1. Chapter 15 Solutions Solution- homogeneous mixture w/ components uniformly intermingled Solute- substance in the smallest amount Solvent- substance in the largest amount Aqueous solution- solution with water as the solvent (tincture- solution with alcohol as the solvent) 15.1 Solubility Dissolution- process of dissolving For ionic substances: NaCl(s)  Na + (aq) + Cl - (aq) For covalent substances: C 12 H 22 O 11 (s)  C 12 H 22 O 11 (aq)

2. Figure 15.1: Dissolving of solid sodium chloride.

3. Solubility of Polar substances (see fig. 15.3, p. 469) Substances insoluble in Water (see fig. 15.6, p. 470) How substances Dissolve- “like dissolves like” Polar solvent easily dissolve polar molecules or ionic (very polar) substances. Nonpolar solvents easily dissolve nonpolar substances.

4. Figure 15.2: Polar water molecules interacting with positive and negative ions of a salt.

5. Figure 15.3: The ethanol molecule contains a polar O—H bond.

6. Figure 15.3: The polar water molecule interacts strongly with the polar O—H bond in ethanol.

7. Figure 15.4: Structure of common table sugar.

8. Figure 15.5: A molecule typical of those found in petroleum.

9. Figure 15.6: An oil layer floating on water.

10. 15.2 Solution Composition Saturated- a solution which contains just the amount of solute that can be dissolved in that amount of solvent @ that temperature Unsaturated- a solution which does not contain the amount of solute that can be dissolved in that amount of solvent @ that temperature; it has less Supersaturated- a solution which contains more solute in a given amount of solvent @ a particular temperature than it should contain Concentrated- a solution w/ a lot of solute Dilute- a solution w/ very little solute

11. 15.3 Factors Affecting the Rate of Dissolution Surface area- dissolution takes place on the outside of particles; increasing the surface area (crushing the crystals) increases the rate (it goes faster) Temperature- particles of solvent move faster when heated; therefore, fresh solvent w/ relatively little solute content moves past the solute particles (gases dissolve poorly in high temperature solvent) Stirring- aids in moving the solvent around, bringing fresh solvent w/ relatively little solute content to the solute particles Pressure (of gases only; little effect on dissolution of liquids & solids)- gases dissolve better under pressure. Note the solubility curve graph

13. 15.4 Solution Composition: Mass Percent Mass% - mass of solute present in a given mass of solution Example: 1.00 g of ethanol in 100. g of water. Mass % = 1.00 g/101. g x 100% = 0.99% Problem: a 135 g sample of seawater is evaporated to dryness, leaving a 4.73 g solid residue. Calculate the % of solute present. Mass % = 4.73 g/ 135 g x 100% = 3.50% Problem: what is the mass by percent if 15.0 g of calcium chloride is dissolved in 285 g of water? (Problem 12.c, p. 499)

14. 15.5 Solution Composition: Molarity Molarity = moles of solute per liter of solution (mol/L) Standard solution- a solution whose concentration is accurately known. Example: calculate molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make a 1.50 L solution. 11.5 g NaOH x 1 mol/40 g = 0.288 mol NaOH M = 0.288 mol/1.50 L = 0.192 M

15. Problem: Calculate the molarity of a solution with 0.101 g NaHCO 3 in a solution of total volume of 11.3 mL (problem 22.b, p. 500) 0.101 g x 1 mol/84 g = 0.00120 mol M = 0.00120 mol/0.0113 L = 0.106 M Problem: If you dissolve 0.50 mol KBr in a total volume 500. mL, what is the molarity? M = 0.50 mol/0.500 L = 1.0 M Problem: What mass (in grams) of NH 4 Cl is needed to prepare 450. mL of 0.251 M solution? (Problem 28, p. 500) moles = 0.450 L x 0.251 mol/1 L = 0.113 mol 0.113 mol x 53.5 g/1 mol = 6.04 g NH 4 Cl needed

16. Figure 15.7: Steps involved in the preparation of a standard aqueous solution.

17. Calculating Ion Concentrations from Molarity Give the concentrations of all the ions in each of the following solutions: a.0.50 M Co(NO 3 ) 2 (0.50 M Co 2+, 1.0 M NO 3 - ) b. 1 M FeCl 3 (1 M Fe 3+, 3 M Cl - ) c. 0.10 M Na 2 CO 3 d. 0.010 M Al 2 (SO 4 ) 3

18. Example: How many moles of Ag +1 ions are present in 25 mL of a 0.75 M AgNO 3 solution? 0.025 L x 0.75 mol/1 L = 1.9 x 10 -2 mol Ag +1 Problem: Calculate the number of moles of Cl -1 ions in 1.75 L of 1.0 x 10 -3 M AlCl 3. 1.75 L x 1.0 x 10 -2 mol/1 L x 3 mol Cl -1 /1 mol AlCl 3 = 5.3 x 10 -3 mol Cl -1 ions Example (Backwards Problem): If you need 1.00 L of an aqueous 0.200 M K 2 Cr 2 O 7 solution, how much solid potassium dichromate must be weighed out to make this solution? 1.0 L x 0.200 mol/1 L = 0.200 mol 0.200 mol x 294.2 g/1 mol = 58.8 g K 2 Cr 2 O 7

