Chapter 12 Solutions

1. Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro Chapter 12Solutions Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall

2. Solutions • soluteis the dissolved substance • seems to “disappear” • “takes on the state” of the solvent • solvent is the substance solute dissolves in • does not appear to change state • when both solute and solvent have the same state, the solvent is the component present in the highest percentage • solutions in which the solvent is water are called aqueous solutions Tro, Chemistry: A Molecular Approach

3. moles of solute liters of solution molarity = Solution ConcentrationMolarity • moles of solute per 1 liter of solution • used because it describes how many molecules of solute in each liter of solution • if a sugar solution concentration is 2.0 M, 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar Tro, Chemistry: A Molecular Approach

4. Molarity and Dissociation • the molarity of the ionic compound allows you to determine the molarity of the dissolved ions • CaCl2(aq) = Ca+2(aq) + 2 Cl-1(aq) • A 1.0 M CaCl2(aq) solution contains 1.0 moles of CaCl2 in each liter of solution • 1 L = 1.0 moles CaCl2, 2 L = 2.0 moles CaCl2 • Because each CaCl2 dissociates to give one Ca+2 = 1.0 M Ca+2 • 1 L = 1.0 moles Ca+2, 2 L = 2.0 moles Ca+2 • Because each CaCl2 dissociates to give 2 Cl-1 = 2.0 M Cl-1 • 1 L = 2.0 moles Cl-1, 2 L = 4.0 moles Cl-1 Tro, Chemistry: A Molecular Approach

5. Solution ConcentrationMolality, m • moles of solute per 1 kilogram of solvent • defined in terms of amount of solvent, not solution • like the others • does not vary with temperature • because based on masses, not volumes Tro, Chemistry: A Molecular Approach

6. Percent • parts of solute in every 100 parts solution • mass percent = mass of solute in 100 parts solution by mass • if a solution is 0.9% by mass, then there are 0.9 grams of solute in every 100 grams of solution • or 0.9 kg solute in every 100 kg solution Tro, Chemistry: A Molecular Approach

7. Using Concentrations asConversion Factors • concentrations show the relationship between the amount of solute and the amount of solvent • 12%(m/m) sugar(aq) means 12 g sugar  100 g solution • or 12 kg sugar  100 kg solution; or 12 lbs.  100 lbs. solution • 5.5%(m/v) Ag in Hg means 5.5 g Ag  100 mL solution • 22%(v/v) alcohol(aq) means 22 mL EtOH  100 mL solution • The concentration can then be used to convert the amount of solute into the amount of solution, or vice versa Tro, Chemistry: A Molecular Approach

8. Preparing a Solution • need to know amount of solution and concentration of solution • calculate the mass of solute needed • start with amount of solution • use concentration as a conversion factor • 5% by mass Þ 5 g solute  100 g solution • “Dissolve the grams of solute in enough solvent to total the total amount of solution.” Tro, Chemistry: A Molecular Approach

9. Example • How would you prepare 250.0 g of 5.00% by mass glucose solution (normal glucose)?

10. grams solute grams solution x 106 Solution ConcentrationPPM • grams of solute per 1,000,000 g of solution • mg of solute per 1 kg of solution • 1 liter of water = 1 kg of water • for water solutions we often approximate the kg of the solution as the kg or L of water mg solute L solution mg solute kg solution

11. Example • What volume of 10.5% by mass soda contains 78.5 g of sugar? Density of glucose is 1.04g/ml

12. moles of components A total moles in the solution mole fraction of A = XA= Solution ConcentrationsMole Fraction, XA • the mole fractionis the fraction of the moles of one component in the total moles of all the components of the solution • total of all the mole fractions in a solution = 1 • unitless • the mole percentageis the percentage of the moles of one component in the total moles of all the components of the solution • = mole fraction x 100% Tro, Chemistry: A Molecular Approach

13. Example • What is the molarity of a solution prepared by mixing 17.2 g of C2H6O2 with 0.500 kg of H2O to make 515 mL of solution? • What is the percent by mass of a solution prepared by mixing 17.2 g of C2H6O2 with 0.500 kg of H2O to make 515 mL of solution? • What is the mole fraction of a solution prepared by mixing 17.2 g of C2H6O2 with 0.500 kg of H2O to make 515 mL of solution?

