1. Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

2. Chapter 12 Slide 2 of 49 Solute and Solvent Solutions Are homogeneous mixtures of two or more substances. Consist of a solvent and one or more solutes. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

3. Chapter 12 Slide 3 of 49 Solutes Spread evenly throughout the solution. Cannot be separated by filtration. Can be separated by evaporation. Are not visible, but can give a color to the solution. Nature of Solutes in Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

4. Chapter 12 Slide 4 of 49 Examples of Solutions The solute and solvent can be a solid, liquid, and/or a gas. Table 12.3 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

5. Chapter 12 Slide 5 of 49 Identify the solute in each of the following solutions: 1. 2 g sugar (A) and 100 mL water (B) 2. 60.0 mL of ethyl alcohol (A) and 30.0 mL of methyl alcohol (B) 3. 55.0 mL water (A) and 1.50 g NaCl (B) 4. Air: 200 mL O 2 (A) and 800 mL N 2 (B) Learning Check P-1

6. Chapter 12 Slide 6 of 49 Water Is the most common solvent. Is a polar molecule. Forms hydrogen bonds between the hydrogen atom in one molecule and the oxygen atom in a different water molecule. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

7. Chapter 12 Slide 7 of 49 Formation of a Solution Na + and Cl - ions On the surface of a NaCl crystal are attracted to polar water molecules. In solution are hydrated as several H 2 O molecules surround each. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

8. Chapter 12 Slide 8 of 49 When NaCl(s) dissolves in water, the reaction can be written as: H 2 O NaCl(s) Na + (aq) + Cl - (aq) solid separation of ions The Na + ions are attracted to the oxygen atom (  - ) of water. The Cl - ions are attracted to the hydrogen atom (  + ) of water. Equations for Solution Formation

9. Chapter 12 Slide 9 of 49 Two substances form a solution When there is an attraction between the particles of the solute and solvent. When a polar solvent such as water dissolves polar solutes such as sugar and ionic solutes such as NaCl. When a nonpolar solvent such as hexane (C 6 H 14 ) dissolves nonpolar solutes such as oil or grease. Like Dissolves Like

10. Chapter 12 Slide 10 of 49 Water and a Polar Solute Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

11. Chapter 12 Slide 11 of 49 Like Dissolves Like Solvents Solutes Water (polar) Ni(NO 3 ) 2 CH 2 Cl 2 (nonpolar) (polar) I 2 (nonpolar) Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

12. Chapter 12 Slide 12 of 49 Electrolytes and Nonelectrolytes Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

13. Chapter 12 Slide 13 of 49 In water, Strong electrolytes produce ions and conduct an electric current. Weak electrolytes produce a few ions. Nonelectrolytes do not produce ions. Solutes and Ionic Charge Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

14. Chapter 12 Slide 14 of 49 Strong electrolytes Dissociate in water producing positive and negative ions. Produce an electric current in water. In equations show the formation of ions in aqueous (aq) solutions. H 2 O 100% ions NaCl(s) Na + (aq) + Cl − (aq) H 2 O CaBr 2 (s) Ca 2+ (aq) + 2Br − (aq) Strong Electrolytes

15. Chapter 12 Slide 15 of 49 A weak electrolyte Dissociates only slightly in water. In water forms a solution of only a few ions and mostly undissociated molecules. HF(g) + H 2 O(l) H 3 O + (aq) + F - (aq) NH 3 (g) + H 2 O(l) NH 4 + (aq) + OH - (aq) Weak Electrolytes

16. Chapter 12 Slide 16 of 49 Nonelectrolytes Dissolve as molecules in water. Do not produce ions in water. Do not conduct an electric current. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

17. Chapter 12 Slide 17 of 49 Chapter 12 Solutions Solubility Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

18. Chapter 12 Slide 18 of 49 Solubility Is the maximum amount of solute that dissolves in a specific amount of solvent. Can be expressed as grams of solute in 100 grams of solvent, usually water. g of solute 100 g water Solubility

19. Chapter 12 Slide 19 of 49 Effect of Temperature on Solubility Solubility Depends on temperature. Of most solids increases as temperature increases. Of gases decreases as temperature increases. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

20. Chapter 12 Slide 20 of 49 Unsaturated Solutions Unsaturated solutions Contain less than the maximum amount of solute. Can dissolve more solute. Dissolved solute Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

21. Chapter 12 Slide 21 of 49 Saturated Solutions Saturated solutions Contain the maximum amount of solute that can dissolve. Have undissolved solute at the bottom of the container. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

