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Solutions and Their Properties

Solutions and Their Properties. shrinks(crenates). Hemolysis(bursts). Solutions. Solution : A homogeneous mixture (  uniform in composition ) . Solvent : The major component. Solute : A minor component. Solution Formation 01.

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Solutions and Their Properties

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  1. Solutions and Their Properties shrinks(crenates) Hemolysis(bursts)

  2. Solutions Solution: A homogeneous mixture ( uniform in composition ). Solvent: The major component. Solute: A minor component. Chapter 11

  3. Solution Formation 01 • Saturated: Contains the maximum amount of solute that will dissolve in a given solvent. • Unsaturated: Contains less solute than a solvent has the capacity to dissolve. • Supersaturated: Contains more solute than would be present in a saturated solution. • Crystallization: The process in which dissolved solute comes out of the solution and forms crystals. • Saturated: Contains the maximum amount of solute that will dissolve in a given solvent. • Unsaturated: Contains less solute than a solvent has the capacity to dissolve. • Supersaturated: Contains more solute than would be present in a saturated solution. • Crystallization: The process in which dissolved solute comes out of the solution and forms crystals. Chapter 11

  4. Solution Formation 02 Chapter 11

  5. Solution Formation 02 Chapter 11

  6. How Does a Solution Form? As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them. Chapter 11

  7. How Does a Solution Form If an ionic salt is soluble in water, it is because the ion-dipole interactions are strong enough to overcome the lattice energy of the salt crystal. Chapter 11

  8. Entropy of Solution Formation 03 ∆S >0 Chapter 11

  9. Energy Changes and the Solution Process DG = DH - TDS endothermic: exothermic: +DH -DH spontaneous: nonspontaneous: -DG +DG

  10. Solution Formation 04 • Exothermic ∆Hsoln: • The solute–solvent interactions are stronger than solute–solute or solvent–solvent. Chapter 11

  11. Solution Formation 05 • Endothermic ∆Hsoln: • The solute–solvent interactions are weaker than solute–solute or solvent–solvent. Chapter 11

  12. Why Do Endothermic Processes Occur? Things do not tend to occur spontaneously (i.e., without outside intervention) unless the energy of the system is lowered. ΔG<0 Chapter 11

  13. Why Do Endothermic Processes Occur? Yet we know that in some processes, like the dissolution of NH4NO3 in water, heat is absorbed, not released. Chapter 11

  14. Enthalpy Is Only Part of the Picture The reason is that increasing the disorder or randomness (known as entropy) of a system tends to lower the energy of the system. Chapter 11

  15. Enthalpy Is Only Part of the Picture So even though enthalpy may increase, the overall energy of the system can still decrease if the system becomes more disordered. Chapter 11

  16. Terms used : 06 • Solubility: A measure of how much solute will dissolve in a solvent at a specific temperature. • Miscible: Two (or more) liquids that are completely soluble in each other in all proportions. • Solvation: The process in which an ion or a molecule is surrounded by solvent molecules arranged in a specific manner. Chapter 11

  17. Solution Formation Predict whether each of the following substances is more likely to dissolve in carbon tetrachloride (CCl4) or in water: C7H16, Na2SO4, HCl, and I2. Chapter 11

  18. Solution Formation C7H16 is a hydrocarbon, so it is molecular and nonpolar. Na2SO4, a compound containing a metal and nonmetals, is ionic; HCl, a diatomic molecule containing two nonmetals that differ in electronegativity, is polar; and I2, a diatomic molecule with atoms of equal electronegativity, is nonpolar. We would therefore predict that C7H16 and I2 would be more soluble in the nonpolar CCl4 than in polar H2O, whereas water would be the better solvent for Na2SO4 and HCl. Chapter 11

  19. Solution Formation Arrange the following substances in order of increasing solubility in water: C5H12 < C5H11 Cl < C5H11 OH < C5H10(OH)2 (in order of increasing polarity and hydrogen-bonding ability) Chapter 11

  20. Solution Formation 07 • Is iodine (I2) more soluble in water or in carbon disulfide (CS2)? • Which would have the largest (most negative) hydration energy and which should have the smallest? Al3+, Mg2+, Na+ Chapter 11

  21. Size and Charge of the ion • The smaller the ion and the greater its charge, it will be the more hydrated, therefore more hydration energy Chapter 11

  22. Concentration Units 01 Chapter 11

  23. mass of solute % by mass of solute = 100 % ´ mass of solution mass of solution = mass of solute + mass of solvent Concentration Units 01 • Concentration:The amount of solute present in a given amount of solution. • Percent by Mass (weight percent):The ratio of the mass of a solute to the mass of a solution, multiplied by 100%. Chapter 11

  24. Concentration Units 02 • Parts per Million: • Parts per million (ppm) = • = % mass x 104 • One ppm gives 1 gram of solute per 1,000,000 g of solution or one mg per kg of solution. For dilute aqueous solutions this is about 1 mg per liter of solution. Chapter 11

  25. A solution is made by dissolving 13.5 g of glucose (C6H12O6) in 0.100 kg of water. a) What is the mass percentage of solute in this solution? b) What is the ppm of solute in this solution? Problem: Chapter 11

