Two Ways to Form Solutions. Role of Disorder in Solutions 2/27/2012. Types of Reactions

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1 Role of Disorder in Solutions Disorder (Entropy) is a factor Solutions mix to form maximum disorder Two Ways to Form Solutions 1. Physical Dissolving (Solvation) NaCl(s) Na + (aq) + Cl - (aq) C 12 H 22 O 11 (s) C 12 H 22 O 11 (aq) Particles are surrounded by solvent molecules Can evaporate water/solvent to get original compound back Types of Reactions 2. Chemical reaction Ni(s) + 2HCl(aq) NiCl 2 (aq) + H 2 (g) Evaporating solvent gives the products Solubility Maximum amount of a solute that can dissolve in 100 ml of a solution Ex: NaCl 35.7 g/100ml Saturated solution Contains the max. amount of solute with some undissolved solid, Unsaturated more solute will dissolve. 1

2 Supersaturated More than the max is dissolved by heating and slowly cooling. Like Dissolves Like: Miscibility Polar dissolves polar (dipole-dipole Forces) and ionic (ion:dipole) Water and Ammonia Non-Polar dissolves non-polar (London Forces) Soap and grease Would acetone (shown below) dissolve in water? :O: CH 3 CCH 3 Acetone Using you knowledge of like dissolves like, explain the following trends in solubility. Alcohol CH 3 OH CH 3 CH 2 OH CH 3 CH 2 CH 2 OH Solubility in H 2 O (mol/100 g H 2 O at 20 o C) CH 3 CH 2 CH 2 CH 2 OH 0.11 CH 3 CH 2 CH 2 CH 2 CH 2 OH CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 OH CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 OH Pressure Effects Solubilty of a gas increases with pressure of gas over the liquid (soda bottle) Henry s Law S gas = kp gas 2

3 Henry s Law The Henry s law constant for CO 2 is mol/latm. a. Calculate the concentration of CO 2 in a soda bottle pressurized to 4.00 atm of CO 2. b. After the bottle has been opened, the concentration drops to 9.3 X 10-6 M. Calculate the partial pressure of CO 2 over the soda. Temperature Effects Solubility of most solids increases with temperature Solubility of most gases decreases with temperature (warm soda) Warm water is deoxygenated Problem with thermal pollution of lakes Ways of Expressing Concentration Mass % = mass of compound in soln X 100 total mass of soln Concentration: Ex g of C 6 H 12 O 6 is dissolved in kg of water. Calculate the mass percentage. Parts Per Million mass % = 13.5 g X 100 = 11.9% (100 g g) ppm = mass of component X 10 6 total mass of soln Concentration: Ex 2 A 2.5 g sample of groundwater is found to contain 5.4 g of Zn 2+. What is the concentration of the Zn 2+ ion in ppm. Concentration: Ex 2 A 2.5 g sample of groundwater is found to contain 5.4 g of Zn 2+. What is the concentration of the Zn 2+ ion in ppm. 5.4 g 1X10-6 g = 5.4 X 10-6 g 1 g ppm = mass of component X 10 6 total mass of soln ppm = 5.4 X 10-6 g X 10 6 = 2.2 ppm 2.5g 3

4 Concentration: Ex 3 Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. Concentration: Ex 4 Bleach is 3.62 % NaOCl. What mass of NaOCl is contained in 2500 g of bleach? ANS: 2.91 % ANS: 90.5 g NaOCl Mole Fraction Mole Fraction X = moles of component total moles of all components What is the mole fraction of HCl if 36.5 grams is dissolved in 144 grams of water? ANS: Molality Molality = moles of solute kilograms of solvent Why not use Molarity? Molarity varies with temperature Total volume of a solution changes with temperature (liquid expands) Mass does not change with temperature Molality: Ex 1 A solution is made by dissolving 4.35 grams of C 6 H 12 O 6 in 25.0 ml of water. Calculate the molality of the glucose. Molality: Ex 2 Calculate the molality of a solution made by dissolving 36.5g C 10 H 8 in 425 grams of toluene (solvent) g 1 mol = mol g Molality = mol = m kg ANS: m 4

5 Molality: Ex 3 A solution of HCl contains 36 percent HCl by mass. Calculate the mole fraction and molality of HCl. 36 g HCl 1 mol HCl = 0.99 mol HCl 36.5 g HCl 64 g H2O 1 mol H 2 O = 3.6 mol HCl 18 g H 2 O Pretend 100 g 36 g HCl 64 g H 2 O X HCl = 0.99 mol HCl = mol + 3.6mol Molality = 0.99 mol HCl kg H2O = 15 m Molality: Ex 4 A commercial bleach solution contains 3.62 percent NaOCl by mass. Calculate the mole fraction and molality of NaOCl. Molality: Ex 5 The density of a solution of 5.0 g of toluene (C 7 H 8 ) and 225 g of benzene (C 6 H 6 ) is g/ml. Calculate the molality and molarity of the toluene. ANS: X NaOCl = , m Molality: Ex 5 The density of a solution of 5.0 g of toluene (C 7 H 8 ) and 225 g of benzene (C 6 H 6 ) is g/ml. Calculate the molality and molarity of the solution. Molality 5.0 g 1 mol = mol 92.0 g Molarity D = mass/v V = mass/d V = 230 g = 263 ml g/ml M = mol = 0.21 M L m = mol/ kg = 0.24 m 5

