Solutions and Their Properties

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1 Chapter 11 Solutions and Their Properties Solutions: Definitions A solution is a homogeneous mixture. A solution is composed of a solute dissolved in a solvent. When two compounds make a solution, the solute is the lesser quantity, and the solvent is the greater quantity. Solutions can exist in all three physical states Chapter

2 Solution Formation There are three types interactions among particles in solutions: Solvent solvent Solute solute Solute solvent Like Dissolves Like Chapter 11 3 Miscible & Immiscible Two liquids that completely dissolve in each other are miscible liquids. Two liquids that are not miscible in each other are immiscible liquids. Polar water and nonpolar oil are immiscible liquids and do not mix to form a solution. Chapter

3 Polar & Nonpolar Solvents A liquid composed of polar molecules is a polar solvent. Water and ethanol are polar solvents. A liquid composed of nonpolar molecules is a nonpolar solvent. Hexane is a non-polar solvent Chapter 11 5 Formation of Solutions When a soluble crystal is placed into a solvent, it begins to dissolve. When a sugar crystal is placed in water, the water molecules attack the crystal and begin pulling part of it away and into solution. The sugar molecules are held within a cluster of water molecules called a solvent cage. Chapter

4 Dissolving of Ionic Compounds In an ionic compound, the water molecules pull individual ions off of the crystal. When a sodium chloride crystal is placed in water, the water molecules attack the edge of the crystal. The anions (Cl - ) are surrounded by the positively charged hydrogens on water. The cations (Na + ) are surrounded by the negatively charged oxygen on water. This process is called solvation (hydration when the solvent is water), which diminished the attraction of the Na + for the Cl - Chapter 11 7 Energetics of Solution Formation: Introduction to Entropy We have stated that chemical and physical processes occur spontaneously only if they go downhill energetically (give off energy) so that the final state is more stable and lower in energy than the initial state. However, we know that reactions that require energy (endothermic reactions) will also occur. This is possible because energy is more than just heat. There are other forms of energy present in reactions that allow it to be spontaneous despite absorbing heat. Another factor that effects spontaneity of a reaction is the molecular disorder of the reaction. The amount of molecular disorder (or randomness) in a system is called the system s Entropy (S) Chapter

5 Entropy Entropy has units of J/K (Joules per Kelvin). S = S final S initial Positive value of S indicates increased disorder. Negative value of S indicates decreased disorder. Second Law of Thermodynamics: Reactions proceed in the direction that increases the entropy of the system plus surroundings. Chapter 11 9 Entropy and Enthalpy To decide whether a process is spontaneous, both enthalpy and entropy changes must be considered: Spontaneous process: Decrease in enthalpy ( H). Increase in entropy (+ S). Nonspontaneous process: Increase in enthalpy (+ H). Decrease in entropy ( S). Chapter

6 Introduction to Gibbs Free Energy Gibbs Free Energy Change ( G) weighs the relative contributions of enthalpy and entropy to the overall spontaneity of a process. G = H T S G < 0 G = 0 G > 0 Spontaneous Equilibrium Non-spontaneous Chapter Gibbs Free Energy G = H T S Chapter

7 Energetics of Solution Formation ( S soln ) Chapter Energetics of Solution Formation ( H soln ) Exothermic H soln The solute solvent interactions are stronger than solute solute or solvent solvent. Favorable process Chapter

8 Energetics of Solution Formation ( H soln ) The solute solvent interactions are weaker than solute solute or solvent solvent. Endothermic H soln Unfavorable process. Chapter Formation of Solutions 1. Predict the relative solubilities in the following cases: (a) Br 2 in benzene (C 6 H 6 ) and in water, (b) KCl in carbon tetrachloride and in liquid ammonia, (c) urea (NH 2 ) 2 CO in carbon disulfide and in water. 2. Is iodine (I 2 ) more soluble in water or in carbon disulfide (CS 2 )? 3. Which would have the largest (most negative) hydration energy and which should have the smallest? Al 3+, Mg 2+, Na + Chapter

9 Solubility The solubility of a compound is used to describe the maximum amount of solute that can dissolve in a given amount of solvent. Many factors affect a solute s solubility: Type of solute Type of solvent Temperature Pressure (for solutions with gases) There are three terms we use to describe solubility: Saturated Unsaturated Supersaturated Chapter Saturation of Solutions A solution containing exactly the maximum amount of solute at a given temperature is a saturated solution. A solution that contains less than the maximum amount of solute is an unsaturated solution. Under certain conditions, it is possible to exceed the maximum solubility of a compound. A solution with greater than the maximum amount of solute is a supersaturated solution. dissolves Solid solute Saturate Solution crystallizes Chapter

