Week 3 Sections 13.3-13.5 13.3 Factors Affecting Solubility Solute-Solvent Interactions Pressure Effects Temperature Effects 13.4 Ways of Expressing Concentration Mass Percentage, ppm, and ppb Mole Fraction, Molarity, and Molality Conversion of Concentration Units 13.5 Colligative Properties Lowering the Vapor Pressure Boiling-Point Elevation Freezing-Point Depression Osmosis Determination of Molar Mass Factors that FAVOR solubility: 1. Strong solute-solvent interactions 2. Weak solute-solute interactions 3. Weak solvent-solvent interactions More often we ll settle for the solute-solvent interactions being similar to the solute-solute and solvent-solvent interactions. A general rule: Like dissolves like. i.e. polar and polar non-polar and non-polar [13.6 Colloids] Realize there is an inherent tendency for the two isolated materials to form solution, regardless of the energetics!!! This represents an entropy factor. Consider the Solubilities (mol/100g water) of Alcohols CH 3 OH CH 3 CH 2 OH CH 3 (CH 2 ) 2 OH CH 3 (CH 2 ) 3 OH 0.11 CH 3 (CH 2 ) 4 OH 0.030 CH 3 (CH 2 ) 5 OH 0.0058 CH 3 (CH 2 ) 6 OH 0.0008 Fig 13.12 Structure of glucose note red O atoms in OH groups C 6 H 12 O 6 or C 5 H 5 OCH 2 (OH) 5
Fig. 25.26 Sucrose and Lactose Molecules. Olestra has fatty acids at each of the OH in sucrose. Fig 25.27 The Starch Molecule, note the blue atoms Fig 25.28 Cellulose Molecule, note the blue atoms Table 13.2 Gas solubilities in water at 20 o C with 1 atm gas pressure (Table 13.2) Solubility/Molarity He 0.40 x 10-3 N 2 0.69 x 10-3 CO 1.04 x 10-3 O 2 1.38 x 10-3 Ar 1.50 x 10-3 Kr 2.79 x 10-3 Fig 25.29 Starch (a) and Cellulose (b) in 3-D representation CO 2 3.1 x 10-2 NH 3 ~ 53
Table 13.2 Gas solubilities in water at 20 o C with 1 atm gas pressure (Table 13.2) Solubility/Molarity He 0.40 x 10-3 N 2 0.69 x 10-3 CO 1.04 x 10-3 O 2 1.38 x 10-3 Ar 1.50 x 10-3 Kr 2.79 x 10-3 Fig 13.14 Henry s Law, C g = k P g See Table 13.2 for k in M/atm CO 2 3.1 x 10-2 NH 3 ~ 53 Consider N2 dissolved in water at 4.0 atm. Note k = 0.69 x 10-3 mol/l-atm Also in Table 13.2 Cg = k Pg = (0.69 x 10-3 mol/l-atm)(4.0 atm) = 2.76 x 10-3 mol/l at normal atmospheric conditions, however, Pg = 0.78 atm Cg = (0.69 x 10-3 mol/l-atm)(0.78 atm) = 0.538 x 10-3 mol/l Note that (2.76-0.54) x 10-3 mol/l = 2.22 x 10-3 mol/l Thus for 1.0 L of water, 0.0022 mol of nitrogen would be released = 0.0022 x 22.4L = 0.049 L = 49 ml! To read about nitrogen narcosis, see http://www.gulftel.com/~scubadoc/narked.html and http://www.diversalertnetwork.org/medical/articles/index.asp Fig 13.18 Temperature Effect on Solubility of Gases. Demonstrations The ammonia fountain Add acetone to a sat d CuSO4 sol n Heat water and calcium acetate Heat water and potassium nitrate Fig 13.17 Effect of T on Solubilities of Ionic Cmpds. Are these exothermic or endothermic processes?
