1. SOLUTIONS SOLUTION – A homogeneous mixture SOLVENT – The major component of a solution SOLUTE – The minor component(s) of a solution 3G-1 (of 15)

2. LIKES DISSOLVE LIKES Polar Component with a Nonpolar Component Polar molecules attract with LDF, DDA Nonpolar molecules attract with LDF To make a solution, polar and nonpolar molecules attract with LDF These are too weak to overcome the polar molecules’ LDF, DDA Polar Nonpolar 3G-2 (of 15)

3. LIKES DISSOLVE LIKES Two Nonpolar Components Nonpolar molecules attract with LDF To make a solution, different nonpolar molecules attract with LDF All molecules equally attract each other  a solution forms 3G-3 (of 15)

4. LIKES DISSOLVE LIKES Two Polar Components Polar molecules attract with LDF, DDA To make a solution, different polar molecules attract with LDF, DDA All molecules equally attract each other  a solution forms Unless… 3G-4 (of 15)

5. LIKES DISSOLVE LIKES Hydrogen-Bonding Solvent The solute must be able to H-Bond with the solvent to dissolve H  O H  O H  O    H H H S  Cl  Cl No H-Bonding H-Bonding 3G-5 (of 15)

6. LIKES DISSOLVE LIKES Hydrogen-Bonding Solvent The solute must be able to H-Bond with the solvent to dissolve H  O H  O H  O    H H H H  N  H  H H-Bonding H  H  N  H H-Bonding 3G-6 (of 15)

7. LIKES DISSOLVE LIKES Hydrogen-Bonding Solvent The solute must be able to H-Bond with the solvent to dissolve H  O H  O H  O    H H H H-Bonding H  C  H  O H-Bonding 3G-7 (of 15)

8. SolventSoluteSolubility C 6 H 6 C 8 H 18 C 6 H 6 CH 2 Br 2 CH 2 Cl 2 C 8 H 18 CH 2 Cl 2 CH 2 Br 2 H 2 OC 8 H 18 H 2 OCH 2 Br 2 H 2 OCH 3 OH H 2 OCH 3 CCH 3 ║ O High Low High Low Low (a little) High 3G-8 (of 15)

9. When water-soluble polar molecules dissolve in water, the MOLECULES separate from each other, and exist as intact, neutral molecules These solutions do not conduct electricity because no ions are formed NONELECTROLYTE – A water-soluble compound whose solution does not conduct electricity 3G-9 (of 15)

10. Solute molecules surrounded by water molecules are said to be HYDRATED C 2 H 5 OH (l) →C 2 H 5 OH (aq) H2OH2O C 6 H 12 O 6 (s) →C 6 H 12 O 6 (aq) H2OH2O Alcohols (C x H y OH) and sugars (C x (H 2 O) y ) are nonelectrolytes 3G-10 (of 15)

11. When acid molecules dissolve in water, the waters rip the acid molecules into IONS These solutions conduct electricity because ions are formed ELECTROLYTE – A water-soluble compound whose solution conducts electricity Acids are electrolytes 3G-11 (of 15)

12. IONIZATION – The formation of ions during the dissolving process HCl (g) →H + (aq) + Cl - (aq) H2OH2O Strong acids produce many ions in solution, and their solutions are good conductors HCl, HBr, HI, and acids with at least 2 more O’s than H’s are strong acids Strong acids are STRONG ELECTROLYTES 3G-12 (of 15)

13. Weak acids produce few ions in solution, and their solutions are poor conductors All other acids are weak acids Weak acids are WEAK ELECTROLYTES 3G-13 (of 15) HF (g) →HF(aq) H2OH2O

14. When water-soluble ionic compounds dissolve in water, the IONS separate from each other These solutions conduct electricity because ions are formed Soluble ionic compounds are strong electrolytes ION-DIPOLE ATTRACTIONS between the ions and the water molecules pull the ions into solution 3G-14 (of 15)

15. DISSOCIATION – The separation of ions from an ionic crystal during the dissolving process NaCl (s) →Na + (aq) + Cl - (aq) H2OH2O 3G-15 (of 15)

17. SATURATED – A solution that contains as much dissolved solute as possible SOLUTION EQUILIBRIUM – When the rates of dissolving and crystallizing are equal Dissolving RateCrystallizing Rate A saturated solution is in solution equilibrium SOLUBILITY – The maximum amount of solute that will dissolve in a specific amount of solvent 3H-1 (of 8)