10 Concentration of Solutions The concentration of a solution tells us how much solute is dissolved in a given quantity of solution or solvent. We often hear imprecise terms such as a dilute solution or a concentrated solution. Four precise ways to express the concentration of a solution: Mass percent (m %) Mole Fraction (X) Molality (m) Molarity (M) (See Chapter 3) Chapter Mass Percent Concentration Mass percent concentration comparea the amount (g) of solute to the amount of solution (g). The mass percent (m %) concentration is expressed as the mass of solute dissolved in 100 g of solution. mass of solute mass of solution 100% = m % g solute g solute + g solvent 100% = m % Chapter

11 Mass Percent Concentration: Parts per Million (ppm) Parts per million (ppm) = Mass of component 6 x 10 Total mass of solution = % mass x 10 4 One ppm gives 1 gram of solute per 1,000,000 g or one mg per kg of solution. For dilute aqueous solutions this is about 1 mg per liter of solution. What is a Part per Billion (ppb)? Chapter Other Concentration Units Mole Fraction (X): X A = Moles of A Total number of moles Molarity (M): Molarity = Moles of solute Liters of SOLUTION Molality (m): Moles of solute Molality = Kilograms of SOLVENT Chapter

12 Working Concentration Problems When working these types of problems you MUST first identify the parts of your solution. Make a list of the solute, solvent and solution with any numerical values as given in the problem. This will make your preparation of any conversion factors much easier! For Example: A student prepares a g of a solution of 5.62 g of MgCl 2 dissolved in water. Identify the solute, solvent and solution Chapter Working Concentration Problems A sample of g of potassium chloride (KCl) is dissolved in 54.6 g of water. What is the percent by mass of KCl in this solution? An aqueous solution is 5.50% H 2 SO 4. How many moles of sulfuric acid (MM = g/mol) are dissolved in g of the solution? Chapter

13 Working Concentration Problems Molality from Mass: Calculate the molality of a sulfuric acid solution containing 24.4 g of sulfuric acid in 198 g of water. The molar mass of sulfuric acid is g. Molality from Molarity: Calculate the molality of a 5.86 M ethanol (C 2 H 5 OH) solution whose density is g/ml. Chapter Working Concentration Problems Mole Fraction from Molality: An aqueous solution is m in glucose (MM = g/mol). What is the mole fraction of the glucose? Mass from Molality: What mass (in grams) of a m aqueous solution of urea [(NH 2 ) 2 CO, MM = 60.1 g/mol] would you use to obtain mole of urea? Chapter

14 Rate of Dissolving There are three ways we can speed up the rate of dissolving for a solid compound. Heating the solution: This increases the kinetic energy of the solvent and the solute is attacked faster by the solvent molecules. Stirring the solution: This increases the interaction between solvent and solute molecules. Grinding the solid solute: There is more surface area for the solvent to attack. Chapter Solubility and Temperature For most solids, solubility increases as the temperature increases. For most gases, solubility in a liquid solution decreases as the temperature increases. Chapter

15 Solubility and Pressure Henry s Law: The solubility of a gas (c) is directly proportional to the pressure (P) of the gas over the solution. c = kp Chapter Solubility and Pressure Calculate the molar concentration of O 2 in water at 25 C for a partial pressure of 0.22 atm. The Henry s law constant for O 2 is 3.5 x 10 4 mol/(l atm). The solubility of CO 2 in water is 3.2 x 10 2 M at 25 C and 1 atm pressure. What is the Henry s law constant for CO 2 in mol/(l atm)? Chapter

16 Colligative Properties Colligative Properties: Depend only on the number of solute particles in solution. These affect properties of the solvent. There are four main colligative properties: Vapor pressure lowering Freezing point depression Boiling point elevation Osmotic pressure Chapter Colligative Properties: Vapor Pressure Vapor Pressure is the pressure exerted on the surface of a liquid by the vapor above it. When a nonvolatile solute (i.e. no vapor pressure of its own) is mixed with a solvent, the solute molecules displace solvent molecules at the surface of the solution. As a result, the vapor pressure decreases since fewer gas molecules are needed to equalize the escape rate and capture rates at the liquid surface. Chapter

17 Colligative Properties: Vapor Pressure Raoult s Law describes the relationship between the vapor pressure of the solution (P soln ) and the vapor pressure of the solvent (P solv ): P soln = P solv X solv The change in the vapor pressure ( P soln ) is dependent on the amount of solute that has been added to the solvent: P soln = P solv X solute P soln = P solv P soln Chapter Colligative Properties: Vapor Pressure The vapor pressure of a glucose (C 6 H 12 O 6 ) solution is mm Hg at 20 C, while that of pure water is mm Hg at the same temperature. Estimate the molality of the solution. How many grams of NaBr must be added to 250 g of water to lower the vapor pressure by 1.30 mm Hg at 40 C? The vapor pressure of water at 40 C is 55.3 mm Hg. Chapter