  26. Solution Mass ppm Glucose = 11.9 x 104 Chapter 11

  27. Other Concentration Units 04 • Mole Fraction (X): • Molarity (M): • Molality (m): Chapter 11

  28. Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate mol fraction, molality, and weight % of glycol. Chapter 11

  29. Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate X, m, and % of glycol. 250. g H2O = 13.9 mol X glycol = 0.0672 Chapter 11

  30. Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate X, m, and % of glycol. Calculate molality Calculate weight % Chapter 11

  31. Effect of Temperatureon Solubility of Solids in water 01 heat + Glucose + water → dissolved Glucose Chapter 11

  32. Effect of Temperature on Solubility of Gases in Water 02 gaseous O2 + water → saturated O2 solution + heat Chapter 11

  33. The Effect of Pressure on theSolubility of Gases, Henry's Law 01 • Henry’s Law: • The solubility of a gas is proportional to the pressure of the gas over the solution. S P S = k·P Chapter 11

  34. Henry's Law

  35. The Effect of Pressure on theSolubility of Gases 02 Flash Animation - Click to Continue Chapter 11

  36. Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over the liquid at 25°C. The Henry’s law constant for CO2 in water at this temperature is 3.1  10–2 mol/L-atm. 2 PRACTICE EXERCISE Calculate the concentration of CO2 in a soft drink after the bottle is opened and equilibrates at 25°C under a CO2 partial pressure of 3.0  10–4 atm. Chapter 11

  37. Colligative Properties of Nonvolatile Solutes 01 • Colligative Properties: Depend only on the number of solute particles in solution. Solutions of electrolytes (which dissociate upon dissolving) should show greater changes than those of nonelectrolytes. These changes are : • Vapor pressure lowering • Freezing point depression • Boiling point elevation • Osmotic pressure Chapter 11

  38. Vapor Pressure Because of solute-solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase. Chapter 11

  39. Vapor Pressure Therefore, the vapor pressure of a solution is lower than that of the pure solvent. Chapter 11

  40. Calculating Change of Vapor Pressure when the solute is Nonvolatile . 03 • Raoult’s Law:P1 = x1P°1 where x1 is the solvent molefraction. • For a single solute solution, x1= 1 – x2, where x2 is the solute mole fraction. • We can obtain an expression for the change in vapor pressure of the solvent (the vapor pressure lowering). • P = P°1 – P1 = P°1 – x1 P°1 = P°1 – (1 – x2) P°1 • ∆P = x2P°1 Chapter 11

  41. Assume the solution containing 62.1 g of glycol in 250. g of water . What is the vapor pressure of water over the solution at 30 oC? (The VP of pure H2O is 31.8 mm Hg at 30 °C) Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate X, m, and % of glycol. 250. g H2O = 13.9 mol X glycol = 0.0672 Chapter 11

  42. Raoult’s Law Assume the solution containing 62.1 g of glycol in 250. g of water . What is the vapor pressure of water over the solution at 30 oC? (The VP of pure H2O is 31.8 mm Hg) Solution Xglycol = 0.0672 ( see previous slides) and so Xwater = ? Because Xglycol + Xwater = 1 Xwater = 1.000 - 0.0672 = 0.9328 Pwater = Xwater • Powater = (0.9328)(31.8 mm Hg) Pwater = 29.7 mm Hg Chapter 11

  43. Calculating Change of Vapor Pressure when the solute is Volatile . 01 • What happens if both components are volatile(have measurable vapor pressures)? • The vapor pressure has a value intermediate between the vapor pressures of the two pure liquids. • PT= PA + PB = XAP°A + XBP°B Chapter 11

  44. Vapor Pressure of Two Volatile Liquids 02 Chapter 11

  45. Boiling-Point Elevation and Freezing-Point Depression Chapter 11

  46. Boiling-Point Elevation and Freezing-Point Depression 01 • Boiling-Point Elevation (∆Tb): The boiling point of the solution (Tb) minus the boiling point of the pure solvent (T°b):∆Tb = Tb – T°b • ∆Tb is proportional to concentration in molality:∆Tb = Kbm Kb= molal boiling-point elevation constant of solvent. Chapter 11

  47. Change in Boiling Point Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the BP of the solution? KBP = +0.512 oC.kg/mol for water (see Table 11.4). Solution 1. Calculate solution molality = 4.00 m (see previous slides). 2. ∆TBP = KBP • m ∆TBP = +0.512 oC.kg/mol (4.00 molal ( mol/Kg)) ∆TBP = +2.05 oC BP = 102.05 oC Chapter 11

  48. T = DH DS Freezing-Point Depression DG = DH - TDS At equilibrium, DG = 0.

  49. Freezing-Point Depression • Freezing-Point Depression (∆Tf):The freezing point of the pure solvent (T°f) minus the freezing point of the solution (Tf). ∆Tf = T°f – Tf • ∆Tf is proportional to concentration:∆Tf = Kfm Kf = molal freezing-point depression constant. Chapter 11

  50. Boiling-Point Elevation and Freezing-Point Depression 04 Chapter 11

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