18. (1) Temperature Increasing the temperature generally increases a solid’s solubility Increasing the temperature decreases a gas’s solubility FACTORS AFFECTING SOLUBILITY (2) Pressure No effect on a solid’s solubility Increasing the pressure increases a gas’s solubility HENRY’S LAW – The amount of gas dissolved in a liquid is directly proportional to the pressure of the gas in contact with the liquid p gas = kC gas (C gas is concentration of dissolved gas) 3H-2 (of 8)

19. CONCENTRATION UNITS (1) MASS PERCENT – The mass of solute per mass of solution, times 100 Mass Percent = Mass Solute x 100 ___________________ Mass Solution Find the mass percent of KCl in a solution made with 20.0 g KCl dissolved in 80.0 g H 2 O. = 20.0 % 20.0 g KCl x 100 _____________________ 100.0 g Solution 3H-3 (of 8)

20. (2) MOLE FRACTION (X) – The moles of solute per moles of solution Find the mole fraction of KCl in a solution made with 20.0 g KCl dissolved in 80.0 g H 2 O. x mol KCl _______________ 74.55 g KCl = 0.2683 mol KCl 20.0 g KCl x mol H 2 O _______________ 18.02 g H 2 O = 4.440 mol H 2 O 80.0 g H 2 O = 0.0570 0.2683 mol KCl ____________________________________ 0.2683 + 4.440 mol Solution 3H-4 (of 8)

21. (3) MOLARITY (M) – The moles of solute per liter of solution M = n ____ V Find the molarity of a solution with 6.24 grams of magnesium chloride dissolved in enough water to make 500.0 mL of solution. = 0.131 M MgCl 2 0.06554 mol MgCl 2 _________________________ 0.5000 L Solution n = quantity of matter (moles), V = volume (liters) x mol MgCl 2 _________________ 95.21 g MgCl 2 = 0.06554 mol MgCl 2 6.24 g MgCl 2 3H-5 (of 8) Used to easily calculate moles of solute by only measuring the volume of solution

22. Electrolyte solutions really consist of individual ions Each MgCl 2 formula unit has 1 Mg 2+ ion and 2 Cl - ions  0.131 M x 1 = 0.131 M Mg 2+ 0.131 M x 2 = 0.262 M Cl - Find the molarities of each ion in a 0.10 M Al 2 (SO 4 ) 3 solution 3H-6 (of 8)

23. Find the mass of potassium nitrate needed to prepare 250. mL of a 0.200 M potassium nitrate solution. x 101.11 g KNO 3 ___________________ mol KNO 3 = 5.06 g KNO 3 0.05000 mol KNO 3 M = n ___ V MV= n x 0.250 L solution = 0.05000 mol KNO 3 0.200 mol KNO 3 _____________________ L solution 3H-7 (of 8)

24. (4) MOLALITY (m) – The moles of solute per kilogram of solvent m = mol solute _______________ kg solvent Find the molality of a solution with 11.8 grams of glucose (C 6 H 12 O 6, m = 180.16 g/mol) dissolved in 150.0 grams of water. = 0.437 m C 6 H 12 O 6 0.06550 mol C 6 H 12 O 6 ___________________________ 0.1500 kg H 2 O x mol C 6 H 12 O 6 ______________________ 180.16 g C 6 H 12 O 6 = 0.06550 mol C 6 H 12 O 6 11.8 g C 6 H 12 O 6 _ 3H-8 (of 8) Used because it does not change when the temperature changes

26. COLLIGATIVE PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES – Properties that depend only on the number of dissolved solute particles, not what they are VOLATILE – A substance (or solute) that evaporates easily NONVOLATILE – A substance (or solute) that does not evaporate 3I-1 (of 11)

27. (1) Vapor Pressure Lowering A liquid’s equilibrium vapor pressure is lowered by having a nonvolatile solute dissolved in it The pressure exerted by a vapor in equilibrium with its liquid is called the EQUILIBRIUM VAPOR PRESSURE With solute particles dissolved in the liquid, there are less solvent molecules on the surface, therefore less solvent molecules can evaporate, so the rate of evaporation decreases Now vapor molecules condense faster than liquid molecules evaporate, so the amount of vapor decreases When the two rates are again equal, there are less vapor molecules than before, so a lower equilibrium vapor pressure 3I-2 (of 11)