18 Colligative Properties: Vapor Pressure When you mix two volatile liquids, the overall vapor pressure (P tot ) is the sum of the vapor pressure of the individual components (Dalton s Law!) P tot = P A + P B The individual vapor pressures (P A and P B ) are determined by Raoult s Law: P tot = P A + P B = (P A X A ) + (P B X B ) Chapter Colligative Properties: Vapor Pressure Two miscible liquids, A and B, have vapor pressures of 250 mm Hg and 450 mm Hg, respectively. They were mixed in equal molar amounts. What is the total vapor pressure of the mixture and what are their mole fractions in the vapor phase? What is the vapor pressure (in mm Hg) of a solution prepared by dissolving 25.0 g of ethanol (C 2 H 5 OH) in g of water at 25.0 C? The vapor pressure of pure water is 23.8 mm Hg and the vapor pressure of ethanol is 61.2 mm Hg at 25.0 C. Chapter

19 Colligative Properties: Boiling Point Elevation The vapor pressure decreases when a solution is formed. What do you think should happen to the boiling point? T b = T b soln Tb solv T b = K b m Chapter K b is the molal boiling-point elevation constant. Colligative Properties: Freezing Point Depression The vapor pressure decreases when a solution is formed. What do you think should happen to the freezing point? T f = T f solv Tf soln T f = K f m K f is the molal Freezing-point depression constant. Chapter

20 Boiling Point Elevation & Freezing Point Depression How many grams of ethylene glycol antifreeze, CH 2 (OH)CH 2 (OH), must you dissolve in one liter of water to get a freezing point of 20.0 C. The molar mass of ethylene glycol is g. For water, K f = 1.86 ( C kg)/mol. What will be the boiling point? Chapter The van t Hoff Factor (i) For ionic compounds, dissolving results in ions in solution. The van t Hoff Factor (i) generally equals the number of ions produced from each molecule of a compound upon dissolving. Use the subscripts to determine the number of ions. i = 1 for CH 3 OH i = 3 for CaCl 2 i = 2 for NaCl i = 5 for Ca 3 (PO 4 ) 2 However, this dissolving is not always 100% meaning that you cannot determine the amount of each ion in solution merely from the balanced equation. Therefore, the van t Hoff Factor has also been determined experimentally for many ionic compounds. Chapter

21 The van t Hoff Factor (i) Whenever you are working a problem that involves an ionic compound, you MUST include the van t Hoff factor in your calculations: T b T f = i K b m = i K f m P soln = i P solv X solute Chapter What is the molality of an aqueous solution of KBr whose freezing point is 2.95 C? K f for water is 1.86 ( C kg)/mol. Boiling Point Elevation & Freezing Point Depression What is the freezing point (in C) of a solution prepared by dissolving 7.40 g of K 2 SO 4 in 110 g of water? The value of K f for water is 1.86 ( C kg)/mol. Chapter

22 Boiling Point Elevation & Freezing Point Depression A 7.85 g sample of a compound with the empirical formula C 5 H 4 is dissolved in 301 g of benzene. The freezing point of the solution is 1.05 C below that of pure benzene. What are the molar mass and molecular formula of this compound? Chapter Osmosis and Osmotic Pressure (Π) Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. Osmotic pressure (Π) is the pressure required to stop osmosis. Π = i (MM) R T R = (L atm)/(mol K) Chapter

23 Osmosis and Osmotic Pressure (Π) Isotonic: Solutions have equal concentration of solute, and so equal osmotic pressure. Hypertonic: Solution with higher concentration of solute. Hypotonic: Solution with lower concentration of solute. Chapter Osmosis and Osmotic Pressure (Π) The average osmotic pressure of seawater is about 30.0 atm at 25 C. Calculate the molar concentration of an aqueous solution of urea [(NH 2 ) 2 CO] that is isotonic with seawater. What is the osmotic pressure (in atm) of a M sucrose solution at 16 C? Chapter

24 Osmosis and Osmotic Pressure (Π) A 202 ml benzene solution containing 2.47 g of an organic polymer has an osmotic pressure of 8.63 mm Hg at 21 C. Calculate the molar mass of the polymer. What is the molar mass of sucrose if a solution of g of sucrose in ml of water has an osmotic pressure of 149 mm Hg at 298 K? Chapter Desalination Uses of Colligative Properties Chapter

25 Uses of Colligative Properties Fractional Distillation is the separation of volatile liquid mixtures into fractions of different composition. Chapter Uses of Colligative Properties Fractional distillation can be represented on a phase diagram by plotting temperature against composition. Chapter

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