1. Solutions and Their Behaviors Chapter 14

2. Styrene, the building block of polystyrene, is a hydrocarbon, a compound consisting of only C and H. If 0.438g of styrene is burned in oxygen and produces 1.481g of CO 2 and 0.303g of H 2 O, what is the empirical formula of styrene? But before we begin…a little review Q for your enjoyment.

3. Units of Concentration Molarity (M)  (moles solute / L total sol’n) Molality (m)  (moles solute / kg solvent) Mole Fraction of A (X)  (moles A / moles all) Weight % A  (mass A / mass all)

4. Example Problem Assume you add 1.2kg of ethylene glycol as an antifreeze to 4.0kg of water in the radiator of your car. What is the mole fraction, molality and weight percent of the ethylene glycol? (RMM of the e.g. is 62.1 g/mol). X = 0.080 m = 4.8m Wt % = 23%

5. Units of Concentration A saturated sol’n is a sol’n in which no more solute will break up into its respective ions. An unsaturated sol’n is a sol’n in which is conc. is less than that of its saturation pt. A supersaturated sol’n is a sol’n in which more of the solute has been made to dissolve than what should under normal conditions. Solubility is the ability of a solute to go into solution.

6. Units of Concentration If two liquids mix to form a solution, they are said to be miscible. In contrast, immiscible liquids do not mix to form a solution. They exist in separate layers. Remember the phrase “like dissolves like”. This means that polar substances dissolve in polar substances and non-polar substance dissolve in non-polar substances. Polar not it non-polar and non-polar not in polar.

7. Colligative Properties Colligative properties of solutions depend on the relative amounts of solute and solvent particles in the solution. Vapor pressure changes Boiling point elevation Freezing point depression Osmotic pressure

8. Vapor Pressure Vapor pressure is how badly liquid molecules want to turn in to gas molecules and leave the container. If VP is high then molecules want to escape badly. If VP is low then they cannot escape easily.

9. Vapor Pressure The equilibrium vapor pressure at a particular temperature is the pressure of the vapor when the rate at which molecules escape the liquid and enter the gas phase equals the rate at which the gas molecules condense to re-form the liquid.

10. Changes in Vapor Pressure: Raoult’s Law The vapor pressure measured above a solution is less than the vapor pressure measure above a pure solvent. The vapor pressure of the solvent, P solv, is proportional to the relative number of solvent molecules in the solution; that is the solvent vapor pressure is proportional to the solvent mole fraction, P solv α X solv. proportional to

11. Changes in Vapor Pressure: Raoult’s Law P solv = X solv Pº solv Raoult’s Law tells us that the VP of solvent over a solution (P solv ) is some fraction(X solv ) of the pure solvent equilibrium vapor pressure (Pº solv ). For example is 95% of the molecules in a solution are solvent particles, then the vapor pressure of the solvent in solution is 95% of the pure solvent’s VP.

12. Pure solvent VP Solvent in solution VP

13. Example Problem Suppose 651g of ethylene glycol (MW = 62.1g/mol) is dissolved in 1.50kg of water. What is the VP of the water over the solution at 90ºC. VP of pure water at 90ºC is 525.8mmHg (Appendix G).

14. Example Problem A 35.0g sample of ethylene glycol, HOCH 2 CH 2 OH, is dissolved in 500.0g of water. The VP of water at 32ºC is 35.7mmHg. What is the VP of the water- ”ethgly” solution at 32ºC?

15. Boiling Point Elevation The VP lowering caused by a nonvolatile solute leads to an increase in the boiling point for the solvent. Adding a solute to solvent will lower the VP of the solvent as explained by Raoult’s Law. If molecules have a lower VP, they are not pushing as hard to get out. So will need an extra push to get to the boiling point, so BP is elevated by the presence of a solute.

16. Boiling Point Elevation ΔT bp = K bp m solute K bp is the molal boiling point elevation constant. (Table 13.4, pg 577)

17. Example Problem Eugenol, the active ingredient in cloves, has a formula of C 10 H 12 O 2. What is the boiling point of a solution containing 0.144g of this compound dissolved in 10.0g of benzene. The normal boiling point of benzene is 80.10ºC and K bp for benzene is 2.53 (ºC/m).

18. Freezing Point Depression Another consequence of dissolving a solute in a solvent is that the freezing point of a solution is lower than that of the pure solvent. ΔT fp = K fp m solute K fp is the molal freezing point depression constant. (Table 14.4, pg 577)

19. Example Problem What mass of ethylene glycol must be added to 5.50kg of water to lower the freezing point of the water from 0.0 to -10.0ºC? (This is approximately the situation in your car).