28. 10 20 30 9.2 17.5 31.8 Temp (ºC)EVP of pure H 2 O (torr) 9.1 17.4 31.6 EVP of a dilute salt water solution (torr) 3I-3 (of 11) (Temp °C) EVP Pure Water Water Solution

29. RAOULT’S LAW – The equilibrium vapor pressure of a solution is proportional to the mole fraction of the solvent in the solution p solution = X solvent pº solvent p solution =equilibrium vapor pressure of the solution X solvent =mole fraction of the solvent in the solution pº solvent =equilibrium vapor pressure of the pure solvent 3I-4 (of 11) 1882 FRANÇOIS-MARIE RAOULT A solution will obey Raoult’s Law if (1)the solution is dilute (2)the solvent and solute particles are about the same size (3)the solvent and solute particles have about the same attractive forces

30. Find the equilibrium vapor pressure of a solution at 20ºC that is prepared with 0.500 moles of sucrose dissolved in 35.00 moles of water, if the equilibrium vapor pressure of water at 20ºC is 17.5 torr. = 0.9859 35.00 mol H 2 O ____________________________________ 35.00 + 0.500 mol solution p soln = X solv pº solv = (0.9859)(17.5 torr)= 17.3 torr 3I-5 (of 11)

31. Find the EVP of a solution at 20ºC that is prepared with 0.250 moles of Na 2 SO 4 dissolved in 10.00 moles of water, if the EVP of water at 20ºC is 17.5 torr. When a solute is an electrolyte (a salt or acid), the fact that it dissociates or ionizes must be considered n solute = 3 x 0.250 mol 3I-6 (of 11) = 0.9302 10.00 mol H 2 O __________________________________ 10.00 + 0.750 mol solution p soln = X solv pº solv = (0.9302)(17.5 torr)= 16.3 torr = 0.750 mol

32. Find the EVP of a solution at 23ºC that is prepared with 0.300 moles of KCl dissolved in 15.00 moles of water, if the EVP of water at 23ºC is 21.1 torr. 3I-7 (of 11) n solute = 2 x 0.300 mol = 0.9615 15.00 mol H 2 O __________________________________ 15.00 + 0.600 mol solution p soln = X solv pº solv = (0.9615)(21.1 torr)= 20.3 torr = 0.600 mol

33. In a solution, if both the solvent and solute are volatile, vapor pressure is produced from both components in the solution p solution = p solvent + p solute p solution = X solvent pº solvent + X solute pº solute 3I-8 (of 11)

34. Find the EVP of a solution prepared with 0.300 moles of acetone and 0.100 moles of chloroform if the EVP of acetone is 293 torr and the EVP of chloroform is 345 torr. p solution = X solvent pº solvent + X solute pº solute (0.7500)(293 torr) + (0.2500)(345 torr) 219.8 torr+86.25 torr = 306.05 torr = 306 torr 3I-9 (of 11) 0.300 mol acetone ________________________ 0.400 mol solution = 0.7500 0.100 mol chloroform ___________________________ 0.400 mol solution = 0.2500

35. If the EVP of a solution equals the sum of the EVP’s of the 2 components, the solution obeys Raoult’s Law IDEAL SOLUTION – A solution that obeys Raoult’s Law 3I-10 (of 11)

36. POSITIVE DEVIATION – Occurs when the attractions between solvent and solute molecules are WEAKER than the attractions in either pure component NEGATIVE DEVIATION – Occurs when the attractions between solvent and solute molecules are STRONGER than the attractions in either pure component Solution formation is endothermicSolution formation is exothermic 3I-11 (of 11)

38. (2) Boiling Point Elevation A liquid’s boiling point is raised by having a nonvolatile solute dissolved in it A solvent boils when its EVP equals the prevailing atmospheric pressure With solute particles dissolved in the solvent, its EVP is lowered Only when the temperature increases will the EVP again equal atmospheric pressure  the boiling point of the solution is now higher than the boiling point of the solvent Atmospheric Pressure 100ºC101ºC102ºC 3J-1 (of 16)

39. (Temp °C) EVP 100 Pure Water Water Solution 3J-2 (of 16)