22. Chapter 12 Slide 22 of 49 Soluble and Insoluble Salts Ionic compounds that Dissolve in water are soluble salts. Do not dissolve in water are insoluble salts. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

23. Chapter 12 Slide 23 of 49 Solubility Rules Soluble salts Typically contain at least one ion from Groups 1A(1) or NO 3 −, or C 2 H 3 O 2 − (acetate). Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings Table 12.3

24. Chapter 12 Slide 24 of 49 Using the Solubility Rules The solubility rules predict if a salt Is soluble or Insoluble in water. Table 12.4 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

25. Chapter 12 Slide 25 of 49 Formation of a Solid When solutions of salts are mixed, A solid forms if ions of an insoluble salt are present. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

26. Chapter 12 Slide 26 of 49 Equations for Forming Solids A molecular equation shows the formulas of the compounds. Pb(NO 3 )(aq) + 2NaCl(aq) PbCl 2 (s) + 2NaNO 3 (aq) An ionic equation shows the ions of the compounds. Pb 2+ (aq) + 2NO 3 − (aq) + 2Na + (aq) + 2Cl − (aq) PbCl 2 (s) + 2Na + (aq) + 2NO 3 − (aq) A net ionic equation shows only the ions that form a solid. Ions remaining in solution are spectator ions. Pb 2+ (aq) + 2Cl − (aq) PbCl 2 (s)

27. Chapter 12 Slide 27 of 49 Equations for the Insoluble Salt STEP 1 Observe the ions in the reactants. Pb 2+ (aq) + 2NO 3 − (aq) 2Na + (aq) + 2Cl − (aq) STEP 2 Determine if any new ion combinations are insoluble salts. Yes. PbCl 2 (s) STEP 3 Ionic equation with insoluble salt product. Pb 2+ (aq) + 2NO 3 − (aq) + 2Na + (aq) + 2Cl − (aq) PbCl 2 (s) + 2Na + (aq) + 2NO 3 − (aq) STEP 4 Net ionic equation. Pb 2+ (aq) + 2Cl − (aq) PbCl 2 (s)

28. Chapter 12 Slide 28 of 49 Chapter 12Solutions Percent Concentration Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

29. Chapter 12 Slide 29 of 49 The concentration of a solution Is the amount of solute dissolved in a specific amount of solution. amount of solute amount of solution As percent concentration describes the amount of solute that is dissolved in 100 parts of solution. amount of solute 100 parts solution Percent Concentration

30. Chapter 12 Slide 30 of 49 The mass percent (%m/m) Concentration is the percent by mass of solute in a solution. mass percent (%m/m) = g of solute x 100 (g of solute + g of solvent) Is the g of solute in 100 g of solution. mass percent = g of solute x 100 100 g of solution Mass Percent

31. Chapter 12 Slide 31 of 49 Mass of Solution Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

32. Chapter 12 Slide 32 of 49 Mass percent (%m/m) is calculated from the grams of solute (g KCl) and the grams of solution (g KCl solution). g of KCl = 8.00 g g of solvent (water) = 42.00 g g of KCl solution = 50.00 g 8.00 g KCl (solute) x 100 = 16.0% (m/m) 50.00 g KCl solution Calculating Mass Percent

33. Chapter 12 Slide 33 of 49 Molarity and Dilution Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

34. Chapter 12 Slide 34 of 49 Molarity (M) Molarity (M) is A concentration term for solutions. The moles of solute in 1 L solution. moles of solute liter of solution

35. Chapter 12 Slide 35 of 49 Preparing a 1.0 Molar Solution A 1.00 M NaCl solution is prepared By weighing out 58.5 g NaCl (1.00 mol) and Adding water to make 1.00 liter of solution. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

36. Chapter 12 Slide 36 of 49 What is the molarity of 0.500 L NaOH solution if it contains 6.00 g NaOH? STEP 1 Given 6.00 g NaOH in 0.500 L solution Need molarity (mol/L) STEP 2 Plan g NaOH mol NaOH molarity STEP 3 Conversion factors 1 mol NaOH = 40.00 g 1 mol NaOH and 40.00 g NaOH 40.00 g NaOH 1 mol NaOH Calculation of Molarity

37. Chapter 12 Slide 37 of 49 STEP 4 Calculate molarity. 6.00 g NaOH x 1 mol NaOH = 0.150 mol 40.00 g NaOH 0.150 mol = 0.300 mol = 0.300 M NaOH 0.500 L 1 L Calculation of Molarity (cont.)