Ways of expressing concentration: Prepare 0.500 L of a 0.100 M aqueous solution of KHCO 3 (MW = 100.12 g) a) percent, ppm, ppb usually m/m b) mole fraction = X A, X B sum of X i = 1 c) molarity = M or mol/l solution depends on T and density of soln preparation requires dilution d) molality = m or mol/kg solvent independent of T easily prepared Prepare 0.500 L of a 0.100 M aqueous solution of KHCO 3 (MW = 100.12 g) Prepare 0.500 L of a 0.100 M aqueous solution of KHCO 3 (MW = 100.12 g) (0.100 mol KHCO 3 /L of soln)(0.500 L) = 0.0500 mol KHCO 3 (0.100 mol KHCO 3 /L of soln)(0.500 L) = 0.0500 mol KHCO 3 Note: molarity = M = n/v therefore n = MV (0.0500 mol KHCO 3 )(100.12g/mol KCHCO 3 ) = 5.01 g KHCO 3 (a) Prepare 0.500 L of a 0.100 M aqueous solution of KHCO 3 (MW = 100.12 g) requires 5.01 g of KHCO 3 dissolved and diluted to 0.500 L (a) Prepare 0.500 L of a 0.100 M aqueous solution of KHCO 3 (MW = 100.12 g) requires 5.01 g of KHCO 3 dissolved and diluted to 0.500 L (b) Use this solution as a stock solution to prepare a final solution of 0.0400 M concentration. What is the final volume of this solution? (b) Use this solution as a stock solution to prepare a final solution of 0.0400 M concentration. What is the final volume of this solution? Since n = MV, M 1 V 1 = M 2 V 2 so, (0.100 M)(0.500 L) = n = (0.0400 M)( V 2 ) or M1V 1 (0.100M )(0.500L) V2 = = = 1. 25L M (0.0400M ) 2
Consider a solution prepared by dissolving 22.4 g MgCl 2 in 0.200 L of water. Assume the density of water is 1.000 g/cm 3 and the density of the solution is 1.089 g/cm 3. Calculate mole fraction molarity molality A 9.386 M solution of H 2 SO 4 has a density of 1.509 g/cm 3. Calculate molality % by mass mole fraction of H 2 SO 4 Week 3 Sections 13.3-13.5 13.3 Factors Affecting Solubility Solute-Solvent Interactions Pressure Effects Temperature Effects 13.4 Ways of Expressing Concentration Mass Percentage, ppm, and ppb Mole Fraction, Molarity, and Molality Conversion of Concentration Units 13.5 Colligative Properties Lowering the Vapor Pressure Boiling-Point Elevation Freezing-Point Depression Osmosis Determination of Molar Mass Colligative Properties Solution properties that depend only on the total # of particles present. Vapor Pressure Boiling Point Freezing Point Osmotic Pressure [13.6 Colloids] Note that VP of solution is lower than that of pure solvent.
Raoult s Law P A = X A P A o Vapor Pressure lowering P A = vapor pressure over solution X A = mole fraction of component A (solvent) P Ao = vapor pressure of pure component A (solvent) also P A = (1 X B ) P A o where X B = mol fraction of B (solute) At 25 o C, the vapor pressure of benzene is 0.1252 atm, i.e. P Ao = 0.1252 atm. If 6.40 g of naphthalene (C 10 H 8, 128.17 g/mol) is dissolved in 78.0 g of benzene (78.0 g/mol), calculate the vapor pressure of benzene over the solution At 25 o C, the vapor pressure of benzene is 0.1252 atm, i.e. P Ao = 0.1252 atm. If 6.40 g of naphthalene (C 10 H 8, 128.17 g/mol) is dissolved in 78.0 g of benzene (78.0 g/mol), calculate the vapor pressure of benzene over the solution At 25 o C, the vapor pressure of benzene is 0.1252 atm, i.e. P Ao = 0.1252 atm. If 6.40 g of naphthalene (C 10 H 8, 128.17 g/mol) is dissolved in 78.0 g of benzene (78.0 g/mol), calculate the vapor pressure of benzene over the solution Raoult s Law says P A = X A P A o Raoult s Law says P A = X A P A o therefore we need X A therefore we need X A let A = benzene n A =? n B =? n A = 78.0 g benzene (1 mol/78.0 g) = 1.00 mol benzene n B = 6.40 g naph. (1 mol/128.17 g) = 0.0499 mol naph. and P A = P benzene = (1.00/1.0499)(0.125) = 0.1193 atm note: P = P A P Ao = X A P Ao P Ao = P Ao (X a -1) or P = -X B P A o
Boiling Point Elevation and Freezing Point Depression Boiling Point Elevation T b = T b T b = K b m where m is the molal concentration Freezing Point Depression T f = T f T f = - K f m where m is the molal concentration. [note the definition and the negative sign!!!] for H 2 O, K b = 0.052 o C/m and K f = 1.86 o C/m Consider a water solution which has 0.500 mol of sucrose in 1.000 kg of water. Therefore it has a concentration of 0.500 molal or 0.500 mol/kg. (a) When 5.50 g of biphenyl (C 12 H 10, 154.2 g/mol) is dissolved in 100 g benzene (C 6 H 6, 78.0 g/mol), the BP increases by 0.903 o C. Calculate K b for benzene. recall K b = 0.52 o C/m and K f = 1.86 o C/m What is the boiling point and freezing point of this solution?
