Solutions

Occur in all phases u The solvent does the dissolving. u The solute is dissolved. u There are examples of all types of solvents dissolving all types of solvent. u We will focus on aqueous solutions.

Energy of Making Solutions  Heat of solution (  H soln ) is the energy change for making a solution. u Most easily understood if broken into steps. u 1.Break apart solvent u 2.Break apart solute u 3. Mixing solvent and solute

1. Break apart Solvent  Have to overcome attractive forces.  H 1 >0 2. Break apart Solute.  Have to overcome attractive forces.  H 2 >0

3. Mixing solvent and solute   H 3 depends on what you are mixing.  If molecules can attract each other  H 3 is large and negative.  Molecules can’t attract-  H 3 is small and negative. u This explains the rule “Like dissolves Like”

EnergyEnergy Reactants Solution H1H1 H2H2 H3H3 Solvent Solute and Solvent  Size of  H 3 helps determine whether a solution will form H3H3 Solution

Ways of Measuring u Molarity = moles of solute Liters of solution u % mass = Mass of solute x 100 Mass of solution  Mole fraction of component A  A = n A n A + n B

u Molality = moles of solute Kilograms of solvent  Molality is abbreviated m u Normality - read but don’t focus on it. u It is molarity x number of active pieces Ways of Measuring

Vapor Pressure of Solutions u A nonvolatile solute lowers the vapor pressure of the solution. u The molecules of the solvent must overcome the force of both the other solvent molecules and the solute molecules.

Raoult’s Law:  P soln =  solvent x P solvent u Vapor pressure of the solution = mole fraction of solvent x vapor pressure of the pure solvent u Applies only to an ideal solution where the solute doesn’t contribute to the vapor pressure.

Aqueous Solution Pure water u Water has a higher vapor pressure than a solution

Aqueous Solution Pure water u Water evaporates faster from for water than solution

u The water condenses faster in the solution so it should all end up there. Aqueous Solution Pure water

Practice Problem u A solution of cyclopentane with a nonvolatile compound has vapor pressure of 211 torr. If vapor pressure of the pure liquid is 313 torr, what is the mole fraction of the cyclopentane? u P soln = X cp P cp u 211 torr = X cp (313 torr) u.674

Try one on your own u Determine the vapor pressure of a solution at 25  C that has 45 grams of C 6 H 12 O 6, glucose, dissolved in 72 grams of H 2 O. The vapor pressure of pure water at 25  C is 23.8 torr. u P solution = X solvent P solvent u P solution =.941(23.8 torr) u P solution = 22.4 torr

u Liquid-liquid solutions where both are volatile. u Modify Raoult’s Law to  P total = P A + P B =  A P A 0 +  B P B 0 u P total = vapor pressure of mixture u P A 0 = vapor pressure of pure A u If this equation works then the solution is ideal. Ideal solutions

χbχb χAχA Vapor Pressure P of pure A P of pure B Vapor Pressure of solution

Deviations u If solvent has a strong affinity for solute (H bonding). u Lowers solvent’s ability to escape. u Lower vapor pressure than expected. u Negative deviation from Raoult’s law.   H soln is large and negative exothermic.  Endothermic  H soln indicates positive deviation.

χbχb χAχA Vapor Pressure Positive deviations- Weak attraction between solute and solvent Positive ΔH soln

χbχb χAχA Vapor Pressure Negative deviations- Strong attraction between solute and solvent Negative ΔH soln

Problem #1 u The vapor pressure of a solution containing 53.6g of glycerin C 3 H 8 O 3 in133.7g ethanol C 2 H 5 OH is 113 torr at 40C. Calculate the vapor pressure of pure ethanol at 40C assuming that the glycerin is a non volatile, nonelectrolyte solute in ethanol.

Answer to #1 P soln = X eth P eth 113torr = 2.90mol/3.48mol (P eth ) torr = P eth

Problem #2 u At a certain temperature, the vapor pressure of pure benzene C 6 H 6 is 0.930atm. A solution was prepared by dissolving 10.0g of a nondissociating, nonvolatile solute in 78.11g of benzene at that temperature. The vapor pressure was found to be 0.900atm. Assuming the solution behave ideally, determine the molar mass of the solute.

