1. Solutions Chapter 13 Properties of Solutions Adapted by SA Green from: John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc. Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

2. Solutions Solutions are homogeneous mixtures of two or more pure substances. In a solution, the solute is dispersed uniformly throughout the solvent.

3. Solutions How does a solid dissolve into a liquid? What ‘drives’ the dissolution process? What are the energetics of dissolution?

4. Solutions How Does a Solution Form? 1.Solvent molecules attracted to surface ions. 2.Each ion is surrounded by solvent molecules. 3.Enthalpy (  H) changes with each interaction broken or formed. Ionic solid dissolving in water

5. Solutions How Does a Solution Form? 1.Solvent molecules attracted to surface ions. 2.Each ion is surrounded by solvent molecules. 3.Enthalpy (  H) changes with each interaction broken or formed.

6. Solutions How Does a Solution Form The ions are solvated (surrounded by solvent). If the solvent is water, the ions are hydrated. The intermolecular force here is ion- dipole.

7. Solutions Energy Changes in Solution To determine the enthalpy change, we divide the process into 3 steps. 1.Separation of solute particles. 2.Separation of solvent particles to make ‘holes’. 3.Formation of new interactions between solute and solvent.

8. Solutions Enthalpy Changes in Solution The enthalpy change of the overall process depends on  H for each of these steps. Start End Start

9. Solutions Enthalpy changes during dissolution The enthalpy of solution,  H soln, can be either positive or negative.  H soln =  H 1 +  H 2 +  H 3  H soln (MgSO 4 )= -91.2 kJ/mol --> exothermic  H soln (NH 4 NO 3 )= 26.4 kJ/mol --> endothermic

10. Solutions Dissolution vs reaction Dissolution is a physical change—you can get back the original solute by evaporating the solvent. If you can’t, the substance didn’t dissolve, it reacted. Ni(s) + HCl(aq)NiCl 2 (aq) + H 2 (g) NiCl 2 (s) dry

11. Solutions Degree of saturation Saturated solution  Solvent holds as much solute as is possible at that temperature.  Undissolved solid remains in flask.  Dissolved solute is in dynamic equilibrium with solid solute particles.

12. Solutions Degree of saturation Unsaturated Solution  Less than the maximum amount of solute for that temperature is dissolved in the solvent.  No solid remains in flask.

13. Solutions Degree of saturation Supersaturated  Solvent holds more solute than is normally possible at that temperature.  These solutions are unstable; crystallization can often be stimulated by adding a “seed crystal” or scratching the side of the flask.

14. Solutions Degree of saturation Unsaturated, Saturated or Supersaturated?  How much solute can be dissolved in a solution? More on this in Chap 17 (solubility products, p 739)

15. Solutions Factors Affecting Solubility Chemists use the axiom “like dissolves like”:  Polar substances tend to dissolve in polar solvents.  Nonpolar substances tend to dissolve in nonpolar solvents.

16. Solutions Factors Affecting Solubility The stronger the intermolecular attractions between solute and solvent, the more likely the solute will dissolve. Example: ethanol in water Ethanol = CH 3 CH 2 OH Intermolecular forces = H-bonds; dipole-dipole; dispersion Ions in water also have ion-dipole forces.

17. Solutions Factors Affecting Solubility Glucose (which has hydrogen bonding) is very soluble in water. Cyclohexane (which only has dispersion forces) is not water- soluble.

18. Solutions Factors Affecting Solubility Vitamin A is soluble in nonpolar compounds (like fats). Vitamin C is soluble in water.

19. Solutions Factors Affecting Solubility Surface Area The larger the surface area of the solute, the faster the dissolution, because of a larger contact with solvent molecules. Hence, CRUSHING a solid solute will cause it to dissolve faster

20. Solutions Temperature Generally, the solubility of solid solutes in liquid solvents increases with increasing temperature.

21. Solutions Gases in Solution In general, the solubility of gases in water increases with increasing mass. Why? Larger molecules have stronger dispersion forces.

22. Solutions Gases in Solution The solubility of liquids and solids does not change appreciably with pressure. But, the solubility of a gas in a liquid is directly proportional to its pressure. Increasing pressure above solution forces more gas to dissolve.

23. Solutions Henry’s Law S g = kP g where S g is the solubility of the gas; k is the Henry’s law constant for that gas in that solvent; P g is the partial pressure of the gas above the liquid.

24. Solutions Temperature The opposite is true of gases. Higher temperature drives gases out of solution.  Carbonated soft drinks are more “bubbly” if stored in the refrigerator.  Warm lakes have less O 2 dissolved in them than cool lakes.