19. Mole Fraction (X) - relates the number of moles of a solute to the total number of moles in the solution (another unit of concentration- NOT in TEXTBOOK!) X = n solute /n total Example: What is the mole fraction of NaCl if 1 mole of NaCl is dissolved in 3 moles of water? X = 1mol/(1 mol + 3 mol) = ¼ = 0.25 No unit; report as decimal instead of fraction

20. 15.6 Dilution Dilution- Process of adding more solvent to a solution Moles of solute before dilution = moles of solute after dilution M 1 V 1 = M 2 V 2 Where M 1 = initial conc. V 1 = initial volume M 2 = final conc. V 2 = final volume Example: What volume of 16 M sulfuric acid must be used to prepare 1.5 L of of a 0.10 M H 2 SO 4 solution? V = [(0.10 M)(1.5 L)]/(16 M) = 9.4 x 10 -3 L = 9.4 mL Problem: What volume of 12 M HCl must be taken to prepare 0.75 L of 0.25 M HCl? V = [(0.25 mol/1 L)(0.75 L)]/(12 mol/1 L) = 0.16 L = 16 mL

21. Figure 15.8: Process of making 500 mL of a 1.00 M acetic acid solution.

22. 15.7 Stoichiometry of Solution Reactions Write a balanced equation Calculate the # of moles of reactants Determine limiting reactant Calculate # of moles of other reactant or product, as required Convert to grams (or other units) Example: Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100 M AgNO 3 solution to precipitate all of the Ag + ions in the form of AgCl. And, calculate the mass of AgCl formed. AgNO 3 + NaCl  AgCl(s) + NaNO 3 1.50 L x 0.100 mol/1 L = 0.150 mol Ag + ions We need 0.150 mol Cl - ions. We need 0.150 mol of NaCl(s): 0.150 mol NaCl x 58.4 g/1 mol = 8.76 g NaCl needed 0.150 mol AgCl x 143.3 g AgCl/1 mol = 21.5 g AgCl will precipitate

23. (Note: copy error in this problem!!) Problem: When Ba(NO 3 ) 2 and K 2 CrO 4 react in aqueous solution, the yellow solid BaCrO 4 is formed. Calculate the mass of BaCrO 4 that forms when 3.50 x 10 -3 mole of Ba(NO 3 ) 2 is dissolved in 265 mL of 0.0100 M K 2 CrO 4 solution. Ba(NO 3 ) 2 + K 2 CrO 4  BaCrO 4 (s) + 2KNO 3 (aq) 3.50 x 10 -3 mol Mol of CrO 4 2- : 0.265 L x 0.0100 mol/1 L = 0.00265 Reacts in a 1:1 ratio (hint: which is the LR?) You will make 2.65 x 10 -3 mol of BaCrO 4. 2.65 x 10 -3 mol x 253.3 g/1 mol = 0.671 g

24. 15.8 Neutralization Reactions # of moles of hydrogen ion = # of moles of hydroxide ion Problem: What volume of a 0.100 M HCl solution is needed to neutralize 25.0 mL of a 0.350 M NaOH solution? V = [(0.025 L x 0.350 mol/1 L)]/0.100 mol/1 L) = 0.0875 L HCl Problem: Calculate the volume of 0.10 M HNO 3 needed to neutralize 125 mL of 0.050 M KOH? V = [(0.125 L x 0.050 mol/1 L)]/(0.10 mol/1 L) = 0.625 L

25. 15.9 Solution Composition: Normality Equivalents of an acid- amount of acid that can furnish 1 mol of H + ions Equivalents of a base- amount of base that can furnish 1 mol of OH- ions Equivalent weight (mass)- mass in grams of 1 equivalent of that acid or base Normality- number of equivalents of solute per liter of solution N = number of equivalents/1 liter of solution = equiv/L

26. This unit is useful when you are working with solution that produce different amounts of ions significant in the reaction you are studying. For example, in a neutralization reaction, 0.1 M H 2 SO 4 (aq) produces twice the H + ions as an equal amount of 0.1 M HCl(aq) The sulfuric acid is 0.1 M, but is 0.2 N. The hydrochloric acid is 0.1 M, and is also 0.1 N. Example: a solution of H 2 SO 4 contains 86 g of sulfuric acid per liter of solution. Calculate normality. Equiv. mass = 98.0 g/2 = 49.0 g 86 g x 1 equiv./49.0 g = 1.8 equiv. N = 1.8 equiv./1.0 L = 1.8 N H 2 SO 4

27. 15.10 The Properties of Solutions: Boiling Point & Freezing Point Colligative Properties- a property that depends on the number of solute particles present in the solution. Need concentration unit of molality = moles of solute per kilogram of solvent (volume changes w/ temperature change; mass does not) Boiling point elevation (why we salt the pasta water) Freezing point depression (why we salt roads in the winter)(ethylene glycol in radiator of car) Osmotic Pressure (important in kidney dialysis) Draw the graph comparing freezing of the pure solvent and freezing of a solution.

28. Figure 15.9: A bubble in the interior of liquid water surrounded by solute particles and water molecules.

29. Figure 15.10: Pure water.

30. Figure 15.10: Solution (contains solute).