Answer #2 P soln = X benzene P benzene.900atm = X benzene (.930atm) X benzene =.9677 X solute =.0323 MM = 10.0g/.0323mol = 310g/mol

Problem #3 A solution of NaCl in water has a vapor pressure of 19.6 torr at 25C. What is the mol fraction of solute particle in this solution if the vapor pressure of water is 23.8 torr at 25C?

Answer #3 P soln = X water P water 19.6torr = X water (23.8torr).824 = X water therefore X solute =.176

Problem #3 For the same problem as #3: What is the vapor pressure of the solution at 45C if the vapor pressure of water is 71.9 torr at 45C?

Answer #3 P soln =.824(71.9torr) P soln = 59.2 torr

Problem #4 A solution is made from mol CH 2 Cl 2 and mol CH 2 Br 2 at 25C. Assuming that the solution is ideal, calculate the % composition of the vapor at 25C. P CH2Cl2 = 133 torr P CH2Br2 = 11.4 torr

Answer #4 P soln = X CH2Cl2 P + X CH2Br2 P P soln = (.03/.08)(133torr) + (.05/.08)(11.4 torr) P soln = 57.0 torr X CH2Cl2 = 49.9 / 57 =.875 = 87.5% X CH2Br2 = 7.13 / 57 =.125 = 12.5%

Colligative Properties u Because dissolved particles affect vapor pressure - they affect phase changes. u Colligative properties depend only on the number - not the kind of solute particles present u Useful for determining molar mass

Water

1 atm Vapor Pressure of solution Vapor Pressure of pure water

1 atm Freezing and boiling points of solvent

1 atm Freezing and boiling points of solution

1 atm TfTf TbTb

Boiling point Elevation u Because a non-volatile solute lowers the vapor pressure it raises the boiling point.  The equation is:  T = K b m solute   T is the change in the boiling point u K b is a constant determined by the solvent.  m solute is the molality of the solute

Freezing point Depression u Because a non-volatile solute lowers the vapor pressure of the solution it lowers the freezing point. Freezing occurs when Pv (l) = Pv(s). Because the Pv(soln) is reduced, the temp must be lowered to form a solid.  The equation is:  T = -K f m solute   T is the change in the freezing point u K f is a constant determined by the solvent  m solute is the molality of the solute

Electrolytes in solution u Since colligative properties only depend on the number of molecules. u Ionic compounds should have a bigger effect. u When they dissolve they dissociate. u Individual Na and Cl ions fall apart. u 1 mole of NaCl makes 2 moles of ions. u 1mole Al(NO 3 ) 3 makes 4 moles ions.

u Electrolytes have a bigger impact on on melting and freezing points per mole because they make more pieces.  Relationship is expressed using the van’t Hoff factor i i = Moles of particles in solution Moles of solute dissolved u The expected value can be determined from the formula of the compound.

u The actual value is usually less because u At any given instant some of the ions in solution will be paired up. u Ion pairing increases with concentration.  i decreases with increasing concentration. u We can change our formulas to  = iKm

Problem #1 A 2.00 gram sample of a large biomolecule was dissolved in 15.0 grams of CCl 4. The boiling point of this solution was found to be 77.85C. Calculate the molar mass of the biomolecule. K b = 5.03C kg / mol BP CCl4 = 76.5C

Answer #1  T = K b m (77.85 – 76.5C) = 5.03C Kg/mol x m.268mol/Kg = m.268 mol/Kg = Xmol /.015kg = mol 2.00g /.00402mol = 498g/mol

Problem #2 The freezing point of t-butanol is 25.5C and K f = 9.1C Kg/mol. Usually t-butanol absorbs water on exposure to air. If the freezing point of a 10.0 gram sample of t- butanol is 24.59C, how many grams of water are present in the sample?

Answer #2  T = -K f m (24.59 – 25.5C) = -9.1 C Kg/mol x m.1 mol/Kg = m.1 m = X mol H 2 O /.01kg =.001mol H 2 O.018g H 2 O

Problem #3 Calculate the freezing point and the boiling point for each:.050m MgCl 2.050m FeCl 3 K f = 1.86C Kg/mol K b = 0.51C Kg/mol Predict which will have higher BP and which will have the lower FP

Answer #3 Dissociate each: MgCl 2 Mg Cl - i= 2.7 FeCl 3 Fe Cl - i= 3.4 MgCl 2 T f = -.25C T b = C FeCl 3 T f = -.32C T b = C