20. If 85.00g of sugar (C 12 H 22 O 11 ) are dissolved in 392.0g of water, what will be the boiling point, freezing point and vapor pressure of the resulting solution at 22.00C? Also determine the weight percent and molarity of the solution if the density is 1.03g/mL. Partner Problem work in pairs for 10min work in pairs for 10min

21. Colligative Properties and Molar Mass Determination Earlier you learned how to calculate a molecular formula from an empirical formula when given the molar mass. How do you know the molar mass of an unknown compound? One way is to perform an experiment using the idea of colligative properties by putting the unknown compound into a known solvent and see what happens to the bp or fp.

22. Example Problem A solution prepared from 1.25g of oil of wintergreen (methyl salicylate) in 99.0g of benzene has a boiling point of 80.31ºC. Determine the molar mass of this compound.

23. Think back… In a 2 nd experiment it was found that this compound has a percent composition of 63.14%C, 5.31%H and 31.55%O. What are the empirical & molecular formulas of methyl salicylate?

24. Think back… If 20.0g of methyl salicylate were used up in a combustion reaction, how many liters of carbon dioxide were formed at STP, and how many grams of O 2 were used?

25. Colligative Properties of Solutions Containing Ions Up until this point we have only been using molecular compounds as our examples for colligative properties. What happens when you add an ionic compound that breaks down into ions when dissolved in the solvent? It is almost true that the boiling point and freezing point change in proportion to the ions they bring to the solution.

26. Practice… When a clear, yellow solution of potassium chromate is added to a clear, colorless solution of silver nitrate, a brick- red precipitate forms. A laboratory technician added 30.0 mL of 0.150 M potassium chromate to 45.0 mL of 0.145 M silver nitrate. a) Write the chemical equation for the reaction. b) Perform calculations to show which of the two reagents is limiting. c) Calculate the theoretical yield of the precipitate in grams. d) Only 0.220 g of dry precipitate is collected. What is the percent yield of the reaction?

27. Colligative Properties of Solutions Containing Ions We use the van’t Hoff factor to our previous equations to correct for this. So, ΔT bp = iK bp m solute ΔT fp = iK fp m solute You would expect the van’t Hoff factor for NaCl to be 2, but in fact it is 1.85. You would expect the van’t Hoff factor for Na 2 SO 4 to be 3, but in fact it is 2.8.

28. Example Problem Calculate the freezing point of 525g of water that contains 25.0g of NaCl. Assume i is 1.85 for NaCl.

29. Osmosis Osmosis is the movement of solvent molecules through a semi permeable membrane from a region of low solute concentration to a region of higher solute concentration. Π = cRT, where Π is the osmotic pressure, c in the molar concentration, R is the gas law constant of 0.0821 L·atm/mol·K, and T is the temp in K.

30. Example Problem Beta-carotene is the most important of the A vitamins. Its molar mass can be determined by measuring the osmotic pressure generated by a given mass of B-carotene dissolved in the solvent chloroform. Calculate the molar mass of B-carotene if 10.0mL of a solution containing 7.68mg has an osmotic pressure of 26.57 mmHg at 25.0ºC

31. Osmosis Π = cRT, where R = 0.0821 L·atm/mol·K, m = 0.00768g T = 298K V = 0.010L Π = 26.57 mmHg = 0.03496atm

32. Osmosis Π = cRT or… Π = MRT (M is n/V) so… Π = nRT/V If Π is pressure then… P = nRT/V or… PV = nRT How do we change this equation to solve for MM of a compound? PV = mRT / MM or MM = mRT / PV

33. Factors Affecting Solubility The solubility of a gas in a liquid is proportional to the gas pressure. The higher the pressure, the higher its solubility. Henry’s Law (Dissolving gases into liquids) S g = k H P g S g is the gas solubility, K H is a constant and P g is the partial pressure of the gas solute.

34. Example Problem What is the concentration of O 2 in a fresh water stream in equilibrium with the air at 25ºC and 1.0 atmosphere? Express the answer in grams per liter of water. Mole fraction of O 2 in air is 0.21 k H is 1.66 x 10 -6 M / mmHg

35. Answer Since the mole fraction of O 2 in the air is 0.21 and the pressure is 1.0 atmosphere, the partial pressure of O 2 in the stream is 0.21atm. Convert this to mmHg and we get 160 mmHg. Using Henry’s Law, we get Sol of O 2 = (1.66 x10 -6 M/mmHg)(160mmHg) Sol of O 2 = 2.66 x 10 -4 M Factor label this to g/L using MW to get… 0.00850 g/L of O 2 in the stream