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(a). (b). (c). Chapter 13: Properties of Solutions. Solvation: interactions between solvent and solute. Hydration: solvation by water. Note: In order for something to dissolve, the IMFs for the solute/solvent have to be stronger than the IMFs for the pure solute and solvent. Exothermic.

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  1. (a) (b) (c) Chapter 13: Properties of Solutions Solvation: interactions between solvent and solute Hydration: solvation by water Note: In order for something to dissolve, the IMFs for the solute/solvent have to be stronger than the IMFs for the pure solute and solvent.

  2. Exothermic Endothermic DHsoln = DH1 + DH2 + DH3

  3. Remember that enthalpy changes only involve transfer of thermal energy between substances. • While most exothermic reactions are spontaneous, not all endothermic reactions are nonspontaneous. • Reaction spontaneity is determined by: Gibb’s Free Energy DG = DH - TDS An endothermic reaction will be spontaneous if there is a large increase in entropy.

  4. Hydrates: a crystalline material that contains a fixed percentage of water molecules within its crystal structure. An anhydrous crystal contains no water Solids can have a variable amount of water adsorbed onto their surfaces. • Hygroscopic = adsorbs a significant amount of water (causes a change in mass) • Deliquescent = adsorbs enough water to dissolve itself A crystal hydrate has a constant amount of water molecules within its crystal lattice.

  5. Naming Hydrates #Waters Prefix Example Hydrate Name 1 mono- CaCl2·H2O Calcium chloride monohydrate 2 di- CaCl2·2H2O Calcium chloride dihydrate 3 tri- Al2O3·3H2O Aluminum oxide trihydrate 4 tetra- Co(C2H3O2)2·4H2O Cobalt(II) acetate tetrahydrate 5 penta- CuSO4·5H2O Copper(II) sulfate pentahydrate 6 hexa- CoCl2·6H2O Cobalt(II) chloride hexahydrate 7 hepta- FeSO4·7H2O Iron(II) sulfate heptahydrate

  6. Ex 1: 122 g of a hydrate of barium chloride (BaCl2) are heated until the compound is anhydrous. What is the formula of the hydrate if the final weight of the compound is 104 g? 104 g Wt of anh. BaCl2 = gfm BaCl2 = 208.233 g/mol Moles BaCl2 = 0.50 mol BaCl2 104 g/(208.233 g/mol) = 1 mol BaCl2 122 g – 104 g = 18 g Wt of H2O in hydrate = gfm H2O = 18.015 g/mol Moles H2O = 18 g/(18.015 g/mol) = 1.00 mol H2O Formula of hydrate = BaCl2·H2O 2 2 mol H2O Name: Barium chloride dihydrate • Recall that we had to divide through by the smallest molar amount when finding the empirical formula of a compound.

  7. Ex 2: A hydrate of MgSO4 is heated strongly until a constant weight is obtained. What is the formula of the hydrate if the %water in the hydrate is 51.2%? % H2O = 51.2% gfm H2O = 18.015 g/mol Mol H2O = 51.2 g/(18.015 g/mol) = 2.84 mol H2O 0.41 = 6.9 mol H2O gfm MgSO4 = 120.369 g/mol %MgSO4 = 100% – 51.2% = 48.8% Mol MgSO4 = 48.8 g/(120.369 g/mol) = 0.41 mol MgSO4 0.41 7 Formula of hydrate = MgSO4·H2O = 1 mol MgSO4 Name: Magnesium sulfate heptahydrate

  8. Solute + solvent solution dissolve crystallize • Unsaturated soln: rate of dissolution > rate of crystallization • Saturated soln: rate of dissolution = rate of crystallization • Supersaturated soln: a metastable solution

  9. Solubilities of some alcohols in water and hexane Solubility in H2O Solubility in C6H14 Alcohol 0.12  C H O H (methanol) 3   C H C H O H (ethanol) 3 2    0.11  0.030 C H C H C H O H (propanol) 3 2 2 0.0058  C H C H C H C H O H (butanol) 3 2 2 2 0.0008  C H C H C H C H C H O H (pentanol) 3 2 2 2 2 C H C H C H C H C H C H O H (hexanol) 3 2 2 2 2 2 C H C H C H C H C H C H C H O H (heptanol) 3 2 2 2 2 2 2 • This series is a good demonstration of the saying that “like dissolves like.” • As the substance becomes less polar, its solubility in the nonpolar solvent, hexane, increases. Miscible = two liquids that will mix/dissolve each other Immiscible = two liquids that do not mix

  10. Henry’s Law: The solubility of a gas is directly proportional to the partial pressure of the gas above the solvent. Sg = kPg Sg = solubility of the gas, k = proportionality constant Pg = partial pressure of the gas above the solvent Because k is a constant for a particular gas/solvent system, Henry’s law can be rewritten as:

  11. 13.22 The partial pressure of O2 in air at sea level is 0.21 atm. Using the data in Table 13.2, together with Henry’s law, calculate the molar concentration of O2 in the surface water of a mountain lake saturated with air at 20°C and an atmospheric pressure of 665 torr. S1 P1 P2 From Table 13.2, the solubility of oxygen gas is 1.38 x 10-3 M under 1 atm of O2 Convert final pressure to atm = 665 Torr/760 = 0.875 atm 21% of the total pressure is due to O2: PO2 = (0.875 atm)0.21 = 0.184 atm S2 = 2.5 x 10-4 M

