Chapter 11 “Solutions”

1. Honors Chemistry 2 Chapter 11 “Solutions”

2. Solutions Solutions, as we saw in Chapter 4, are homogeneous mixtures of two or more substances. The substance present in greatest amount is called the solvent, and the other components present are called solutes. Although solutions consisting of a liquid solvent and a solid solute (dissolved in the solvent) are most familiar, solutions can involve many different combinations of the three states of matter.

3. Types of solutions

4. Solvation Intermolecular forces also act between solvent and solute particles in solutions. When sodium chloride dissolves in water, the sodium and chloride ions and the polar water molecules are powerfully attracted to one another by ion-dipole interactions. Polar water molecules surround the ions in solution in a process called solvation. (When water is the solvent, the process is called specifically hydration.)

5. Enthalpy of solution Solution formation is accompanied by a change in energy referred to as the enthalpy of solution, Hsoln. The process of solution formation can be thought of as occurring in three distinct steps.

6. Steps 1 and 2 In the first step the intermolecular forces holding the solute particles together must be overcome, resulting in separation of the solute molecules. This step is always endothermic. In the second step the intermolecular forces holding the solvent particles must be overcome. This, too, is an endothermic step.

7. The sum of all steps. • In the third step the solute and solvent particles are attracted to one another, causing formation of solute-solvent interactions. • The last of the three steps is exothermic. • The sum of all three steps may result in an exothermic process overall or an endothermic process overall.

8. Exo on left, endo on right.

9. Exo or endothermic? Whether a substance is soluble in a particular solvent and whether the dissolution process will be exothermic or endothermic depends on the relative magnitudes of the energy changes for the three steps. In general, a substance will be soluble unless the energy expended to break apart the solvent and solute particles is significantly greater than the energy given off when solute-solvent interactions are established.

10. Tending towards disorder. The driving force behind solution formation is the increased disorder achieved when solvent and solute are mixed. In general, processes that increase disorder tend to be spontaneous. This is how we can explain the fact that some spontaneous dissolution processes, such as that of ammonium nitrate, are endothermic.

11. Saturated solutions As a solid dissolves in water, the number of aqueous solute particles increases. As it dissolves, though, the reverse process (recrystallization of the solute) also occurs. As the concentration of dissolved solute increases, so does the rate of recrystallization. After a period of time the rate of recrystallization eventually equals the rate of dissolution, and a dynamic equilibrium is achieved. No further net increase in the amount of solute in solution occurs.

12. Equilibrium

13. Solubility When this dynamic equilibrium has been established, we call the solution saturated. (By definition, a saturated solution must be in contact with undissolved solid.) Furthermore, the concentration of solute in the saturated solution is the solubility of that solute in that solvent and at that temperature.

14. Unsaturated/supersaturated solutions Unsaturated solutions contain less dissolved solute than is needed to form a saturated solution. Under certain conditions it is possible to prepare a supersaturated solution. A supersaturated solution contains a greater amount of solute than that needed to form a saturated solution. One way to prepare a supersaturated solution is to warm the solution and saturate it at a higher temperature and then to cool it gradually. A supersaturated solution is unstable and can be made to crystallize by the addition of a single crystal of the solid.

15. Factors that affect solubility. The solubility of one substance in another depends in part on the nature of the particles involved. In a hypothetical three-step process if the energy return in step 3 of the dissolution process does not at least equal, or nearly equal, the energy that was expended in steps 1 and 2, the dissolution will not happen spontaneously. The energy return in step 3 depends on the magnitude of attractive forces between solvent and solute particles.

16. Solubilities of gases in liquids. The solubilities of some gases in water are given in Figure 11.7. Note that the solubility increases with increasing molar mass. The primary interactions between gas molecules and water molecules in solutions are dispersion forces. Recall that dispersion forces increase with increasing molar mass. For two gases of roughly equal molar mass (N2 and CO), the more polar molecule (CO) will be more soluble in water because of a dipole-dipole component in addition to the solute-solvent attractions.

17. Gases in water

18. “Like dissolves like” Liquids that mix with each other in all proportions are termed miscible, and those that do not mix with one another are immiscible. Polar and ionic substances tend to be soluble in polar solvents. Nonpolar substances tend to be soluble in nonpolar solvents. This observation is summarized in the expression, "Like dissolves like."

19. Best of both worlds. The –OH group is polar and can engage in hydrogen bonding with water. Carbon chains, though, are nonpolar and typically insoluble in water. Note that the longer the carbon chain gets, the more nonpolar character the molecule exhibits. With very long carbon chains, the polarity of the –OH group becomes less and less significant in the behavior of the molecule.

20. Why we want water in our gasoline.

21. Henry’s Law • The solubilities of liquids and solids in water are not appreciably affected by increased pressure. • The solubilities of gases are significantly affected by pressure. • The solubility of a gas is directly proportional to the partial pressure of the gas over the solution. • This relationship is known as Henry's law.

22. Usually, hot is better. In familiar examples, such as stirring sugar into tea before the ice is added versus after the ice is added, we recognize that solubility of a solid in water typically increases with increasing temperature. However, there are some solids that actually become less soluble at higher temperatures. Sodium hydroxide is an example.

23. Exo or endo again? What governs the way increased temperature impacts solubility is whether the dissolution process for the substance is endothermic or exothermic. When the enthalpy of dissolution is endothermic, a temperature increase will increase solubility. When the enthalpy of dissolution is exothermic, a temperature increase will decrease solubility.