38. Chapter 12 Slide 38 of 49 Molarity Conversion Factors The units of molarity are used as conversion factors in calculations with solutions. Table 2.6 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

39. Chapter 12 Slide 39 of 49 Molarity in Calculations How many grams of KCl are needed to prepare 125 mL of a 0.720 M KCl solution? STEP 1 Given 125 mL (0.125 L) of 0.720 M KCl Need Grams of KCl STEP 2 Plan L KCl mol KCl g KCl

40. Chapter 12 Slide 40 of 49 Molarity in Calculations (cont.) STEP 3 Conversion factors 1 mol KCl = 74.55 g 1 mol KCl and 74.55 g KCl 74.55 g KCl 1 mol KCl 1 L KCl = 0.720 mol KCl 1 L and 0.720 mol KCl 0.720 mol KCl 1 L STEP 4 Calculate grams. 0.125 L x 0.720 mol KCl x 74.55 g KCl = 6.71 g KCl 1 L 1 mol KCl

41. Chapter 12 Slide 41 of 49 Dilution In a dilution Water is added. Volume increases. Concentration decreases. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

42. Chapter 12 Slide 42 of 49 Comparing Initial and Diluted Solutions In the initial and diluted solution The moles of solute are the same. The concentrations and volumes are related by the equation M 1 V 1 = M 2 V 2 initial diluted

43. Chapter 12 Slide 43 of 49 Dilution Calculations What is the molarity if 0.180 L of 0.600 M KOH is diluted to a final volume of 0.540 L? STEP 1 Prepare a table: M 1 = 0.600 MV 1 = 0.180 L M 2 = ?V 2 = 0.540 L STEP 2 Solve dilution equation for unknown. M 1 V 1 = M 2 V 2 M 1 V 1 / V 2 = M 2 STEP 3 Set up and enter values: M 2 = M 1 V 1 = (0.600 M)(0.180 L) = 0.200 M V 2 0.540 L

44. Chapter 12 Slide 44 of 49 Chapter 12 Solutions Solutions in Chemical Reactions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

45. Chapter 12 Slide 45 of 49 Molarity in Chemical Reactions In a chemical reaction, The volume and molarity of a solution are used to determine the moles of a reactant or product. volume (L) x molarity ( mol ) = moles 1 L If molarity (mol/L) and moles are given, the volume (L) can be determined mol x 1 L = volume (L) mol

46. Chapter 12 Slide 46 of 49 Using Molarity of Reactants How many mL of 3.00 M HCl are needed to react 4.85 g CaCO 3 ? 2HCl(aq) + CaCO 3 (s) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) STEP 1 Given 3.00 M HCl; 4.85 g CaCO 3 Need volume in mL STEP 2 Plan g CaCO 3 mol CaCO 3 mol HCl mL HCl

47. Chapter 12 Slide 47 of 49 Using Molarity of Reactants (cont.) 2HCl(aq) + CaCO 3 (s) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) STEP 3 Equalitites 1 mol CaCO 3 = 100.09 g; 1 mol CaCO 3 = 2 mol HCl 1000 mL HCl = 3.00 mol HCl STEP 4 Set Up 4.85 g CaCO 3 x 1 mol CaCO 3 x 2 mol HCl x 1000 mL HCl 100.09 g CaCO 3 1 mol CaCO 3 3.00 mol HCl = 32.3 mL HCl required

48. Chapter 12 Slide 48 of 49 Learning Check How many mL of a 0.150 M Na 2 S solution are needed to react 18.5 mL of 0.225 M NiCl 2 solution? NiCl 2 (aq) + Na 2 S(aq) NiS(s) + 2NaClaq) 0.0185 L x 0.225 mol NiCl 2 x 1 mol Na 2 S x 1000 mL 1 L 1 mol NiCl 2 0.150 mol = 27.8 mL Na 2 S solution

49. Chapter 12 Slide 49 of 49 Learning Check What is the molarity if 15.0 mL of AgNO 3 solution reacts with 22.8 mL of 0.100 M MgCl 2 ? MgCl 2 (aq) + 2AgNO 3 (aq) 2AgCl(s) + Mg(NO 3 ) 2 (aq) 0.0228 L x 0.100 mol MgCl 2 x 2 mol AgNO 3 x 1 = 1 L 1 mol MgCl 2 0.0150 L = 0.304 mol/L = 0.304 M AgNO 3