1. 1 Solutions Chapter 14

2. 2 Solutions Solutions are homogeneous mixtures Solute is the dissolved substance –Seems to “disappear” or “Takes on the state” of the solvent Solvent is the substance the solute dissolves in –Does not appear to change state –When both solute and solvent have the same state, the solvent is the component present in the highest percentage Solutions in which the solvent is water are called aqueous solutions –Water is often called the universal solvent Solutions that contain metal solutes and solvent are called alloys

3. 3 The Solution Process - Ionic Compounds When ionic compounds dissolve in water they dissociate into ions –ions become surrounded by water molecules - hydrated When solute particles are surrounded by solvent molecules we say they are solvated

4. 4 Figure 14.1: When solid sodium chloride dissolves, the ions are dispersed randomly throughout the solution

5. 5 Figure 14.2: Polar water molecules interact with the positive and negative ions of a salt

6. 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. 7 Figure 14.3: Polar O—H bond similar to those in the water molecule

8. 8 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 they are said to be insoluble –Oil is insoluble in Water 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. 9 Solutions & Solubility Molecules that are similar in structure tend to form solutions –Like dissolves like The solubility of the solute in the solvent depends on the temperature –Higher Temp = Larger solubility of solid in liquid –Lower Temp =Larger solubility of gas in liquid The solubility of gases depends on the pressure –Higher pressure = Larger solubility

10. 10 Figure 14.4: The structure of common table sugar (called sucrose)

11. 11 Figure 14.5: A molecule typical of those found in petroleum

12. 12 Figure 14.6: An oil layer floating on water

13. 13 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 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

14. 14 Describing Solutions Quantitatively Solutions have variable composition To describe a solution accurately, you need to describe the components and their relative amounts Concentration = amount of solute in a given amount of solution –Occasionally amount of solvent

15. 15 Figure 14.7: Preparation of a standard aqueous solution

16. 16 Solution Concentration Percentage Parts Per Hundred % = grams of solute per 100 g of solution –Mass Percent or Percent by Mass –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%

17. 17 Solution Concentration Molarity moles of solute per 1 liter of solution used because it describes how many molecules 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

18. 18 Molarity & Dissociation the molarity of the ionic compound allows you to determine the molarity of the dissolved ions 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 –1 L = 1.0 moles CaCl 2, 2 L = 2.0 moles CaCl 2, 0.5 L = 0.5 moles CaCl 2 Because each CaCl 2 dissociates to one Ca +2, 1.0 M CaCl 2 = 1.0 M Ca +2 –1 L = 1.0 moles Ca +2, 2 L = 2.0 moles Ca +2, 0.5 L = 0.5 moles Ca +2 Because each CaCl 2 dissociates to 2 Cl -1, 1.0 M CaCl 2 = 2.0 M Cl -1 –1 L = 2.0 moles Cl -1, 2 L = 4.0 moles Cl -1, 0.5 L = 1.0 moles Cl -1

19. 19 Dilution Dilution is adding solvent to decrease the concentration of a solution The amount of solute stays the same, but the concentration decreases Dilution Formula M 1 x V 1 = M 2 x V 2 Concentrations and Volumes can be most units as long as consistent

20. 20 Figure 14.8: Acetic acid dilution

21. 21 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 Basic strategy is the same 1.Balance the Equation 2.Change Given Amounts to Moles 3.Determine Limiting Reactant 4.Calculate Moles of Required Substance 5.Convert Moles of the Required Substance into the Desired Unit

22. 22 Example 1. Write and Balance the Reaction The reaction is a Precipitation Reaction. The reaction 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) After Balancing we get Ag +1 (aq) + Cl -1 (aq)  AgCl(s) 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. 23 2. 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 The concentration of Ag +1 = 0.100 M 1 L Solution = 0.100 mol Ag +1 Example 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

24. 24 3. 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 4. 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 Example 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

25. 25 5. 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 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. 26 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 visa versa

27. 27 Normality Normality is a 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 The equivalent weight is the mass of 1 equivalent of an acid or base

28. 28 Equivalents 1 mol HCl = 1 mol H +1 = 1 equivalent HCl –Therefore the equivalent weight of HCl = the Molar Mass of HCl = 36.45 g 1 mol H 2 SO 4 = 2 mol H +1 = 2 equiv H 2 SO 4 –Therefore the equivalent weight of H 2 SO 4 = one-half the Molar Mass of H 2 SO 4 = ½(98.07 g) = 49.04 g 1 mol NaOH = 1 mol OH -1 = 1 equivalent NaOH –Therefore the equivalent weight of NaOH = the Molar Mass of NaOH = 40.00 g 1 mol KOH = 1 mol OH -1 = 1 equiv KOH –Therefore the equivalent weight of KOH = the Molar Mass of KOH = 56.11 g

29. 29 Solution Concentration Normality equivalents of solute per 1 liter of solution used because it describes how many H +1 or OH -1 in each liter of solution 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

30. 30 Normality Normality = equivalents of solute liters of solution Liters x Normality =Equivalents of Solute

31. 31 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 The volumes can be most any unit, as long as they are consistent

32. 32 Example ¬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 What volume of 0.075 N KOH is required to Neutralize 0.135 L of 0.45 N H 3 PO 4

33. 33 ®Plug the Values into the Equation and Solve Acid SolutionBase Solution Normality0.45 N0.075 N Volume0.135 L? L Example What volume of 0.075 N KOH is required to Neutralize 0.135 L of 0.45 N H 3 PO 4