25. Solutions Ways of Expressing Concentrations of Solutions Chap 13: Ways of Expressing Concentrations of Solutions

26. Solutions Mass Percentage Mass % of A = mass of A in solution total mass of solution  100

27. Solutions Parts per Million and Parts per Billion ppm = mass of A in solution total mass of solution  10 6 Parts per Million (ppm) Parts per Billion (ppb) ppb = mass of A in solution total mass of solution  10 9

28. Solutions moles of A total moles in solution X A = Mole Fraction (X) In some applications, one needs the mole fraction of solvent, not solute— make sure you find the quantity you need!

29. Solutions Mass Concentration C m = Unit = g/L mass of solute volume of solution

30. Solutions mol of solute L of solution [ ] =C = Molarity (M) You will recall this concentration measure from Chapter 4. Because volume is temperature dependent, molarity can change with temperature.

31. Solutions mol of solute kg of solvent m = Molality (m) Because neither moles nor mass change with temperature, molality (unlike molarity) is not temperature dependent.

32. Solutions Mass/Mass Moles/Moles Moles/Mass Moles/L

33. Solutions Changing Molarity to Molality If we know the density of the solution, we can calculate the molality from the molarity, and vice versa.

34. Solutions SAMPLE EXERCISE 13.4 Calculation of Mass-Related Concentrations (a) A solution is made by dissolving 13.5 g of glucose (C 6 H 12 O 6 ) in 0.100 kg of water. What is the mass percentage of solute in this solution? (b) A 2.5-g sample of groundwater was found to contain 5.4  g of Zn 2+ What is the concentration of Zn 2+ in parts per million? PRACTICE EXERCISE (a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution? PRACTICE EXERCISE A commercial bleach solution contains 3.62 mass % NaOCl in water. Calculate (a) the molality and (b) the mole fraction of NaOCl in the solution.

35. Solutions SAMPLE EXERCISE 13.4 Calculation of Mass-Related Concentrations (a) A solution is made by dissolving 13.5 g of glucose (C 6 H 12 O 6 ) in 0.100 kg of water. What is the mass percentage of solute in this solution? (b) A 2.5-g sample of groundwater was found to contain 5.4  g of Zn 2+ What is the concentration of Zn 2+ in parts per million? PRACTICE EXERCISE (a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution? Answers: (a) 2.91%, (b) 90.5 g of NaOCl Solution (a)Analyze: We are given the number of grams of solute (13.5 g) and the number of grams of solvent (0.100 kg = 100 g). From this we must calculate the mass percentage of solute. Plan: We can calculate the mass percentage by using Equation 13.5. The mass of the solution is the sum of the mass of solute (glucose) and the mass of solvent (water). Solve: Comment: The mass percentage of water in this solution is (100 – 11.9)% = 88.1%. (b) Analyze: In this case we are given the number of micrograms of solute. Because 1  g is 1  10 –6 g, 5.4  g = 5.4  10 –6 g. Plan: We calculate the parts per million using Equation 13.6. Solve: Answers: (a) 0.505 m, (b) 9.00  10 –3

36. Solutions SAMPLE EXERCISE 13.8 Calculation of Vapor-Pressure Lowering Glycerin (C 3 H 8 O 3 ) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25°C. Calculate the vapor pressure at 25°C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25°C is 23.8 torr (Appendix B). Solution Analyze: Our goal is to calculate the vapor pressure of a solution, given the volumes of solute and solvent and the density of the solute. Plan: We can use Raoult’s law (Equation 13.10) to calculate the vapor pressure of a solution. The mole fraction of the solvent in the solution, X A, is the ratio of the number of moles of solvent (H 2 O) to total solution (moles C 3 H 8 O 3 + moles H 2 O). Solve: To calculate the mole fraction of water in the solution, we must determine the number of moles of C 3 H 8 O 3 and H 2 O:

37. Solutions van’t Hoff Factor One mole of NaCl in water does not really give rise to two moles of ions.

38. Solutions van’t Hoff Factor Some Na + and Cl − reassociate as hydrated ion pairs, so the true concentration of particles is somewhat less than two times the concentration of NaCl.

39. Solutions Colloids: Suspensions of particles larger than individual ions or molecules, but too small to be settled out by gravity.

40. Solutions Tyndall Effect Colloidal suspensions can scatter rays of light. This phenomenon is known as the Tyndall effect.

41. Solutions END Chap 13