1. by Steven S. Zumdahl & Donald J. DeCoste University of Illinois Introductory Chemistry: A Foundation, 6 th Ed. Introductory Chemistry, 6 th Ed. Basic Chemistry, 6 th Ed.

2. Chapter 15 Solutions

3. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 3 Solutions Solutions are homogeneous mixtures. Solvent: the substance present in the highest percentage Solute: the dissolved substance, which is present in lesser amount

4. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 4 Solutions (cont.)

5. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 5 The Solution Process: Ionic Compounds When ionic compounds dissolve in water they dissociate into ions and become hydrated. When solute particles are surrounded by solvent molecules we say they are solvated.

6. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 6 The Solution Process: Covalent Molecules Covalent molecules that are small and have “polar” groups tend to be soluble in water. The ability to H-bond with water enhances solubility. O H H C O H H H H O H H

7. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 7 Solubility When one substance (solute) dissolves in another (solvent), it is said to be soluble. –Salt is soluble in water –Bromine is soluble in methylene chloride When one substance does not dissolve in another, it is said to be insoluble. –Oil is insoluble in water

8. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 8 Solubility (cont.) There is usually a limit to the solubility of one substance in another. –Gases are always soluble in each other –Some liquids are always mutually soluble

9. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 9 Solutions & Solubility Molecules that are similar in structure tend to form solutions: “like dissolves like”

10. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 10 Solutions & Solubility (cont.) The solubility of the solute in the solvent depends on the temperature. –Higher temp = greater solubility of solid in liquid –Lower temp = greater solubility of gas in liquid The solubility of gases depends on the pressure. –Higher pressure = greater solubility

11. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 11 Describing Solutions Qualitatively A concentrated solution has a high proportion of solute to solution. A dilute solution has a low proportion of solute to solution.

12. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 12 Describing Solutions Qualitatively (cont.) A saturated solution has the maximum amount of solute that will dissolve in the solvent. –Depends on temp An unsaturated solution has less than the saturation limit. A supersaturated solution has more than the saturation limit. –Unstable

13. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 13 Describing Solutions Quantitatively (cont.) Solutions have variable composition. To describe a solution accurately, you need to describe the components and their relative amounts. Concentration: the amount of solute in a given amount of solution –Occasionally amount of solvent

14. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 14 Solution Concentration Percentage Mass percent = grams of solute per 100 g of solution –5.0% NaCl has 5.0 g of NaCl in every 100 g of solution Mass of solution = mass of Solute + mass of solvent Divide the mass of solute by the mass of solution and multiply by 100% to get mass percent.

15. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 15 Solution Concentration Molarity Moles of solute per 1 liter of solution Used because it describes how many moles of solute in each liter of solution If a sugar solution concentration is 2.0 M, 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar, etc. molarity = moles of solute liters of solution

16. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 16 Molarity & Dissociation The molarity of the ionic compound allows you to determine the molarity of the dissolved ions.

17. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 17 Molarity & Dissociation (cont.) CaCl 2 (aq) = Ca +2 (aq) + 2 Cl -1 (aq) A 1.0 M CaCl 2 (aq) solution contains 1.0 moles of CaCl 2 in each liter of solution Because each CaCl 2 dissociates to one Ca +2, 1.0 M CaCl 2 = 1.0 M Ca +2 Because each CaCl 2 dissociates to 2 Cl -1, 1.0 M CaCl 2 = 2.0 M Cl -1

18. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 18 Dilution Dilution: adding solvent to decrease the concentration of a solution The amount of solute stays the same, but the concentration decreases.

19. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 19 Dilution (cont.) Dilution Formula M 1 x V 1 = M 2 x V 2 # Moles/L · # L = # moles –In dilution we take a certain number of moles of solute and dilute to a bigger volume. Concentrations and volumes can be most units as long as they are consistent.

20. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 20 Solution Stoichiometry Many reactions occur in solution. Therefore you need to be able to predict amounts of reactants and products in terms of concentrations and volumes as well as masses.

21. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 21 Solution Stoichiometry (cont.) Basic strategy is the same: 1. Balance the equation 2. Change given amounts to moles (M x V = #moles) 3. Determine limiting reactant 4. Calculate moles of required substance 5. Convert moles of the required substance into the desired unit

22. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 22 Example #1: Calculate the mass of solid NaCl required to precipitate all the Ag +1 ions from 1.50 L of a 0.100 M AgNO 3 solution.

23. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 23 Example #1 (cont.) Write and balance the reaction: –The reaction is a precipitation reaction. It involves Cl -1 ions from NaCl reacting with Ag +1 ions from AgNO 3 to form AgCl(s). Therefore we get: Ag +1 (aq) + Cl -1 (aq)  AgCl(s) (balanced)

24. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 24 Change the given amounts to moles: –We are given 1.50 L of 0.100 M AgNO 3. Since 1 AgNO 3 dissociates into 1 Ag +1 Example #1 (cont.)

25. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 25 Example #2: Calculate the mass of solid NaCl required to precipitate all the Ag +1 ions from 1.50 L of a 0.100 M AgNO 3 solution.

26. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 26 Example #2 (cont.) Determine the limiting reactant: –Since we are going to precipitate all the Ag +1 by adding Cl -1, the Ag +1 is the limiting reactant Determine the number of moles of the required substance: –We need to calculate the moles of Cl -1 required to precipitate 0.150 moles of Ag +1

27. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 27 Convert moles of the required substance into the desired unit: –We need 0.150 moles of Cl -1. Since 1 NaCl dissociates into 1 Cl -1, the moles of NaCl needed = 0.150 moles. 1 mol NaCl = 58.44 g NaCl Example #2 (cont.)

28. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 28 Neutralization Reactions Acid-Base reactions are also called neutralization reactions. Often we use neutralization reactions to determine the concentration of an unknown acid or base. The procedure is called a titration. With this procedure we can add just enough acid solution to neutralize a known volume of a base solution. –Or vice-versa

29. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 29 Normality Normality: concentration unit used mainly for acids and bases One equivalent of an acid is the amount of acid that can furnish 1 mol of H +1 One equivalent of a base is the amount of base that can furnish 1 mol of OH -1 Equivalent weight: the mass of 1 equivalent of an acid or base

30. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 30 Equivalents

31. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 31 Solution Concentration Normality Equivalents of solute per 1 liter of solution Used because it describes how many H + or OH - in each liter of solution

32. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 32 Solution Concentration Normality (cont.) If an acid solution concentration is 2.0 N, 1 liter of solution contains 2.0 equiv. of acid - which means 2 mol H +1 –2 liters = 4.0 equiv acid = 4.0 mol H +1 –0.5 liters = 1.0 equiv acid = 1.0 mol H +1

33. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 33 Normality Normality = equivalents of solute liters of solution Liters x normality =equivalents of solute

34. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 34 Normality and Neutralization One equivalent of acid exactly neutralizes one equivalent of base. Can be used to simplify neutralization stoichiometry problems to the equation N acid x V acid = N base x V base # equiv/L x # L = # equiv # eqiv acid = # equiv base The volumes can be most any unit, as long as they are consistent.

35. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 35 Example #3: What volume of 0.075 N KOH is required to neutralize 0.135 L of 0.45 N H 3 PO 4 ?

36. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 36 Example #3 (cont.) Determine the quantities and units in the problem Acid SolutionBase Solution Normality0.45 N0.075 N Volume0.135 L? L Solve the formula for the unknown quantity

37. Copyright © Houghton Mifflin Company. All rights reserved. 15 | 37 Plug the values into the equation and solve Acid SolutionBase Solution Normality0.45 N0.075 N Volume0.135 L? L Example #3 (cont.)