  12. Solution concentration calculations Mass/Volume% (m/v) = Mass% (m/m) = Volume% (v/v) = Example: What is the m/m% of NaCl if 5 g of NaCl is dissolved in 95 g of water? 5% by mass Example: What is the v/v% if 20 mL of ethanol is dissolved into 80 mL of water? 20% by volume Example: What is the m/v% if 15 g of NaCl is dissolved into 100 mL of water? 15% mass/volume

  13. Parts per thousand (ppt): (grams solute/grams solution)*103 Parts per million (ppm): (grams solute/grams solution)*106 Parts per billion (ppb): (grams solute/grams solution)*109 13.24 (b) Seawater contains 0.0079 g Sr2+ per kilogram of water. What is the concentration of Sr2+ measured in ppm? ppm Sr2+ = (0.0079 g Sr2+/1000 g H2O)*106 = 7.9 ppm

  14. Molarity = molality = Example: What is the molarity of a solution prepared by dissolving 1.5 moles of NaOH in 500 mL of water? Example: What is the molality of a solution prepared by dissolving 3.0 moles of KCl in 750 g of water?

  15. Colligative properties: properties of solutions that depend only on the NUMBER of solute particles present. The chemical nature of the solute particles does not affect the colligative properties. = = Solution properties that are affected: freezing point, boiling point and osmotic pressure of the SOLVENT.

  16. Particles in solution Total #particles C6H12O6(s) C6H12O6(aq) 1 Remember that molecular (covalent) compounds exist as discrete particles…they don’t break up when they dissolve. + NaCl (s) Na+1 (aq) Cl-1 (aq) 2 If an ionic compound is used as the solute, it will break up into its ions. + 2 MgCl2(s) Mg+2 (aq) Cl-1 (aq) 3 + 3 AlCl3(s) Al+3 (aq) Cl-1 (aq) 4 Notice how the subscripts in the formula for the ionic solids become coefficients for the ions.

  17. Colligative properties Freezing point depression: The normal freezing point of the solvent decreases as the number of solute particles increases. Example: antifreeze Freezing point DTf = Kfm DTf = the change in the freezing point of the solvent Kf = molal freezing point constant (depends on solvent) Moles of solute particles m = MOLALITY of the solution = Kg of solvent

  18. Which solute would cause the greatest freezing point depression? A) 1 g of NaCl B) 2 g of NaCl C) 3 g of NaCl D) 4 g of NaCl A) 1 mol NaCl B) 1 mol MgCl2 C) 1 mol KCl D) 1 mol AlCl3 A) 1 mol of a nonelectrolyte* B) 0.4 mol MgCl2 C) 0.25 mol KCl D) 0.4 mol NaCl *a nonelectrolyte is one that doesn’t break apart in solution.

  19. DT = 176.7°C – 179.8°C = 3.1°C Determination of molar mass by freezing point depression Ex: Camphor (C10H16O) melts at 179.8°C, and it has a particularly large freezing-point-depression constant, Kf = 40.0°C/m. When 0.186 g of an organic substance of unknown molar mass is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7°C. What is the molar mass of the solute? DT = Kfm molality = DT/Kf = 3.1°C/ 40.0°C/m = 0.0775 m nUNK = (molality)(kg of camphor) = (0.0775 m)(0.02201 kg) nUNK = 1.71 x 10-3 moles MM = grams/moles = 0.186 g/(1.71 x 10-3 mol) = 109 g/mol MM = 110 g/mol

  20. Boiling point Boiling point elevation: The normal boiling point of the solvent increases as the number of solute particles increases. Example: engine coolant DTb = Kbm DTb = the change in the boiling point of the solvent Kb = molal boiling point constant (depends on solvent) Ex: What will be the boiling point of a 0.1 m aqueous solution of a nonelectrolyte solute? (Kb of H2O = 1.86 C/m) DTb = Kbm DTb = (1.86 C/m)(0.1 m) DTb = 0.186 C New boiling point = 100C + 0.186C = 100.186C

  21. Determination of molar mass by boiling point elevation Ex: A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.250 g of the substance in 40.0 g of CCl4. The boiling point of the resultant solution was 0.357°C higher than that of the pure solvent. Calculate the molar mass of the solute if Kb for CCl4 is 5.02°C/m. DTb = 0.357°C DTb = Kbm Molality = (DTb)/Kb = 0.357°C/(5.02°C/m) = 0.07112 molal nUNK = (molality)(kg CCl4) = (0.07112 m)(0.0400 kg) nUNK = 2.845 x 10-3 mol MM=grams/moles = 0.250 g/(2.845 x 10-3 moles) = 87.89 g/mol MM = 87.9 g/mol

  22. The vapor pressure of the solution, , can be calculated using Raoult’s Law: Where is the mole fraction of the solvent and is the vapor pressure of the neat solvent. Vapor Pressure lowering: Addition of a nonvolatile solute lowers the vapor pressure of the solvent. Vapor Pressure • Vapor pressure lowering and boiling point elevation are linked colligative properties. Can you explain why? • Because the boiling point of a solution is defined as being the point at which the vapor pressure of the solvent = the exterior pressure. As vapor pressure , boiling point must .

  23. Q: How do you find the vapor pressure of the solution when both the solvent AND the solute are volatile? Example: What is the vapor pressure of a solution consisting of 1.0 mol benzene and 2.0 mol toluene? Xbz = 1.0mol/3mol total = 0.33 Xtol = 2.0/3.0 = 0.67 Ptot = (0.33)(75 Torr) + (0.67)(22 Torr) = 39 Torr Q: What is the mole fraction of the benzene in the vapor? = 0.64

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