Understanding Solution Properties: Concentration, Colligative Properties, and Formation Factors

1. Chapter Twelve Physical Properties Of Solutions

2. Outline of the Chapter • The fundamental properties of Solutions • Working with solutions • Concentration Units • Saturated Solutions and Equilibrium • Colligative Properties • Freezing point depression • Boiling point elevation

3. Some Types Of Solutions

4. Solution Concentration • Molarity is the moles of solute per liter of solution. moles of solute (mol) Molarity (M) = _____________________________ volume of solution (L) • Molarity is the most frequently used concentration in chemical situations. Needed in many calculations to determine how many moles are present in a certain volume of solution: moles of solute = volume of solution x molarity

5. More Concentration Units amount ofsolute (g) Percent by mass = ____________________________ x 100% amount of solution (g) volume ofsolute (L) Percent by volume = ____________________________ x 100% volume of solution (L) amount ofsolute (g) Mass/volume percent = ___________________________ x 100% volume of solution (L)

6. Solutions By “Parts” • Parts per million is the number of particles of solute per one-million particles of solution. Fluoridated drinking water contains ~ 1 ppm of fluoride ion. 1 ppm = 1 mg/L (10-3) • Parts per billion is the number of particles of solute per one-billion particles of solution. 1 ppb = 1 µg/L (10-6) • Parts per trillion is the number of particles of solute per one-trillion particles of solution. 1 ppt = 1 ng/L (10-9)

7. Molality • Molarity varies with temperature due to the expansion or contraction in the volume of the solution. • To be independent of temperature, a concentration unit must be based on mass only, not volume. • Molality (m) is the number of moles of solute per one kilogram of solvent (not solution!). Molality does not vary with temperature. moles of solute (mol) M = _____________________________ volume of solution (L)

8. An Example How many milliliters of water (d = 0.998 g/mL) are required to dissolve 25.0 g of urea and thereby produce a 1.65 m solution of urea, CO(NH2)2?

9. Mole Fraction And Mole Percent • The mole fraction (xi) of a solution component i is the fraction of all the molecules in the solution that are molecules of i. amount of component i (mol) xi = = ni / ntotal total amount of solution components (mol) xi< 1; x1 + x2 + x3 + … = 1 • The mole percent of a solution component is its mole fraction multiplied by 100%.

10. An Example Example 12.6 (p.516) An aqueous solution of ethylene glycol 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 mole fraction of HOCH2CH2OH.

11. Formation Of Solution Consider two important factors in solution formation: • Enthalpy of solution • Intermolecular forces in mixtures

12. Enthalpy Of Solution • Solution formation can be considered to take place in three steps: • Move the molecules of solvent apart to make room for the solute molecules. H1 > 0 • Separate the molecules of solute to the distances found between them in the solution. H2 > 0 • Allow the separated solute and solvent molecules to mix randomly. H3 < 0 Hsoln = H1 + H2 + H3

13. Illustrating enthalpy terms when solute dissolves in solvent

14. Intermolecular Forces In Solution Formation • If all intermolecular forces are of comparable strength, this type of solution is called an ideal solution and Hsoln = 0. • If the intermolecular forces between solute and solvent molecules are stronger than other intermolecular forces, H3 > H1 +H2, Hsoln < 0 - exothermic. • If the intermolecular forces between solute and solvent molecules are weaker than other intermolecular forces, H3 < H1 +H2, Hsoln > 0 – endothermic. • If the intermolecular forces between solute and solvent are much weaker than other intermolecular forces, the solute does not dissolve in the solvent. The compound is described as relatively insoluble in the solvent.

15. Intermolecular Forces In Solution

16. Aqueous Solutions Of Ionic Compounds • Two forces - inter-ionic attractions that hold ions together in crystal - ion-dipole forces, the attraction of water dipoles for cations and anions, that cause an ionic solid to dissolve in water • The extent to which an ionic solid dissolves in water is determined largely by the competition between inter-ionic attractions and the ion-dipole attractions.