6 Molality: Ex 6 A solution containing equal masses of glycerol (C 3 H 8 O 3 ) and water has a density of 1.10 g/ml. Calculate a) molality (10.9 m) b) mole fraction (X C3H8O3 = 0.163) c) molarity of glycerol in the solution (5.97 M) Colligative Properties: Vapor Pressure Lowering Non-volatile solutes lower the vapor pressure of the solvent Raoult s law P A = X A P o A P A = Vapor pressure X A = Mole fraction of solvent P o A = Pressure of pure solvent Raoult s Law: Ex 1 What is the vapor pressure of a solution made by adding 50.0 ml of glycerin (C 3 H 8 O 3 ) to ml of water? The density of glycerin is 1.26 g/ml and the vapor pressure of pure water is 23.8 torr. X H2O = 27.8 mol = (27.8 mol mol) P A = X A P o A P A = (0.976)(23.8 torr) = 23.2 torr Mass C3H8O3 = (50.0 ml)(1.26 g/ml ) = 63.0 g Moles C3H8O3 = 63.0 g/92.1 g/mol = mol Moles H2O = g/18 g/mol = 27.8 mol Raoult s Law: Ex 2 The vapor pressure of water at 110 o C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1 atm at 110 o C. What is the mole fraction of ethylene glycol in the solution? ANS: Colligative Properties: Boiling Point Elevation Non-volatile solute raises the boiling point of a solution Shifts the phase diagram The pressure of the solution reaches atmospheric pressure at a higher temp. T b = ik b m 6

7 Colligative Properties: Freezing Point Depression Solutions freeze at a lower temperature than pure solvent Salt water freezes lower (-2 o C) than distilled water (0 o C) T f = ik f m The more ions produced, the greater the freezing point depression or boiling point elevation C 12 H 22 O 11 (i=1) NaCl = Produces two ions (i=2) CaCl 2 = Produces three ions (i=3) i = Van t Hoff factor m = molality of the nonvolatile solute Pure Solvent Boiling Point Solution Boiling Point Colligative: Ex 1 Ethylene Glycol, C 2 H 6 O 2, is used in antifreeze. What will be the freezing and boiling point of a 25.0 mass percent solution of ethylene glycol and water? Freezing Point Freezing Point Pretend 100 grams of solution 25 grams of C 2 H 6 O 2 75 grams of H 2 O (0.075 kg) 25 grams of C 2 H 6 O 2 = moles Colligative: Ex 2 m = moles = 5.37 m kg H 2 O T b = ik b m = (1)(0.52 o o C/m)(5.37 m) = 2.8 o C T f = ik f m = (1)( 1.86 o C/m)(5.37 m) =10.0 o C Calculate the freezing point of a solution containing kg of CHCl 3 and 42.0 g of C 10 H 18 O. K f for CHCl 3 is 4.68 o C/m and the normal freezing point is o C. Boiling Point = o C Freezing Point = o C ANS: o C 7

8 Colligative: Ex 3 Rank the following aqueous solutions in order of their expected freezing points: m CaCl m NaCl 0.10 m HCl m HC 2 H 3 O 2 (acetic acid) 0.10 m C 12 H 22 O 11 (sugar) m CaCl 2 (0.15 m in particles) 0.15 m NaCl (0.30 m in particles) 0.10 m HCl (0.20 m in particles) m HC 2 H 3 O 2 (just above 0.05 m) 0.10 m C 12 H 22 O 11 (0.10 m in particles) Lowest FP Highest FP NaCl < HCl < CaCl 2 < C 12 H 22 O 11 < HC 2 H 3 O 2 Colligative: Ex 4 Rank the following in order of the increase in boiling point that they will produce in 1 kg of water 1 mol Co(NO 3 ) 2 2 mol KCl 3 mol C 2 H 6 O 2 (a very, very weak electrolyte(acidic)) 1 mol Co(NO 3 ) 2 (3 mol particles) 2 mol KCl (4 mol of particles) 3 mol C 2 H 6 O 2 (3+ mol of particles) Lowest BP Highest BP Co(NO 3 ) 2 < C 2 H 6 O 2 < KCl Freezing Pt Depression: Ex 5 What would be the molality of salt water if it freezes at 0 o F? K f = 1.86 o C/m. Colligative Properties: Osmotic Pressure Osmosis movement of solvent from high concentration to low concentration semipermeable membrane allows to passage of some particles but not others Cucumber Skin cell after in salt water soaking in a tub ANS: 4.78 m 8

