1 / 50

Solutions

Solutions. The Nature of Aqueous Solutions. A solution is a homogeneous mixture. The substance that is dissolved is the solute. The substance that does the dissolving is the solvent. (Water is known as the universal solvent due to its ability to dissolve a wide range of solutes).

iola
Download Presentation

Solutions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Solutions

  2. The Nature of Aqueous Solutions • A solution is a homogeneous mixture. • The substance that is dissolved is the solute. • The substance that does the dissolving is the solvent. (Water is known as the universal solvent due to its ability to dissolve a wide range of solutes)

  3. How do solutes dissolve? • Ionic solids are all soluble in water. • The “positive” ends of a water molecule are attracted to the anion of the solid. • The “negative” ends of a water molecule are attracted to the cation of the solid. • This process is called hydration and results in the ionic solid being pulled apart, or dissolved.

  4. What is an electrolyte? • Solutions that conduct an electric current efficiently contain strong electrolytes. • Strong electrolytes are substances that completely ionize in water. • Three types of compounds are classified as strong electrolytes.

  5. Strong Electrolytes • The following are classified as strong electrolytes: • Soluble salts (see the handout of the solubility rules) • Strong acids (HCl, HBr, HI, H2SO4, HNO3, HClO3, HClO4) • Strong bases (group 1A hydroxides, and Ba(OH)2

  6. What is an acid? • According to the Arrhenius definition, an acid is any substance that produces H+ ions when it is dissolved in water. • Strong acids are those acids in which every molecule ionizes into H+ ions and anions.

  7. What is a strong base? • According to the Arrhenius definition, a strong base is any soluble ionic compound containing the hydroxide ion (OH-). • When these compounds dissolve in water, the cations and OH- ions separate and move independently.

  8. What is a weak electrolyte? • Weak electrolytes are substances that exhibit a small degree of ionization in water and therefore conduct an electric current inefficiently. • The most common weak electrolytes are weak acids and weak bases. • An example of a weak acid is acetic acid (HC2H3O2). An example of a weak base is ammonium hydroxide (NH4OH)

  9. What is a nonelectrolyte? • Nonelectrolytes are substances that dissolve in water but do not produce any ions. • Organic/covalent molecules are nonelectrolytes. Table sugar (C12H22O11) is an example of a soluble compound that is a nonelectrolyte. This is because, when dissolved, the sugar molecule remains intact.

  10. Concentration • Concentrations of solutions are expressed qualitatively and quantitatively • Qualitative expressions include the terms saturated, unsaturated, and supersaturated.

  11. Concentration (continued) • Saturated solution-a solution in which no more solute will dissolve at a given temperature (the maximum has been dissolved) • Unsaturated solution-a solution in which more solute will dissolve at a given temperature (less than the maximum has been dissolved) • Super-saturated solution-a solution that contains higher-than-saturation concentration of solute (a slight disturbance will cause crystallization)

  12. Quantitative Expressions of Concentration • Percent by mass • Molarity • Molality • Dilution of solutions

  13. Percent by Mass • Percent by mass gives the mass of solute per 100 mass units of solution • Percent solute=mass of solute x 100% mass of solution • Mass of solution = mass of solute + mass of solvent

  14. Percent by Mass Problems • Calculate the mass of NiSO4 contained in 200. g of a 6.00% solution of NiSO4 • A 6.00% solution of NiSO4 contains 40.0 g of NiSO4. Calculate the mass of the solution. • Calculate the mass of NiSO4 present in 200. mL of a 6.00% solution of NiSO4. The density of the solution is 1.06 g/mL at 25oC. • What is the volume of a solution that is 15.0% Fe(NO3)3 and contains 30.0 g of the solute? The density of the solution is 1.16 g/mL at 25oC.

  15. Molarity • Molarity-the moles of solute pervolume of solution in liters. M = molarity = moles of solute liters of solution (solution = solvent + solute)

  16. Molarity Problems • Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution. • Calculate the mass of Ba(OH)2 required to prepare 2.50 L of a 0.0600 M solution of Ba(OH)2.

  17. Molality • Molality-the number of moles of solute per kilogram of solvent • Molality = moles of solute kilogram of solvent

  18. Molality Problems • A solution is prepared by mixing 1.00 g of ethanol with 100.0 g of water to give a final volume of 101 mL. What is the molality? What is the molarity? What is the mass percent? What is the mole fraction?