(a) When 5.50 g of biphenyl (C 12 H 10, 154.2 g/mol) is dissolved in 100 g benzene (C 6 H 6, 78.0 g/mol), the BP increases by 0.903 o C. Calculate K b for benzene. K b = 2.53 K kg/mol (b) When 6.30 g of an unknown hydrocarbon is dissolved in 150.0 g of benzene, the BP of the solution increases by 0.597 o C. A sample of sea water contains the following in 1.000 L of solution. Estimate the freezing point of this solution. Na + = 4.58 mol Cl- = 0.533 mol Mg 2+ = 0.052 mol SO 2-4 = 0.028 mol Ca 2+ = 0.010 mol HCO - 3 = 0.002 mol K + = 0.010 Br - = 0.001 mol neutral species = 0.001 mol Sum of species = 1.095 mol What is the MW of the unknown substance? Consider Exercise 13.9 List the following aqueous solutions in increasing order of their expected freezing points. 0.050 m CaCl 2 0.15 m NaCl 0.10 m HCl 0.050 m HOAc 0.10 m C12H 22 O 11 Consider Exercise 13.9 List the following aqueous solutions in increasing order of their expected freezing points. 0.050 m CaCl 2 x 3 = 0.150 0.15 m NaCl x 2 = 0.30 0.10 m HCl x 2 = 0.20 0.050 m HOAc x 1 = 0.050 0.10 m C 12 H 22 O 11 x 1 = 0.10 These calculations assume total dissociation of the salts and zero dissociation of the last two. This effect of the dissociation of electrolytes is usually taken into account through the van Hoff i factor. T b = i K b m where it is defined as Osmosis (get started) T f(actual) K f m effective m effective i = -------------- = ------------- = ------------ T f(ideal) K f m ideal m ideal In real systems, these i factors are NOT integers, but rather fractions whose values depend on concentration.
Osmotic Pressure: a fascinating behavior. Yet it is the result of a very simple tendency to equalize the concentrations of solutions. The critical part is the membrane!!! Osmotic Pressure or π = (n/v) R T π V = n R T or π = M R T π = ρ g h, where ρ = density of solution g = 9.807 m s -2 h = height of column π = ρ g h, where ρ = density of solution g = 9.807 m s -2 h = height of column If h = 0.17 m of a dilute aqueous soln with ρ = 1.00 g/cm 3 π = [(1.00g/cm 3 )(10-3 kg/g)(106 cm 3 /m 3 )](9.807 m/s 2 )(0.17 m) = 1.7 x 103 kg m -1 s -1 = 1.7 x 10 3 Pa A chemist dissolves 2.00 g of protein in 0.100 L of water. The observed osmotic pressure is 0.021 atm at 25 o C. What is the MW of the protein? or = (1.7 x 10 3 Pa) / (1.013 x 10 5 Pa/atm) = 0.016 atm
Fractional Distillation How it s done Real solutions do not behave ideally!