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  1. 1. Solutions<br />Chapters 12 and 13<br />Holt Modern Chemistry<br />Lisa Allen, Stonington High School<br />
  2. 2. Concepts to consider<br />Solubility<br />Energy changes in dissolvingConcentrationsPrecipitation reactions<br />Net ionic equationsEffects on vapor pressure<br />
  3. 3. Solubility Vocabulary<br />Solvent<br />Solute<br />Electrolyte<br />Saturated<br />Unsaturated<br />Solution equilibrium<br />Supersaturated<br />
  4. 4. I put some salt in water. The water is the solvent, and the salt is a solute. The solution was dilute, because I only used a little solute. I added more, and made a more concentrated solution. Later, by heating the solution, I increased the solubility of the salt, and was able to dissolve much more. The concentration at equilibrium was much greater in hot water. By gently cooling this hot saturated solution, I was able to obtain a supersaturated solution.<br />How to use these words…<br />
  5. 5. Reading solubility curves<br />Solubility curve<br />Simple questions<br />What is the solubility of potassium nitrate at 60˚C?<br />What is the solubility of sodium chloride at 20˚C?<br />Which of these substances is most soluble at 80 ˚C?<br />
  6. 6. What is the effect of temperature on solubility?<br />See pages 410 and 414. Note the differences between gases and solids!<br />
  7. 7. And what is the effect of dissolving on temperature?<br />Enthalpy of solution: “the net amount of energy absorbed as heat by the solution when a specific amount of solute dissolves in a solvent”<br />HMC, p. 416<br />
  8. 8. Enthalpy of solution<br />Endothermic processes<br />Exothermic processes<br />ABSORBING energy<br />Breaking up the solid solvent<br />Moving molecules of solvent apart to make room for solute<br />RELEASING energy<br />Forming solvent/solute attractions<br />
  9. 9. What is the net change?<br />Endothermic<br />Exothermic<br />If these processes absorb MORE energy than those on the right, the net difference is that the solution TAKES IN energy.<br />The enthalpy of solution is POSITIVE<br />Solutions get COLDER<br />If these processes absorb MORE energy than those on the left, the net difference is that the solution GIVES OFF energy.<br />The enthalpy of solution is NEGATIVE.<br />Solutions get WARMER<br />
  10. 10. Enthalpy of solution problems<br />As always, solving chemistry problems is easy if you use dimensional analysis, and know what the units of your answer should be!<br />How much energy is released by the dissolving of 125 grams of KOH? (p 416)<br />If that energy is used to heat 800 grams of 25 ˚C water, which has a specific heat capacity of 4.18 J/g ˚C, what is the final temperature of the water?<br />
  11. 11. Calculating temperature change<br />
  12. 12. What is the final temperature of 240 grams of water at 25 ˚C to which 35 grams of HCl is added?<br />What is the final temperature of 100. grams of water at 25 ˚C in which 12 grams of KClO3 is dissolved? <br />What is the final temperature of 50. grams of water at 25 ˚C in which 20. grams of NaCl is dissolved?<br />More practice<br />
  13. 13. HOMEWORK<br />Pages 426 – 427<br />#10-14<br />#36, 37<br />
  14. 14. Concentration<br />Molarity and molality:<br />2 ways to define concentrations of solutions<br />
  15. 15. Designated with a capital M<br />Easiest to use of all of the ways to quantify concentrations<br />Allows users to obtain known number of moles of solutes from volumes of solutions<br />2M means that a liter of the solution will contain 2 moles of the solute, 500 mL will contain a mole of it.<br />Molarity = moles solute liters solvent<br />
  16. 16. Algebra in action!<br />
  17. 17. Determine the molarity of a solution prepared by dissolving 141.6 grams of citric acid, C3H5O(COOH)3, in water and then diluting the resultant solution to 3500.0 mL.<br />What mass of ammonium chloride is dissolved in 300. mL of a 0.875M solution?<br />What mass of magnesium bromide would be required to prepare 720. mL of a 0.0939 M aqueous solution?<br />Molarity math practice<br />
  18. 18. Molarity x volume = number of moles<br />This means that if you put a certain number of moles of a solute into a solvent, adding more solvent does not change the number of moles! <br />The number of moles will stay constant as you alter the volume of the solution, and the molarity will change to reflect that.<br />Dilution math M1 xV1 = M2 xV2<br />
  19. 19. Dilutions and mixing solutions<br />HCl is 12M when concentrated. If I need 150 mL of 2.0M solution for a lab, how do I mix it?<br />125 mL of 5.0 M potassium nitrate solution is diluted to 500 mL. What is the new concentration?<br />Rearrange the equation first!<br />Plug in the values with their labels.<br /><ul><li>BTW-Special precautions must be taken when diluting acids! </li></ul>Use moles/liters instead of M to facilitate canceling units.<br />
  20. 20. If I combine 5.00 L of 1.5M HCl with 3.5 L of 2.5M NaOH, which is the limiting reactant?<br />What is the molarity of the resultant salt solution?<br />Don’t panic! <br />Just calculate moles of each reactant, and use the mole ratio to find the moles of product from each. The one that makes the smallest amount of product is the limiting reactant.<br />Molarity and stoichiometry<br />
  21. 21. 50 mL of .065 M silver nitrate is combined with 50 mL of 1.0M NaCl solution. What is the mass of solid silver chloride that precipitates?<br />More practice with molarity math<br />Strategy:<br />Write a balanced chemical equation <br />Convert each reactant to moles<br />Use stoichiometry to find moles of product<br />Establish which reactant is limiting<br />Calculate grams of product<br />
  22. 22. Homework <br />Page 427, 18-28<br />
  23. 23. Aqueous cadmium chloride reacts with sodium sulfide to produce bright yellow cadmium sulfide.<br />How many moles of CdCl2 are in 50.00 mL of 3.91 M solution?<br />If this solution is reacted with excess sodium sulfide, how many moles of CdS would form?<br />If this solution is reacted with 35.00 mL of 4.33 M sodium sulfide solution, which is the limiting reactant?<br />More practice on solution stoichiometry<br />
  24. 24. What is the molarity of a solution of barium nitrate made from 28.83 grams of solid in 250. mL of solution?<br />What is the molarity of sodium sulfate if 7.00 grams are in 35.0 mL of solution?<br />If I combine these solutions, what is the limiting reactant?<br />How many grams of solid product is produced?<br />What is the molarity of the sodium nitrate solution that remains?<br />Barium nitrate + Sodium sulfate ?<br />
  25. 25. Net ionic equations<br />A slightly different way to look at solution reactions.<br />
  26. 26. Writing net ionic equations<br />Step by step<br />Write the dissociation reactions for each of the reactant compounds.<br />Write the reaction of these aqueous compounds with the dissociated ions as reactants.<br />Determine the solubility of the possible products.<br />Insoluble products are written as a compound, soluble ones are called “spectators”<br />NO PRECIPITATE, NO REACTION!!!<br />Practice from page 440<br />Will a precipitate form if solutions of potassium sulfate and barium nitrate are combined? If so, write the net ionic equation for the reaction.<br />
  27. 27. <ul><li>Write the equation that represents the dissolution of barium nitrate in water.