14. A water sample is found to contain the pollutant chlorobenzene with a concentration of 15 ppb (by mass). What volume of this water contains 5.00 x 102 mg of chlorobenzene? Assume density of 1.00 g/ml

15. Spontaneous Mixing Tro, Chemistry: A Molecular Approach

16. Mixing and the Solution ProcessEntropy • formation of a solution does not necessarily lower the potential energy of the system • the difference in attractive forces between atoms of two separate ideal gases vs. two mixed ideal gases is negligible • yet the gases mix spontaneously • the gases mix because the energy of the system is lowered through the release of entropy • entropy is the measure of energy dispersal throughout the system • energy has a spontaneous drive to spread out over as large a volume as it is allowed Tro, Chemistry: A Molecular Approach

17. Intermolecular Forces and the Solution ProcessEnthalpy of Solution • energy changes in the formation of most solutions also involve differences in attractive forces between particles • must overcome solute-solute attractive forces • endothermic • must overcome some of the solvent-solvent attractive forces • endothermic • at least some of the energy to do this comes from making new solute-solvent attractions • exothermic Tro, Chemistry: A Molecular Approach

18. Intermolecular Attractions Tro, Chemistry: A Molecular Approach

19. Relative Interactions and Solution Formation • when the solute-to-solvent attractions are weaker than the sum of the solute-to-solute and solvent-to-solvent attractions, the solution will only form if the energy difference is small enough to be overcome by the entropy Tro, Chemistry: A Molecular Approach

20. Will It Dissolve? • Chemist’s Rule of Thumb – Like Dissolves Like • a chemical will dissolve in a solvent if it has a similar structure to the solvent • when the solvent and solute structures are similar, the solvent molecules will attract the solute particles at least as well as the solute particles to each other Tro, Chemistry: A Molecular Approach

21. Classifying Solvents Tro, Chemistry: A Molecular Approach

22. Example 12.1a  predict whether the following vitamin is soluble in fat or water The 4 OH groups make the molecule highly polar and it will also H-bond to water. Vitamin C is water soluble Vitamin C Tro, Chemistry: A Molecular Approach

23. Example 12.1b  predict whether the following vitamin is soluble in fat or water The 2 C=O groups are polar, but their geometric symmetry suggests their pulls will cancel and the molecule will be nonpolar. Vitamin K3 is fat soluble Vitamin K3 Tro, Chemistry: A Molecular Approach

24. Energetics of Solution Formation • overcome attractions between the solute particles – endothermic • overcome some attractions between solvent molecules – endothermic • for new attractions between solute particles and solvent molecules – exothermic • the overall DH depends on the relative sizes of the DH for these 3 processes DHsol’n = DHsoluteDHsolvent + DHmix Tro, Chemistry: A Molecular Approach

25. Heats of Hydration • for aqueous ionic solutions, the energy added to overcome the attractions between water molecules and the energy released in forming attractions between the water molecules and ions is combined into a term called the heat of hydration • attractive forces in water = H-bonds • attractive forces between ion and water = ion-dipole • DHhydration = heat released when 1 mole of gaseous ions dissolves in water Tro, Chemistry: A Molecular Approach

26. Ion-Dipole Interactions • when ions dissolve in water they become hydrated • each ion is surrounded by water molecules Tro, Chemistry: A Molecular Approach

27. Solution Equilibrium Tro, Chemistry: A Molecular Approach

28. Solubility Limit • a solution that has the maximum amount of solute dissolved in it is said to be saturated • depends on the amount of solvent • depends on the temperature • and pressure of gases • a solution that has less solute than saturation is said to be unsaturated • a solution that has more solute than saturation is said to be supersaturated Tro, Chemistry: A Molecular Approach

29. Temperature Dependence of Solubility of Solids in Water • solubility is generally given in grams of solute that will dissolve in 100 g of water • for most solids, the solubility of the solid increases as the temperature increases • when DHsolution is endothermic • solubility curves can be used to predict whether a solution with a particular amount of solute dissolved in water is saturated (on the line), unsaturated (below the line), or supersaturated (above the line) Tro, Chemistry: A Molecular Approach

30. Temperature Dependence of Solubility of Gases in Water • solubility is generally given in moles of solute that will dissolve in 1 Liter of solution • generally lower solubility than ionic or polar covalent solids because most are nonpolar molecules • for all gases, the solubility of the gas decreases as the temperature increases • the DHsolution is exothermic because you do not need to overcome solute-solute attractions Tro, Chemistry: A Molecular Approach

31. Soap Action • most dirt and grease is made of nonpolar molecules – making it hard for water to remove it from the surface • soap molecules form micelles around the small oil particles with the polar/ionic heads pointing out • this allows the micelle to be attracted to water and stay suspended Tro, Chemistry: A Molecular Approach

32. The polar heads of the micelles attract them to the water, and simultaneously repel other micelles so they will not coalesce and settle out. Tro, Chemistry: A Molecular Approach

33. Tro, Chemistry: A Molecular Approach