  19. Dilution of Solutions • Dilution occurs when more solvent is added to a solution without changing the amount of solute that is present. • V1M1 = V2M2 • For safety purposes when in a lab setting, concentrated solutions of acids and bases are always added slowly to water.

  20. Dilution Problems • How many mL of 18.0 M H2SO4 are required to prepare 1.00 L of a 0.900 M solution of H2SO4?

  21. Review of Solution Concentration • The electrolyte in car batteries is a 3.75 M H2SO4 solution that has a density of 1.230 g/mL. What is the mass %, molality, and mole fraction of the sulfuric acid?

  22. Factors Affecting Solubility Solubility is the maximum amount of solute that will dissolve in a given amount of solvent at a particular temperature.

  23. Molecular Structure • The statement “like dissolves like” summarizes the formation of solutions based on the structure of the molecules. • “Like dissolves like” means that polar solvents dissolve ionic and polar solutes while nonpolar solvents dissolve nonpolar solutes. • The term miscibility is used to describe the ability of one liquid to dissolve in another. (The observation of “like dissolves like” still applies).

  24. Example of Differentiating Solvent Properties Based on Structure • Decide whether liquid hexane or liquid methanol is the more appropriate solvent for the substances grease (C20H42) and potassium iodide.

  25. Effect of Temperature on Solubility • The solubility of most solid solutes increases with increasing temperature. (exceptions include: Na2SO4 and Ce2(SO4)3) • The solubility of gases decreases with increasing temperature. (example of this includes thermal pollution)

  26. Solubility Graphs • Solubility curves are graphs that illustrate the effect of temperature on the solubilities of various salts. • Using the graph on p.496, answer the following questions: *How many grams of KNO3 will dissolve in 200 g of water at 70oC? *At what temperature will 80 g of KBr dissolve in 100 g of water? *Which solute’s solubility is least affected by temperature?

  27. Effect of Pressure on Solubility • Changing the pressure has no effect on the solubilities of either solids or liquids in liquids. • The solubilities of gases in all solvents will increase as the partial pressure of the gases increase.

  28. Henry’s Law • Henry’s law applies to gases that do not react with the solvent in which they dissolve. • The law states that the pressure of a gas above the surface of a solution is proportional to the concentration of the gas in the solution.

  29. Henry’s Law (continued) • The equation for Henry’s law is: Pgas = kCgas Pgas is the pressure of the gas above the solution k is a constant for a particular gas and solvent at a particular temperature Cgas represents the concentration of the dissolved gas; usually expressed as molarity or mole fraction.

  30. Calculations Using Henry’s Law • A certain soft drink is bottled so that at 25oC a bottle contains CO2 gas at a pressure of 5.0 atm over the liquid. Assuming that the partial pressure of CO2 in the atmosphere is 0.0004 atm, calculate the equilibrium concentration of CO2 in the soda both before and after the bottle is opened. The Henry’s law constant for CO2 at 25oC is 32 L. atm/mol at 25oC. How does your answer explain why soft drinks go flat after being opened?

  31. Colligative Properties • Colligative properties are physical properties that depend on the number, not the kind, of solute particles in a given amount of solvent. • There are four important colligative properties: vapor pressure lowering, boiling point elevation, freezing point depression, osmotic pressure

  32. Lowering of Vapor Pressure • Solutions containing a nonvolatile liquid or a solid as a solute always has a lower vapor pressure than the pure solvent. • When a solute is dissolved, there are fewer solvent molecules at the surface of the liquid. • Therefore, the solvent molecules vaporize at a slower rate than if no solute were present. • The lowering of the vapor pressure of a solvent due to the presence of a nonvolatile solute is summarized by Raoult’s Law.

  33. Raoult’s Law • The vapor pressure of a solvent in an ideal solution is directly proportional to the mole fraction of the solvent in the solution. • Psolvent = Xsolvent P0solvent • The lowering of the vapor pressure can be expressed as: ΔPsolvent = XsoluteP0solvent

  34. Sample Problem • Determine the vapor pressure lowering of a sucrose solution made by dissolving 50.0 g of sucrose (C12H22O11) in 117 g of water. The vapor pressure of pure water at 25oC is 23.8 torr.

  35. Sample Problem #2 • Predict the vapor pressure of a solution made by mixing 35.0 g of solid Na2SO4 with 175 g water at 25oC. The vapor pressure of pure water at 25oC is 23.76 torr.