40. To determine the increase in boiling point from the solvent to the solution: ΔT b = K b mi ΔT b =increase in the boiling point K b =molal boiling point constant of the solvent m=molality of the solution i=van’t Hoff factor of the solute = moles of particles in solution _____________________________________ moles of dissolved solute 1 3 C 6 H 12 O 6 MgBr 2 HClAl 2 (SO 4 ) 3 25 i = 3J-3 (of 16)

41. Find the boiling point of a solution that is prepared with 0.150 moles of sodium chloride dissolved in 90.0 grams of water. The boiling point of pure water is 100.00ºC and molal boiling point constant for water is 0.51 Cºkg H 2 O/mol solute. 0.150 mol NaCl ____________________ 0.0900 kg H 2 O = 1.667 m NaCl = 101.7ºC100.00ºC + 1.7 Cº (0.51 Cºkg H 2 O/mol solute)(1.667 mol solute/kg H 2 O)(2) = 1.7 Cº ΔT b = K b mi ΔT b = 3J-4 (of 16)

42. 3J-5 (of 16) All colligative properties can be used to calculate the molar mass of a nonvolatile solute

43. Find the molar mass of a nonelectrolyte (i = 1) if a solution prepared with 4.80 grams of the nonelectrolyte dissolved in 150.0 grams of carbon disulfide has a boiling point of 47.5ºC. molar mass X = grams X _____________ moles X molar mass N.E. = 4.80 grams N.E. ____________________ ? moles N.E. 3J-6 (of 16)

44. Find the molar mass of a nonelectrolyte (i = 1) if a solution prepared with 4.80 grams of the nonelectrolyte dissolved in 150.0 grams of carbon disulfide has a boiling point of 47.5ºC. Pure carbon disulfide has a boiling point of 46.3ºC and molal boiling point constant of 2.37 Cºkg CS 2 /mol solute. ΔT b = K b miΔT b = 47.5ºC – 46.3ºC = 1.2 Cº 3J-7 (of 16) ΔT b = K b (mol solute) i _______________ (kg solvent) (kg solvent) ΔT b = (mol solute) _____________________ K b i = (0.1500 kg CS 2 )(1.2 Cº) _____________________________________ (2.37 Cºkg CS 2 /mol solute)(1) = 0.0759 mol solute

45. Find the molar mass of a nonelectrolyte (i = 1) if a solution prepared with 4.80 grams of the nonelectrolyte dissolved in 150.0 grams of carbon disulfide has a boiling point of 47.5ºC. Pure carbon disulfide has a boiling point of 46.3ºC and molal boiling point constant of 2.37 Cºkg CS 2 /mol solute. 4.80 grams N.E. _______________________ 0.0759 moles N.E. = 63 g/mol 3J-8 (of 16)

46. (3) Freezing Point Depression A liquid’s freezing point is lowered by having a nonvolatile solute dissolved in it A solvent freezes when the EVP of its liquid equals the EVP of its solid With solute particles dissolved in the liquid solvent, its EVP is lowered Only when the temperature decreases will the EVP of the liquid solution equal the EVP of the solid solvent 0ºC-1ºC-2ºC  the freezing point of the solution is now lower than the freezing point of the solvent 3J-9 (of 16)

47. (Temp °C) EVP 0 Pure Water Water Solution Ice 3J-10 (of 16)

48. To determine the decrease in freezing point from the solvent to the solution: ΔT f = K f mi ΔT f =decrease in the freezing point K f =molal freezing point constant of the solvent m=molality of the solution i=van’t Hoff factor of the solute 3J-11 (of 16)

49. Find the freezing point of a solution that is prepared with 0.150 moles of calcium chloride dissolved in 97.0 grams of water. The freezing point of pure water is 0.00ºC and molal freezing point constant for water is 1.86 Cºkg H 2 O/mol solute. 0.150 mol CaCl 2 _______________________ 0.09700 kg H 2 O = 1.546 m CaCl 2 = -8.83ºC0.00ºC - 2.88 Cº (1.86 Cºkg H 2 O/mol solute)(1.546 mol solute/kg H 2 O)(3) = 8.63 Cº ΔT f = K f mi ΔT f = 3J-12 (of 16)

50. Find the molar mass of a nonelectrolyte if a solution prepared with 10.0 grams of the nonelectrolyte dissolved in 100.0 mL of benzene has a freezing point of 3.51ºC. 3J-13 (of 16) molar mass X = grams X _____________ moles X molar mass N.E. = 10.00 grams N.E. ____________________ ? moles N.E.

