Chemistry 201: General Chemistry II - Lecture Dr. Namphol Sinkaset Chapter 14 Study Guide Concepts 1. Solutions are homogeneous mixtures of two or more substances. 2. solute: substance present in smaller amount. 3. solvent: substance present in greater amount. 4. Liquid solutions are easiest to study because: (1) components are localized, (2) components mix well, and (3) they are easy to handle/transfer. 5. Solutions do not have to be liquid based. 6. solubility: amount of substance that will dissolve in a given amount of solvent. 7. Formation of a gas phase solution comprised of two noble gases is driven by entropy, not energy. 8. entropy: measure of energy of randomization or energy dispersal in a system. 9. Formation of liquid solutions are driven by entropy and intermolecular forces. 10. In solution, there are three types of particle interactions. 11. Like dissolves like rule: when solute and solvent have similar intermolecular forces, they tend to form a solution. 12. miscible: term used to describe two liquids that are soluble in all proportions. 13. The relationship between solvent-solute, solvent-solvent, and solute-solute interactions determines when a solution will form. If solvent-solute interactions are about equal to solvent-solvent and solute-solute interactions, entropy will decide whether a solution forms. 14. Energy changes are observed when solutions form. The enthalpy of solution, H soln, is the enthalpy associated with solution creation, and it can be estimated with a Hess s Law type calculation. 15. Formation of a solution can be broken up into three steps: (1) separation of solid into its constituent particles, H solute ; (2) separation of solvent particles to make room for solute, H solvent ; (3) mixing solute particles with solvent particles, H mix. 16. For ionic, water-based solutions, H solvent and H mix can be combined into H hydration. 1
17. heat of hydration: enthalpy change when 1 mole of gaseous solute ions are dissolved in water. 18. For ionic compounds, H solute = H lattice. 19. When excess solid is added to a solution, an equilibrium is set up between the dissolution and deposition processes. If an equilibrium is established, the solution is said to be saturated. 20. If an equilibrium cannot be established, then the solution is unsaturated. 21. Supersaturated solutions hold more solute than is theoretically possible. 22. Generally, higher temperatures result in higher solubility of solids. 23. Henry s Law states that if we push a gas hard enough (with external pressure), more of it should dissolve in solution. Note however that higher temperatures generally decrease gas solubility. 24. Since molarity varies with temperature, sometimes it s necessary to express concentrations in temperature-independent units. 25. mass percent: ratio of mass of component to mass of solution times 100%. 26. molality (m): moles of solute per 1 kg solvent. 27. mole fraction (X): ratio of moles of component to total moles of solution. 28. ppm and ppb are parts by mass concentrations. 29. Raoult s Law expresses how the vapor pressure of a solvent is affected when a nonvolatile solute is added. 30. For solutions of two volatile liquids, Raoult s Law can be combined with Dalton s Law of Partial Pressures to get the total vapor pressure of the solution. 31. Ideal solutions obey Raoult s law at all concentrations for both solute and solvent. Typically ideal solutions have H soln = 0. Physically, this means that the IM forces between solute-solute and solvent-solvent are equal to the IM forces between solventsolute, i.e. a solute particle doesn t feel more or less attraction to a solvent particle and vice versa. 32. Negative deviations from Raoult s Law are due to stronger forces of attraction in solution than in the pure liquid. 33. Positive deviations from Raoult s Law are due to weaker forces of attraction in solution than in the pure liquid. 34. colligative properties: properties that depend on the amount of solute and not the type of solute. 2
35. Freezing point depression and boiling point elevation are two colligative properties that deal with changes in freezing and boiling temperatures as a function of added solute. 36. osmosis: the flow of solvent from a solution of lower solute concentration to one of higher solute concentration. 37. The pressure needed to stop the flow of solvent is called the osmotic pressure. 38. If the solute is an electrolyte (salt that can dissociate), then its effective molality is higher. 39. The formation of ion pairs can lead to deviations from the expected freezing/boiling point depression/elevation. Ion pairs are closely associated ions of opposite charge that act as one species instead of completely dissociating. 40. van t Hoff factor: a number that gives an idea of the amount of dissociation in an electrolyte. Equations 1. H soln = H solute + H solvent + H mix (enthalpy of solution calculation) 2. S gas = k H P gas at constant T (Henry s Law) 3. P solution = X solvent P solvent (Raoult s Law) 4. T f = mk f (freezing point depression) 5. T b = mk b (boiling point elevation) 6. Π = MRT (osmotic pressure) 7. i = moles particles in soln moles formula units dissolved (van t Hoff factor) 8. T f = imk f (freezing point depression, for ionics) 9. T b = imk b (boiling point elevation, for ionics) 10. Π = imrt (osmotic pressure, for ionics) Representative Problems 68. In an aqueous solution of sulfuric acid, the concentration is 1.89 mole % of acid. The density of the solution is 1.0645 g/ml. Calculate the following: (a) the molal concentration of H 2 SO 4, (b) the weight percent of the acid, and (c) the molarity of the solution. 3