  36. Sample Problem #3 • At 40oC, the vapor pressure of pure heptane is 92.0 torr and the vapor pressure of pure octane is 31.0 torr. Consider a solution that contains 1.00 mole of heptane and 4.00 moles of octane. Calculate the vapor pressure of each component and the total vapor pressure above the solution.

  37. Boiling Point Elevation A nonvolatile solute elevates the boiling point of the solvent. • Boiling occurs at a temperature where the vapor pressure is equal to 1 atm. • Addition of a nonvolatile solute lowers vapor pressure. • Therefore, a solution must be heated to a higher temperature than the boiling point of the pure solvent to reach a vapor pressure of 1 atm.

  38. Equation for Boiling Point Elevation ∆T = Kbmsolute • ∆T is the boiling point elevation • Kb is the boiling point elevation constant for the solvent • msolute is the molality of the solute

  39. Calculating Boiling Point Elevation • Find the boiling point of a 1.25 m sucrose solution. • A solution was prepared by dissolving 18.00 g glucose in 150.0 g of water. The resulting solution was found to have a boiling point of 100.34oC. Calculate the molar mass of glucose.

  40. Freezing Point Depression • When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent. • This is the reason that compounds such as sodium chloride and calcium chloride are spread on roadways to prevent ice from forming in freezing weather.

  41. Equation for Freezing Point Depression ∆T = Kfmsolute • ∆T is the freezing point depression • Kf is the freezing point depression constant • msolute is the molality of the solute

  42. Calculating Freezing Point Depression • Calculate the freezing point of a 1.25 m sucrose solution. • What mass of ethylene glycol (C2H6O2 = 62.1 g/mol), the main component of antifreeze, must be added to 10.0 L of water to produce a solution for use in a car’s radiator that freezes at -10.0oF (-23.3oC)? • A sample weighing 0.546 g was dissolved in 15.0 g of benzene, and the freezing point depression was determined to be 0.240oC. Calculate the molar mass of the sample.

  43. The van’t Hoff Factor • Colligative properties depend on the number of solute particles in a given mass of solvent. • The dissociation of electrolytes in aqueous solutions causes a greater change in these properties than a solution containing a nonelectrolyte (as in the previous examples). • One measure of the extent of dissociation of an electrolyte in water the van’t Hoff factor, i, for the solution. • i = moles of particles/moles of solute ( or the number of ions per formula unit) • The observed value of i is always slightly lower than the predicted value due to ion pairing. (see page 513). • For problem solving purposes, use the ideal value of i unless the problem indicates otherwise.

  44. Determining the van’t Hoff factor • What is the value of the van’t Hoff factor, i, for the following strong electrolytes? a) Na2SO4 b) KOH c) Al2(SO4)3 d) SrSO4

  45. Properties of Electrolyte Solutions • From the following solutions: 0.010 m Na3PO4 0.020 m CaBr2 0.020 m KCl 0.020 m HF in water (HF is a weak acid) • Which would have the same boiling point as 0.040 m C6H12O6 in water? • Which solution would have the highest vapor pressure at 28oC? • Which solution would have the largest freezing point depression?

  46. Properties of Electrolyte Solutions (continued) • From the following: pure water 0.01 m C12H22O11 in water 0.01 m NaCl in water 0.01 m CaCl2 in water Choose the one with the • Highest freezing point • Lowest freezing point • Highest boiling point • Lowest boiling point • Highest vapor pressure

  47. Solving Problems Using the van’t Hoff Factor? • Calculate the freezing point and the boiling point of each of the following aqueous solutions. a) 0.050 m MgCl2 b) 0.050 m FeCl3

  48. Osmotic Pressure • Osmosis is the passage of a solvent through a semi-permeable membrane. • Osmotic Pressure is the minimum pressure that stops osmosis. • The equation for osmotic pressure is π = iMRT π is the pressure in atm M is the molarity of the solution R = 0.08206 L. atm/K. mol T is the Kelvin temperature

  49. Calculating Osmotic Pressure • To determine the molar mass of a certain protein, .00100 g of it was dissolved in enough water to make 1.00 mL of solution. The osmotic pressure was found to be 1.12 torr at 25.0oC. Calculate the molar mass of the protein. • What concentration of NaCl in water is needed to produce an aqueous solution isotonic with blood (π = 7.70 atm at 25oC)?

  50. Calculating Osmotic Pressure • What osmotic pressure would 50.0 g of sucrose in 117 g of water exhibit at 25oC? The density of this solution is 1.34 g/mL.

More Related