  28. 28. Write the equation that represents the dissolution of sodium sulfate in water.
  29. 29. Write the complete ionic equation.
  30. 30. Write the net ionic equation.
  31. 31. Identify the spectator ions.</li></li></ul><li>Will a precipitate form if solutions of potassium sulfate and barium nitrate are combined? If so, identify the spectator ions and write the net ionic equation.<br />Write the net ionic equation for the precipitation of nickel (II) sulfide.<br />Write one possible “regular” equation from which the previous equation could have come.<br />Note: there are several answer to this! Why?<br />Practice from page 440<br />
  32. 32. What will form a precipitate?<br />See page 437 of your text<br />Simple rules? Sodium, potassium, ammonium, nitrates, acetates, chlorates are soluble. Anything else, look up.<br />
  33. 33. Na2CO3(aq) + Pb(NO3)4(aq)????<br />Note that sodium carbonate and lead nitrate are both soluble. (Table 14-1)<br />Write the net ionic equation for the reaction between solutions of lead acetate and ammonium carbonate.<br />Practice writing net ionic equations<br />
  34. 34. Solubility rules: a more accurate view<br />The simple rules are good enough to get through the problems.<br />Substances are considered insoluble if less than .1 g dissolves in 100 g of solvent.<br />Silver chloride has a solubility of <br /> .000 089 g/100g H2O<br />EVERY ionic substance has some level of solubility. EVERY one!<br />
  35. 35. Homework:<br />Page 458, 1, 2, 8-13<br />
  36. 36. The van t’Hoff factor is a fancy name for a simple concept. If a compound dissociates into ions, you get more than one mole of ions for a mole of solute.<br />NaCl Na+ + Cl-<br />For one mole of salt, you get 2 moles of ions.<br />We say that the van t’Hoff factor is 2 for NaCl, or i=2<br />What is the van t’Hoff factor for aluminum nitrate?<br />Van t’Hoff factor<br />
  37. 37. Molality<br />m = Moles of solute<br />Kg of solvent<br />
  38. 38. Calculating molality<br />
  39. 39. What is the molality of a solution of 12.9 g of fructose, C6H12O6, in 31.0 g of water?<br />What is the molality of 71.5 grams of linoleic acid, C18H32O2 in 525 g of hexane, C6H14?<br />What mass of urea, NH2CONH2, must be dissolved in 2250 g of water to prepare a 1.50 m solution?<br />Practice<br />
  40. 40. Colligative properties<br />Properties of materials that change as a result of the addition of a solute<br />
  41. 41. Solutes lower the vapor pressure of solutions.<br />Why is vapor pressure reduced?<br />In a solution, the molecules at the surface are no longer all solvent molecules.<br />Vapor pressure comes from surface molecules establishing an equilibrium concentration with the air above the liquid.<br />Lower concentrations of solvent molecules at the surface means lower vapor pressure.<br />
  42. 42. What is the effect of changing vapor pressure?<br />Boiling point elevation<br />Freezing point depression<br />Consider the graph on page 344<br />Notice that the boiling point is the temperature at which vapor pressure equals atmospheric pressure.<br />If solutes lower vapor pressure for a solution, the solution has to be hotter for the vapor pressure to hit atmospheric pressure.<br />It is easier to think of the effect of solutes on freezing point in a different way<br />Consider that the particles of solute get in the way of crystals of solvent that try to form.<br />It has to be colder to get crystals to form in a solution than in pure solvent.<br />
  43. 43. This alteration in the vapor pressure is not due to the quality or the nature of the particles that make up the solute. It is only a result of the number of those particles.<br />Compare the dissolution of aluminum sulfate to the dissolving of sugar. What is the number of particles produced by the dissolving of 1 mole of each of these substances?<br />THIS is where we use the van t’Hoff factor<br />Important facts about colligative properties<br />
  44. 44. Dt = Kim<br />Kf is molal freezing point constant<br />Kb is molal boiling point constant<br />Dtf= Kfim<br />See page 448 for the constants for various solvents<br />Not every solution is aqueous!<br />Try sample problems on page 449 and 450<br />Dtb= Kbim<br />Try sample problems on page 451<br />Keep in mind, the original boiling point of each specific solvent Is unique.<br />These depend on molality, not molarity<br />
  45. 45. Homework:<br />Page 459, 19-28<br />Study for quiz<br />