9 Note that solvent moves both ways Solute too large to pass through membrane Net movement is to try to dilute the side with solutes Osmotic Pressure ( ) pressure required to prevent osmosis PV = inrt V = inrt = inrt V = imrt M = molarity Osmotic Pressure: Ex 1 The average osmotic pressure of blood is 7.7 atm at 25 o C. What concentration of glucose will be isotonic with blood? Osmotic Pressure: Ex 2 What is the osmotic pressure at 20 o C of a M sucrose, C 12 H 22 O 11, solution? Express your answer both in atmosphere and in torr. (0.31 M) ANS; atm, 37 torr Molar Mass: Ex 1 A solution of an unknown nonelectrolyte was prepared by dissolving g in 40.0 g of CCl 4. The boiling point of the resulting solution was o C higher than that of the pure solvent. K b for CCl 4 is 5.02 o C/m. Calculate the molar mass of the unknown. T b = ik b m m = T b /ik b m = (0.357 o C)/(1 X 5.02 o C/m) = m m = mol of solute kilograms of solvent mol solute = (m)(kg of solvent) mol solute = ( m)( kg) = mol Molar Mass = g = 88.0 g/mol mol 9

10 Molar Mass: Ex 2 Camphor, C 10 H 16 O, melts at o C and has a K f of 40.0 o C/m. When g of an unknown substance is dissolved in g of liquid camphor, the freezing point is o C. What is the molar mass of the solute? Molar Mass: Ex 3 A solution contains 3.50 mg of protein dissolved in water to form 5.00 ml of a solution. The osmotic pressure at 25 o C was found to be 1.54 torr. Calculate the molar mass of the protein torr 1 atm = atm 760 torr ANS: 110 g/mol = imrt M = /irt M = atm = 8.28X10-5 M (1)( L-atm/mol-K)(298) M = moles liter moles = (M)(liters) = (8.28X10-5 M)( L) moles = 4.14 X 10-7 mol Molar Mass: Ex 4 A 2.05 g sample of a plastic was dissolved in enough toluene to form 100 ml of solution. The osmotic pressure of this solution is 1.21 kpa at 25 o C. Calculate the molar mass of the plastic. (1 atm = kpa). Molar mass = 3.50 X 10-3 g = 8454 g/mol 4.14 X 10-7 mol ANS: 42,000 g/mol Solutions/ Homogeneous Ions/Molecular size solute particles Never separate Colloids Colloids Medium particles (10 to 2000 Å) Suspensions/ Heterogeneous Mixture Larger particles (like dirt in water) 2 or more separate phases Separates quickly Examples Fog Smoke Whipped Cream Milk Tyndall effect scattering of light 10

11 Stabilization of Colloids Hydrophobic/hydrophilic imf Biomolecules Emulsifying agents Soap Sodium stearate(used to digest fats) 6. Larger Noble gases have greater London Forces (greater dipole: induced dipole forces) 16.a) Ion:Dipole b) Dipole: induced dipole c) Hydrogen bonding (weakest) b < c < a (strongest) 30.a) glucose (OH s allow h-bonding) b) sodium propionate (has an ion) c) HCl (small and polar) 34.S = kp k = S/P = 1.38 X 10-3 M/0.21 atm k =6.57 X 10-3 M/atm Partial Pressure of O 2 at the higher elevation: 650 torr = atm P O2 = (0.855 atm)(0.21) P O2 = atm S = kp S=(6.57 X 10-3 )(0.180 atm) S = 1.18 X 10-3 M 36a) 7.2% I 2 b) 7.9 ppm Sr 2+ 38a) b) 5.66% c) m 40a) M b) M c) M 42a) 4.34 m b) 3.1 g S8 44a) 27.7% b) c) 2.18 m d) 1.92 M 46a) b) m c) M 48a) mol b) 6.25X10-5 mol c) mol 50a) 21.8 g b) 7.7 g/112.3g c) 209 g d) 11 ml of 6.0 M HCl 52) 15 M NH 3 11

12 62.a) 222 torr b) 150 g 64. a) 0.75 b) % sucrose < 10% glucose < 10% NaNO m phenol < KBr = m glycerin 70.a) , 78.8 b) -78, 72.4 c) -9.3, o C atm g/mol g/mol Write Net Ionic Equations for: MgCO 3 (s) + HNO 3 (aq) H 2 SO 4 (aq) + 2KOH(aq) NaHCO 3 (aq) + HCl(aq) A solution of nickel(ii)sulfate is stored in a zinc coated bucket. a. Write the net ionic reaction that occurs. b. Suppose the nickel(ii)sulfate was stored in a copper bucket. Would this be a better choice? c. Write the net ionic reaction that occurs between nickel(ii)sulfate and barium nitrate. 1. Lots of ice, little water 2. Stir and take temperature (~2 min) 3. Add medicine cup of rock salt 4. Stir and take temperature (~2 min) 5. Clean Up 1. Rinse Foam cup, medicine cup, and thermometer 2. Place in drying rack 12

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