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Properties of Solutions Terminology Solvation Compositions Solubility Rules Heats of Solution Structure/Intermolecular Forces Temperature Pressure Henry's Law Vapor Pressures Raoult's Law Colligative Properties B. pt. Elevation Fr. pt. Depression Osomotic Pressure Terminology • Solution = A homogeneous mixture. • Solute = a substance dissolved in a solvent to form a solution; usually the smaller portion. • Solvent = The dissolving medium of a solution; usually the greater portion. • Solubility = Amount of substance dissolved. • Dilute , Concentrated , Saturated . The Solution Process Figure 13.2: The Solution Process Hydration or Solvation Figure 13.3: Three Steps of Solution Formation The Solution Process Solution Formation, Spontaneity, and Disorder • If the process leads to a greater state of disorder, then the process is spontaneous. • Example: a mixture of CCl4 and C6H14 is less ordered than the two separate liquids. Therefore, they spontaneously mix even though Hsoln is very close to zero. • There are solutions that form by physical processes and those by chemical processes. Figure 13.6: The Solution Process Solution Formation, Spontaneity, and Disorder Factors Affecting Solubility Solute-Solvent Interaction • Miscible liquids: mix in any proportions. • Immiscible liquids: do not mix. [ Oil and Vinegar ] • Generalization: “like dissolves (in) like”. • Polar liquids tend to dissolve in polar solvents. Factors Affecting Solubility Solute-Solvent Interaction Ways of Expressing Concentration Mass Percentage • All methods involve quantifying amount of solute per amount of solvent (or solution). • Generally amounts or measures are masses, moles or liters. mass of component in solution mass % of component 100 total mass of solution Ways of Expressing Concentration • Mole Fraction, Molarity, and Molality moles of component in solution Mole fraction of component total moles of solution moles solute Molarity liters of solution moles solute Molality, m kg of solvent Solution Compositions • s = solute ; A = solvent; V = Tot. Vol. of solution. • Weight %: ws ws % x 100 ws wA • Mole Fraction: ns s ns nA • Molarity: ns Ms V • Molality: ns ms kg A Example of Solution Compositions • A solution is prepared by mixing 78.9 g of ethanol (C2H5OH) with 100.0 g of water to give 190.5 mL of solution. Calculate the solution compositions. • The electrolyte in automobile lead storage batteries is a 3.75 M H2SO4 solution that has a density of 1.230 g/mL. Calculate mass %, molality, and mole fraction in terms of H2SO4 . • [Hint: Assume exactly one liter of solution.] • [Answers: 29.9% , 4.35 molal, 0.0727 ] s = ethanol (solute); A = water (solvent); 78 .9 g ws % x 100 44 .1% 178 .9 g 78 .9 s ns 46 .07 1.71 0.236 ns n A 78 .9 100 .0 1.71 5.55 46 .07 18 .02 ns 1.713 mol Ms 8.99 mol L1 V 190 .5 mL ns 1.713 mol ms 17 .1 mol kg 1 kg A 0.1000 kg water Exam ple 2 (Solution Com positions) The e lectrolyte in autom obile le ad s torage batterie s is a 3.75 M HSO 4 solution that has a 2 -1 de nsity of 1.230 g m L . Calculate w eight%, m olality, and m ole fraction in te rm s of H SO 4 . 2 Hint: As sum e exactly one liter of solution in your calculations. De fine: s = H2 SO 4 A=H 2O In one liter of solution, from 3.75 mol/L solution, there are 3.75 moles of H 2 SO 4 . In one liter of solution, from d=1.230 g/mL, there are 1230 grams of solution. Weight % of H2 SO 4 Calculation: 98.07gm 3.75mol 368gm H2SO 4 1 mol 368 gm Wt % 100 Wt % 29.9 1230gm Molality of H2 SO 4 Calculation : In one liter of solution => 1230 gm soln - 368 gm H 2 SO 4 = 862 gm H 2 O n s 3.75mol kgA 0.862kg ns mol ms ms 4.35 kgA kg Mole Fraction of H2 SO 4 Calculation : 862 n A mol 18.015 ns s s 0.0727 ns nA Concentrations of Solutions In the Dilution process of a more concentrated solution: • The number of moles are the same in diluted and concentrated solutions. • So: MdiluteVdilute = moles = MconcentratedVconcentrated General Properties of Aqueous Solutions Electrolytic Properties • Three types: • Strong electrolytes, • Weak electrolytes, and • Nonelectrolytes. General Properties of Aqueous Solutions Strong and Weak Electrolytes • Strong electrolytes: completely dissociated in