17. Ion-Dipole Forces in Dissolution

18. An Example Example 12.8 Predict whether each of the following is likely to be a solution or a heterogeneous mixture. • ethanol, CH3OH, and water, HOH • Pentane, CH3(CH2)3CH3, and octane, CH3(CH2)6CH3 • Sodium chloride, NaCl, and carbon tetrachloride, CCl4 • 1-decanol, CH3(CH2)8CH2OH, and water, HOH

19. Formation of a Saturated Solution Dynamic equilibrium: rate of crystallization = rate of dissolving

20. Some Solubility Terms • Liquids that mix in all proportions are called miscible. • 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.

21. Solubility As A Function of Temperature • About 95% of all ionic compounds have aqueous solubilities that increase significantly with increasing temperature. • Most of the remainder have solubilities that change little with temperature. • A very few have solubilities that decrease with increasing temperature. • A supersaturated solution is created when a warm, saturated solution is allowed to cool without the precipitation of the excess solute.

22. Some Solubility Curves

23. A Supersaturated Solution

24. Selective Crystallization When KNO3(s) is crystallized from an aqueous solution of KNO3 containing CuSO4 as an impurity, CuSO4 remains in the solution.

25. The Solubilities Of Gases The solubilities of gases depend on temperature and pressure. • Most gases become less soluble in liquids as the temperature increases. • 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 = k Pgas (k > 1) 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.

26. Effect of Temperature on Solubility of Gases

27. Effect of Pressure on Solubility of Gases As a gas is compressed into a smaller volume, increasing the number of molecules per unit volume, the number of dissolved molecules per unit volume – the concentration of the solution – also increases.

28. Aqueous Solubility As AFunction Of Gas Pressure

29. Vapor Pressures of Solutions • Raoult’s law states that the vapor pressure of the solvent above a solution (Psolv) is the product of the vapor pressure of the pure solvent (Posolv) and the mole fraction of the solvent in the solution (xsolv): Psolv = xsolv . Posolv • 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.

30. Fractional Distillation

31. An Interesting Phenomenon Two different aqueous solutions in a closed container. After a time, the levels are unequal. Explain why this spontaneous change occurred.

32. Colligative Properties Colligative Properties – physical properties of solutions that depend on the number of solute particles present but not on the identity of the solute. • Freezing point depression • Boiling point elevation

33. Freezing Point Depressionand Boiling Point Elevation • Consider solutions with - a volatile solvent - asolute: nonvolatile, nonelectrolytic, and soluble in liquid solvent but not frozen solvent • The vapor pressure of the solution is that of the solvent in the solution, and at all temperatures this vapor pressure is lower than that of the pure solvent. • The presence of the solute lowers (depresses) the freezing point of the solvent (Tf) and increases (elevates) the boiling point of the solvent (Tb).

34. Vapor Pressure LoweringBy A Nonvolatile Solute In a solution, the vapor pressure of the solvent is lowered and the fusion curve is displaced to lower temperatures (red curves)

35. Equations for FPD, BPE • Tf = Tf, solution– Tf, solvent = – Kf x m • Tf = Tb, solution– Tb, solvent = Kb x m Tf – fusion temperature (melting) Tb – boiling temperature m – molality of solute Kf, Kb – constants

36. FPD And BPE Constants

37. An Example What mass of sucrose, C12H22O11, should be added to 75.0 g H2O to raise the boiling point to 100.35 0C?

38. Summary • Molarity is expressed as moles per liter and molality is expressed as moles of solute per kilogram of solvent. • Units used for very dilute concentrations of solutes include parts per million (ppm), parts per billion (ppb), and parts per trillion (ppt). • The type and magnitude of intermolecular forces are important in solution composition. • The solubility of a solute is its concentration in a saturated solution.

39. Summary (continued) • Solubility of solids vs. gasses • Solubility of ionic solids increases with increasing temperature • Solubility of gasses decreases with increasing temperature • Solubility of gases in solutions: • decreases with an increase in temperature • increases linearly with pressure – Henry’s Law • Solubility (S) Pgas S = k  P (as long as T is constant)

40. Summary (continued) • Vapor pressure of solutions vs. pure solvents • Raoult’s Law: • Vapor pressure of solvent above solution is equal to the product of the mole fraction of solvent and the vapor pressure of pure solvent: Psolution = xsolvent Psolvent

41. Summary (continued) • Colligative properties depend on the particular solvent and the number of solute particles present. • The presence of solutes lowers the vapor pressure of the solution and causes both a freezing point depression and a boiling point elevation of the solvent.