51. Find the molar mass of a nonelectrolyte if a solution prepared with 10.0 grams of the nonelectrolyte dissolved in 100.0 mL of benzene has a freezing point of 3.51ºC. Pure benzene has a freezing point of 5.48ºC, a molal freezing point constant of 5.12 Cºkg benzene/mol solute, and a density of 0.780 g/mL. ΔT f = K f miΔT f = 5.48ºC – 3.51ºC = 1.97 Cº 3J-14 (of 16) ΔT f = K f (mol solute) i _______________ (kg solvent) x 0.780 g benzene _____________________ mL benzene = 78.00 g benzene100.0 mL benzene = 0.07800 kg benzene

52. Find the molar mass of a nonelectrolyte if a solution prepared with 10.0 grams of the nonelectrolyte dissolved in 100.0 mL of benzene has a freezing point of 3.51ºC. Pure benzene has a freezing point of 5.48ºC, a molal freezing point constant of 5.12 Cºkg benzene/mol solute, and a density of 0.780 g/mL. 3J-15 (of 16) = (0.07800 kg C 6 H 6 )(1.97 Cº) ______________________________________ (5.12 Cºkg C 6 H 6 /mol solute)(1) = 0.03001 mol solute ΔT f = K f miΔT f = 5.48ºC – 3.51ºC = 1.97 Cº (kg solvent) ΔT b = (mol solute) _____________________ K b i ΔT f = K f (mol solute) i _______________ (kg solvent)

53. 10.0 grams N.E. _________________________ 0.03001 moles N.E. = 333 g/mol Find the molar mass of a nonelectrolyte if a solution prepared with 10.0 grams of the nonelectrolyte dissolved in 100.0 mL of benzene has a freezing point of 3.51ºC. Pure benzene has a freezing point of 5.48ºC, a molal freezing point constant of 5.12 Cºkg benzene/mol solute, and a density of 0.780 g/mL. 3J-16 (of 16)

55. Cells in an ISOTONIC SOLUTION have equal flow rates of water in and out Cells in a HYPERTONIC SOLUTION have water flowing out faster than flowing in Cells in a HYPOTONIC SOLUTION have water flowing in faster than flowing out 3K-1 (of 6) (4)OSMOTIC PRESSURE

56. SEMIPERMEABLE MEMBRANE – A membrane that will allow small molecules to pass through but not big ones water / not hydrated salt ions water, salt ions / not proteins Because the membrane blocks salt ions, water goes out of the tube slower than it goes in The extra mass of salt water builds up pressure on the membrane, forcing water through faster OSMOTIC PRESSURE (π) – The pressure on a semipermeable membrane needed to equalize the passage of water across the membrane between two solutions of different concentrations 3K-2 (of 6)

57. To determine the osmotic pressure in a solution: πV = inRT π=Osmotic Pressure (atm) V=Volume of Solution (L) i=van’t Hoff factor n=Difference in Quantity of Solute between the 2 solutions (mol) R=Universal Gas Constant (0.08206 Latm/molK) T=Temperature (K) 3K-3 (of 6)

58. To determine the osmotic pressure in a solution: πV = inRT π= inRT _______ V π = iMRT n/V is molarity 3K-4 (of 6)

59. Find the osmotic pressure that would develop in a solution that is prepared at 22ºC with 0.0300 moles of glucose dissolved in enough water to make 100. mL of solution. π = (1)(0.0300 mol)(0.08206 Latm/molK)(295.2 K) __________________________________________________________ 0.100 L = 7.27 atm π = inRT _______ V 22ºC + 273.2 K = 295.2 K 3K-5 (of 6)

60. Find the molar mass of a protein if a solution is prepared by dissolving 1.00 x 10 -3 grams of the protein in enough water to make 1.00 milliliter of solution, and the osmotic pressure at 25ºC is 1.12 torr. 25ºC + 273.2 = 298.2 K πV = n _____ iRT = (0.001474 atm)(0.00100 L) _________________________________________ (1)(0.08206 Latm/molK)(298.2 K) = 6.024 x 10 -8 mol 1.00 x 10 -3 g protein __________________________________ 6. 024 x 10 -8 moles protein = 16,600 g/mol π = inRT _______ V x 1 atm _____________ 760.0 torr = 0.001474 atm1.12 torr 3K-6 (of 6)