solution. e.g. HCl(aq) H+(aq) + Cl-(aq) • Weak electrolytes: produce small concentration of ions when dissolved. • e.g. HC2H3O2(aq) H+(aq) + C2H3O2-(aq) Precipitation Reactions Precipitation Reactions Exchange (Metathesis) Reactions • Metathesis reactions involve swapping ions in solution: AX + BY AY + BX. Precipitation Reactions Ionic Equations • Ionic equation: used to highlight reaction between ions. • Molecular equation: all species listed as molecules: AgNO3(aq) + NaI(aq) AgI(s) + NaNO3(aq) • Complete ionic equation(CIE): lists all ions: Ag+(aq) + NO3-(aq) + Na+(aq) + I-(aq) AgI(s) + Na+(aq) + NO3-(aq) • Net ionic equation: cancel spectator ions from CIE Ag+(aq) + I-(aq) AgI(s) Figure 13.14: Factors Affecting Solubility Intermolecular Forces Pressure Temperature Figure 13.14: Factors Affecting Solubility Pressure Effects Factors Affecting Solubility Pressure Effects • If Sg is the solubility of a gas, k is a constant, and Pg is the partial pressure of a gas, then Henry’s Law gives: S g k Pg • Carbonated beverages are bottled with a partial pressure of CO2 >1 atm, ( pressure inside can ~4 atm above liq). • As bottle is opened, partial pressure of CO2 decreases and solubility of CO2 decreases. • Therefore, bubbles of CO2 escape from solution. CyberChem Diving Gases Factors Affecting Solubility Temperature Effects: Solids in Liquids • Generally, as temperature increases, solubility of solids generally increases, BUT • Sometimes, solubility decreases as temperature increases (e.g. Ce2(SO4)3). Figure 13.17 Factors Affecting Solubility Temperature Effects: Gases in Liquids • Gases get less soluble as temperature increases. • Thermal pollution: if lakes get too warm, CO2 and O2 become less soluble and are not available for plants or animals. Figure 13.18 Colligative Properties • Colligative properties depend on quantity of solute molecules. (e.g. freezing point depression and boiling point elevation.) Lowering Vapor Pressure • Non-volatile solutes reduce the ability of the surface solvent molecules to escape the liquid. • Therefore, vapor pressure is lowered. • The amount of vapor pressure lowering depends on the amount of solute. Figure 11.22: Vapor Pressure Vapor Pressure on the Molecular Level Figure 11.24 Figure 11.26: Phase Diagrams Figure 11.27: Phase Diagrams The Phase Diagrams of H2O and CO2 Figure 13.20: Colligative Properties Lowering Vapor Pressure Colligative Properties Lowering Vapor Pressure • Raoult’s Law: PA A P A • Where: PA = vapor pressure with solute, • PA = vapor pressure without solute (pure solvent), and • A = mole fraction of A. Vapor Pressure Examples • Calculate the expected vapor pressure at 25oC for a solution prepared by dissolving 158.0 g of common table sugar (sucrose – MW=342.3) in 643.5 mL of water. At 25oC, the density of water is 0.9971 g/mL and the vapor pressure is 23.76 torr. PA A P A • A solution was prepared by adding 20.0 g of urea to 125 g of water at 25oC, a temperature at which pure water has a vapor pressure of 23.76 torr. The observed vapor pressure of the solution was found to be 22.67 torr. Calculate the molecular weight of urea. [Answer: 60. g/mol ] Example 1: Sugar in Water 1 s = sugar A = w ater ws 158.0 gm MW s 342.3gm mol 3 3 1 VA 643.5cm d A 0.9971gm cm PoA 23.76t orr MW A 18.015gm mol VA d A 643.5cm 0.9971gm 3 1 mol n A MW A n A 35.62mol nA 3 18.015gm cm ws 158.0gm n s n s 0.4616mol 1 mol MW s ns 342.3gm nA A A 0.9872 35.62 35.62 ns nA A 0.4616 35.62 36.08 PA A PoA PA 23.46t orr PA 0.9872( 23.76t orr) Example 2: Urea Molecular Weight • o A solution was prepared by adding 20.0 g of urea to 125 g of water at 25 C, a temp erature at which pure water has a vapor pressure of 23.76 torr. T he observed vapor p ressure of the solution was found to be 22.67 torr. Calculate the molecular weight of urea. [Answer: 60. g/mol ] s = urea A = w ater ws 20.0gm wA 125 gm PoA 23.76t orr PA 22.67t orr PA 1 PA A PoA A A 0.9541 MW A 18.015gm mol PoA nA wA A n A n A 6.939mol ns nA MW A nA solving for n has solution of: s A 1 A n A ns nA A n s nA 1 A n s 0.334mol A ws ws 1 ns MW s MW s 60gm mol MW s ns Figure 13.22 Colligative Properties Boiling-Point Elevation • Molal boiling-point-elevation constant, Kb, expresses how much Tb changes with molality, mS : Tb K b mS • Decrease