24. Solubility curves for solids in water.

25. All gases behave the same. The solubilities of all gases decrease at higher temperatures because the enthalpies of dissolution for all gases are exothermic. If you have ever left a glass of tap water on the kitchen counter for a period of time, you may have noticed the formation of bubbles on the inside of the glass. These bubbles are dissolved gases coming out of the water as the temperature increases.

26. Solubility curves for gases.

27. Expressions of concentration.First there’s mass percent… • Qualitatively, we refer to solutions as being concentrated (containing a relatively large amount of solute) or dilute (containing a relatively small amount of solute). • Usually, though, it is necessary to specify the concentration of a solution in a more quantitative way. Concentrations of solutes can be expressed in a number of different ways. • Mass percentage is simply the ratio of solute mass to total mass, times 100.

28. ppm • For very dilute solutions the concentration might be expressed in parts per million (ppm). A 1 ppm aqueous solution contains approximately 1 milligram of solute per liter of solution.

29. ppb • For even more dilute solutions the term parts per billion (ppb) may be used. A 1 ppb aqueous solution contains 1 microgram of solute per liter of solution.

30. Molarity • Molarity, originally discussed in Chapter 4.5, is the ratio of moles of solute to liters of solution. Molarity (M) can vary slightly with temperature because the density of a solution can vary slightly.

31. Molality • Molality, symbolized m, is the ratio of moles of solute to kilograms of solvent. (For dilute aqueous solutions, molarity and molality are roughly equal.) Molality does not vary with temperature, because mass is not temperature-dependent

32. Mole fraction • Mole fraction is the ratio of moles of a solution component to total moles of all components in the solution. Often symbolized , mole fraction has no units.

33. Vapor Pressure Solutions have properties that differ from those of the pure solvent. For example, vapor pressure is altered. Raoult's law (for nonvolatile liquids) Psoln = Xsolvent P0solvent Psoln = vapor pressure of solution Xsolvent = mole fraction of solvent P0solvent = original vapor pressure of solvent

34. Example Calculate the expected vapor pressure at 25° C for a solution prepared by dissolving 158.0 g of common table sugar (sucrose, molar mass = 342.3 g/mol) in 643.5 cm 3 of water. At 25° C, the density of water is 0.9971 g/cm3 and the vapor pressure is 23.76 torr.

35. Answer 23.46 torr

36. Volatile Situations When the solute is volatile, Raoult's law is altered to accommodate the contribution of the solute to the total vapor pressure and becomes PTOTAL = PA + PB = XA P0A + XB P0B PTOTAL = total vapor pressure of solution PA and PB = partial pressures of A and B XA and XB = mole fractions of A and B P0A and P0B = vapor pressures of pure A and B.

37. Ideal or Nonideal Raoult's law functions much like the ideal gas law. A liquid-liquid solution that obeys Raoult's law is dubbed an ideal solution. Solutions are never perfectly ideal.

38. Example A solution is prepared by mixing 5.81 g acetone (C3H6O, molar mass = 58.1 g) and 11.9 g chloroform (HCCl3, molar mass = 119.4 g/mol). At 35° C, this solution has a total vapor pressure of 260. torr. Is this an ideal situation? The vapor pressures of pure acetone and pure chloroform at 35° C are 345 and 293 torr, respectively.

39. Answer The solution does not behave ideally as the observed value of 260. torr is much lower than the theoretical value of 319 torr.

40. Colligative Properties Changes of state depend on vapor pressure, so boiling point and freezing point is depressed by the addition of a nonvolatile solute to a solvent. Change in boiling point is determined by ΔT = Kb msolute ΔT = boiling-point elevation Kb = constant dependent on solvent (Table 11.5) msolute = molality of solute

41. Example A solution was prepared by dissolving 18.00 g glucose in 150.0 g water. The resulting solution was found to have a boiling point of 100.34º C. Kb for water is 0.51. Calculate the molar mass of glucose. Glucose is a molecular solid that is present as individual molecules in solution.

42. Answer 180 g/mol

43. Freezing-Point Depression The change in freezing point is calculated by ΔT = Kf msolute ΔT = freezing-point depression Kf = molal freezing-point depression constant; depends on solvent (Table 11.5) msolute = molality of solute

44. Osmotic Pressure Consider an apparatus minimum in which a solution and a pure solvent are divided by a semipermeable membrane. The solvent will flow across the semipermeable membrane into the solution by osmosis. At some point the liquid levels will stop changing as equilibrium is reached. There is greater hydrostatic pressure on the solution than on the solvent and this is known as Osmotic pressure.

45. Osmotic Pressure Osmotic pressure can also be defined as the minimum amount of pressure that when applied to the solution surface prevents osmosis from occurring. Osmotic pressure depends on molarity by Π = MRT Π = osmotic pressure M = molarity R = gas law constant (0.0821 L*atm/mol*K T = temperature

46. Example To determine the molar mass of a certain protein, 1.00 x 10-3 g of it was dissolved in enough water to make 1.00 mL of solution. The osmotic pressure of this solution was found to be 1.12 torr at 25.0º C. Calculate the molar mass of the protein.

47. Answer 1.66 x 104 g/mol

48. Colloids A colloid is a mixture in which particles are suspended in a medium and do not settle out, but which are not considered solutions because the particles are too large. A colloid can be distinguished from a solution by the Tyndall effect. When a beam of light is passed through a colloid, the particles scatter the light and cause the beam to be visible as it passes through the mixture.