Osmotic Pressure For dilute solutions of nonelectrolytes: πV = nRT n π= RT = MRT V Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 1 of 46 Osmotic Pressure • 2. 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? Henry’s Law • 4. 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 Solutions • A 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 13 Slide 4 of 46 ΔTf = -Kf m ΔTb = +Kb m FIGURE 13-19 Vapor-pressure lowering by a nonvolatile solute Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 5 of 46 ΔTf = -Kf m Copyright © 2011 Pearson Canada Inc. ΔTb = +Kb m General Chemistry: Chapter 13 Slide 6 of 46 Practical Applications Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 7 of 46 Solutions of Electrolytes • Svante Arrhenius –Nobel Prize 1903. –Ions form when electrolytes dissolve in solution. –Explained anomalous colligative properties. Compare 0.0100 m aqueous urea to 0.0100 m NaCl (aq) ΔTf = -Kf m = -1.86°C m-1 0.0100 m = -0.0186°C Freezing point depression for NaCl is -0.0361°C. Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 8 of 46 Van’t Hoff i= measured ΔTf expected ΔTf = 0.0361°C = 1.98 0.0186°C π = i M RT ΔTf = -i Kf m ΔTb = i Kb m Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 9 of 46 Interionic Interactions •Arrhenius theory does not correctly predict the conductivity of concentrated electrolytes. Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 10 of 46 Interionic Interactions • The 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 van’t 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. Van’t Hoff “i Factors”: Chemical Name Chemical Formula Ionic/Covalent ? Ideal “i” value Sucrose C12H22O11 Covalent 1 Potassium nitrate KNO3 Ionic 2 Urea (NH2)2C=O Covalent 1 Magnesium chloride MgCl2 Ionic 3 Aluminum nitrate Al(NO3)3 Ionic 4 Lead (II) iodide PbI2 Ionic 3 Copper (II) nitrate hexahydrate Cu(NO3)2∙6H2O Ionic Hydrochloric acid HCl(aq) Potassium hydroxide Boiling Pt. and Freezing Pt. Changes • Ex. 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 Points • Ex. 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?