is 1.94 instead of 2.
Arrheniuss theory of electrolytic dissociation allows us to explain different values of the vant Hoff factor i for different solutes. For such solutes as urea, glycerol, and sucrose (all nonelectrolytes), i = 1. For a strong electrolyte such as NaCl, which produces two moles of ions in solution per mole of solute dissolved, we would expect the effect on freezing-point depression to be twice as great as for a nonelectrolyte. We would expect that i = 2. Similarly, for MgCl2, our expectation would be that i = 3. For the weak acid (acetic acid), which is only slightly ionized in aqueous solution, we expect i to be slightly larger than one but not nearly equal to two. Thus the equations presented for osmotic pressure, freezing point depression and boiling point elevation need to be modified.
Chemistry 140 Fall 20029Interionic InteractionsCopyright 2011 Pearson Canada Inc.Slide 10 of 46General Chemistry: Chapter 13Arrhenius theory does not correctly predict the conductivity of concentrated electrolytes.
Interionic InteractionsThe data on the previous slide show that solutions of electrolytes behave less ideally as (1) ionic charges increase and (2) electrolyte concentraion increases. Are there correspondences to nonideal gas behaviour? The latter manifests itself at high P and low T (high gas concentration) and as molecules become more polar or polarizable.Ideal vant Hoff Factors (Aqueous Solutions):We need to (a) identify covalent and ionic substances by inspecting their chemical formulas and (b) for ionic compounds (and strong acids) determine how many ions are formed in solution per formula unit of solute dissolved. Examples are given on the next slide.Vant Hoff i Factors: Chemical NameChemical FormulaIonic/Covalent ?Ideal i valueSucroseC12H22O11Covalent1Potassium nitrateKNO3Ionic2Urea(NH2)2C=O Covalent1Magnesium chlorideMgCl2Ionic3Aluminum nitrateAl(NO3)3Ionic4Lead (II) iodidePbI2Ionic3Copper (II) nitrate hexahydrateCu(NO3)26H2OIonicHydrochloric acidHCl(aq)Potassium hydroxideBoiling Pt. and Freezing Pt. ChangesEx. 2. If 3.30 g of ammonium nitrate were dissolved in 475 g of water what would you expect the boiling point and freezing point of the solution to be? (Kb and Kf values for water are 0.512 0C m-1 and 1.86 0C m-1 .) (Mention different form of freezing pt. equation in some texts?) Freezing and Boiling PointsEx. 3. Which aqueous solution would have a higher boiling point? (a) 1.00 m sucrose (C12H22O11) or (b) 0.400 m calcium nitrate? Ex. 4. Which aqueous solution would have the lowest freezing point? (a) 1.20 m sucrose, (b) 0.600 m sodium chloride (c) 0.400 m magnesium nitrate or (d) 0.700 m copper (II) sulfate? Why?