in freezing point (Tf) is directly proportional to molality (Kf is the molal freezing-point-depression constant): Tf K f mS Colligative Properties Applications of Colligative Properties • A solution was made by dissolving 18.00 g of glucose in 150.0 g of water. The resulting solution was found to have a boiling point of 100.34oC. Calculate the molecular weight of glucose. Tb K b mS • How many kg of ethylene glycol (C2H6O2, MW=62.1), antifreeze, must be added to 10.0 L of water to produce a solution for use in a car’s radiator that freezes at –23.3oC? Assume that the density of water is 1.00 g/mL. [Answer: 7.78 kg (with d = 1.18 g/mL => 6.59 L)] • A solution was made by dissolving 18.00 g of glucose in 150.0 g of water. The resulting solution was found to have a boiling point of 100.34oC. Calculate the molecular weight of glucose. Tb K b mS How many kg of ethylene glycol (C2H6O2, MW=62.1), antifreeze, must be added to 10.0 L of water to produce a solution for use in a car’s radiator that freezes at –23.3oC? Assume that the density of water is 1.00 g/mL. Tf K f mS Free zing Point De pre ssion Example • How many kg of ethylene glycol (C2H6O2, M W=62.1), antifreeze, must be added to 10.0 L of water to produce a solution for use in a car’s radiator that freezes at –23.3 oC? Assume that the density of water is 1.00 g/mL. [Answer: 7.78 kg (with d = 1.18 g/mL => 6.59 L)] Let: s = ethylene glycol A=water 1 gm 1 mL 10 kg 3 1 Kf 1.86K kg mol Tf 23.3 kg_A 10.0 1 mL 10 3 L 1 gm K L 1 kg_A 10.0kg MW s 62.1gm mol ns ws Tf Kf ms Kf Kf kg_A MW s kg_A Tf MW s kg_A ws Kf 1 23.3K 62.1 gm mol 10.0 kg ws 1 ws 7.78kg 1.86K kg mol 1 ws d s 1.18gm mL Vs Vs 6.59L ds Colligative Properties Osmosis • movement of a solvent from low solute concentration to high solute concentration across a semipermeable membrane. Figure 13.23 Colligative Properties Osmosis • Osmotic pressure, , is the pressure required to stop osmosis: V n R T n R T V M S R T Application of Osmotic Pressure • To determine the molecular weight of a certain protein, 1.00 mg of it was dissolved in enough water to make 1.00 mL of solution. The osmotic pressure of this solution was found to be 1.12 torr at 25.0oC. Calculate the molecular weight of the protein. [Answer: 1.66x104 g/mol] M S R T To determine the molecular weight of a certain protein, 1.00 mg of it was dissolved in enough water to make 1.00 mL of solution. The osmotic pressure of this solution was found to be 1.12 torr at 25.0 oC. Calculate the molecular weight of the protein. M S R T Colligativ e Prope rties: Osmotic Pressure Example • To determine the molecular weight of a certain protein, 1.00 mg of it was dissolved in enough water to make 1.00 mL of solution. T he osmotic pressure of this solution was found to be 1.12 torr at 25.0 oC. Calculate the molecular weight of the protein. [Answer: 1.66x104 g/mol] Def ine: s = protein A = w ater 3 3 1 1 ws 1.0010 gm V 1.0010 L 1.12 orr t R 0.08206L at m mol K 1.12t orr 1at m 3 T ( 25.0 273.15 K ) 1.474 10 at m 760 t orr ns ws M s R T R T R T V MW s V The only unknow n is the molecular w eight of the protein ( MW s ). MW s ws R T MW s 1.0010 3 gm 0.08206L at m mol 1 K 1 298.15K V 1.47410 3 3 at m 1.0010 L 4 1 MW s 1.66 10 gm mol Colligative Properties Osmosis • Isotonic solutions: two solutions with the same separated by a semipermeable membrane. • Osmosis is spontaneous. • Red blood cells are surrounded by semipermeable membranes. Figure 13.25: Colligative Properties Osmosis Crenation: hypertonic solution Hemolysis: hypotonic solution IV(intravenous) fluids must be Isotonic. CyberChem: Desalination Colloids • Colloids are suspensions in which the suspended particles are larger than molecules but too small to drop out of the suspension due to gravity. • Particle size: 10 to 2000 Å. • Tyndall effect: ability of a Colloid to scatter light. The beam of light can be seen through the colloid. Colloids Figure 13.31: Colloids Properties of Solutions Terminology Solvation Compositions Solubility Rules Heats of Solution Structure/Intermolecular Forces Temperature Pressure Henry's Law Vapor Pressures Raoult's Law PA A PA o Colligative Properties nA A nS nA B. pt. Elevation Fr. pt. Depression Osomotic Pressure nS mS kg A Tb, f Kb, f mS M S R T MS nS V