Whenever you re given mole percent or weight percent, the best starting point is the assume that you have 100 units of it. If we do this for 1.89 mole % sulfuric acid, then it means we have: 1.89 mole H 2 SO 4 100 total moles = 1.89 mole H 2 SO 4 1.89 mole H 2 SO 4 + 98.11 mole H 2 O From here, it s probably easiest to go the weight percent, which is mass H 2 SO 4 divided by the total mass and then multiplied by 100%. To get there, we need to calculate the masses from the moles that we got above. 1.89 mole H 2 SO 4 98.0716 g 1 mole H 2 SO 4 = 185. 3 g For (b), we have: 98.11 mole H 2 O 18.015 g 1 mole H 2 O = 1767. 4 g weight % H 2 SO 4 = mass H 2SO 4 total mass 100% = 185. 3 g H 2 SO 4 185. 3 + 1767. 4 g 100% = 0.0948 9 100% = 9.49% For (a), we want to calculate the molality, which is mole per kg solvent. We already have all the necessary numbers, so we just plug it in. m H 2 SO 4 = mole H 2SO 4 kg H 2 O 1.89 mole = 1.767 4 kg = 1.07 For (c), we need to use the density. However, note that it s the density of the sulfuric acid solution, not just plain water. Therefore, we need to have total mass in the denominator. Recall that molarity is in mole per L. 1.89 mole H 2 SO 4 1952. 7 g 1.0645 g 1 ml 1000 ml 1 L = 1.03 0 mole/l = 1.03 M 72. Pentane (C 5 H 12 ) and heptane (C 7 H 16 ) are two hydrocarbon liquids present in gasoline. At 20 C, the vapor pressure of pentane is 420 torr and the vapor pressure of heptane is 36.0 torr. What will be the total vapor pressure (in torr) of a solution prepared by mixing equal masses of the two liquids? 4
This a Raoult s Law and Dalton s Law of Partial Pressures problem. First it says that we re mixing equal masses of the two liquids, so we need to pick something. I will choose a total mass of 100.0 g which means that each component will have a mass of 50.0 g. Raoult s Law deals with mole fractions, so we need to calculate the mole fractions of each component. We start by finding the moles of each component. 50.0 g pentane 1 mole pentane 72.15 g = 0.693 mole pentane 50.0 g heptane Now we calculate mole fraction of each. 1 mole heptane 100.2 g = 0.499 mole heptane mole pentane X pentane = total moles 0.693 mole = 1.192 mole = 0.581 Since the mole fractions of all components must add up to one, we know that the mole fraction of heptane must be 0.419 (1.000-0.581). Now we calculate the new vapor pressures of the components using Raoult s Law. P pentane = X pentane Ppentane = (0.581)(420 torr) = 244 torr P heptane = X heptane Pheptane = (0.419)(36.0 torr) = 15.1 torr Now we use Dalton s Law of Partial Pressures to get the total vapor pressure of the solution. P total = P pentane + P heptane = 244 torr + 15.1 torr = 259.1 torr = 259 torr 81. A solution of 12.00 g of an unknown nondissociating compound dissolved in 200.0 g of benzene freezes at 3.45 C. Calculate the molecular mass of the unknown. 5
We use the freezing point depression equation for this problem. Benzene freezes at 5.45 C and has a K f of 5.07 C/m. T f = K f m 5.45 C - 3.45 C = (5.07 C/m)m 2.00 C = (5.07 /m)m m = 0.394 Now, we need to find out how many moles of the compound exist in this solution. The problem states that we have only 200.0 g of benzene, or 0.2000 kg. 0.2000 kg 0.394 mole compound 1 kg = 0.0788 mole compound Molecular mass is expressed in g/mole. We re given that 12.00 g of the unknown was added, and from out calculation above, this corresponds to 0.0788 mole. MW unknown = 12.00 g 0.0788 mole = 152 g/mole 89. What is the osmotic pressure, in torr, of a 2.0% solution of NaCl in water when the temperature of the solution is 25 C? Assume the density of the solution is 1.00 g/ml. This is an osmotic pressure problem that involves an ionic compound. Colligative property problems with an ionic compound involve using the van t Hoff factor. Looking up the van t Hoff factor for NaCl, we see that i = 1.9. We plug in what we know into the osmotic pressure equation. Π = im RT = (1.9) M (0.082056 L atm/k mole) (298.15 K) We are given the concentration in % mass, so we need to convert to molarity. 2.0 g NaCl 1 mole NaCl 1.00 g solution 1000 ml = 0.34 223 M NaCl 100.0 g solution 58.44 g NaCl 1.00 ml solution 1 L We substitute into the osmotic pressure equation we ve started: Π = (1.9) M (0.082056 L atm/k mole) (298.15 K) = (1.9) (0.34 223 M) (0.082056L atm/k mole) (298.15 K) = 15. 908 atm 6
We finish the problem by converting to torr: 15. 908 atm 760 torr 1 atm = 1.2 104 torr 7