1. INTRODUCTORY CHEMISTRY INTRODUCTORY CHEMISTRY Concepts and Critical Thinking Sixth Edition by Charles H. Corwin 1 Chapter 14 © 2011 Pearson Education, Inc. Chapter 14 Solutions by Christopher Hamaker 1 Chapter 14

2. 2 © 2011 Pearson Education, Inc. Solutions A solution is a homogeneous mixture. A solution is composed of a solute dissolved in a solvent. Solutions exist in all three physical states:

3. 3 Chapter 14 © 2011 Pearson Education, Inc. Gases in Solution Temperature affects the solubility of gases. The higher the temperature is, the lower the solubility of a gas is in solution. An example is carbon dioxide in soda: –Less CO 2 escapes when you open a cold soda than when you open a warm soda.

4. 4 Chapter 14 © 2011 Pearson Education, Inc. Pressure and Gas Solubility Pressure also influences the solubility of gases. According to Henry’s law, the solubility of a gas is directly proportional to the partial pressure of the gas above the liquid. If we double the partial pressure of a gas, we double the solubility.

5. 5 Chapter 14 © 2011 Pearson Education, Inc. Henry’s Law We can calculate the solubility of a gas at a new pressure using Henry’s law. = new solubility solubility x new pressure old pressure What is the solubility of oxygen gas at 25  C and a partial pressure of 1150 torr if the solubility of oxygen is 0.00414 g/100 mL at 25  C and 760 torr? 1150 torr 760 torr = 0.00626 g/100 mL 0.00414 g/100 mL x

6. 6 Chapter 14 © 2011 Pearson Education, Inc. Polar Molecules When two liquids make a solution, the solute is the lesser quantity, and the solvent is the greater quantity. Recall that a net dipole is present in a polar molecule. Water is a polar molecule.

7. 7 Chapter 14 © 2011 Pearson Education, Inc. Polar and Nonpolar Solvents A liquid composed of polar molecules is a polar solvent. Water and ethanol are polar solvents. A liquid composed of nonpolar molecules is a nonpolar solvent. Hexane is a nonpolar solvent.

8. 8 Chapter 14 © 2011 Pearson Education, Inc. Like Dissolves Like Polar solvents dissolve in one another. Nonpolar solvents dissolve in one another. This is the like dissolves like rule. Methanol dissolves in water, but hexane does not dissolve in water. Hexane dissolves in toluene, but water does not dissolve in toluene.

9. 9 Chapter 14 © 2011 Pearson Education, Inc. Miscible and Immiscible Two liquids that completely dissolve in each other are miscible liquids. Two liquids that are not miscible in each other are immiscible liquids. Polar water and nonpolar oil are immiscible liquids and do not mix to form a solution.

10. 10 Chapter 14 © 2011 Pearson Education, Inc. Solids in Solution When a solid substance dissolves in a liquid, the solute particles are attracted to the solvent particles. When a solution forms, the solute particles are more strongly attracted to the solvent particles than other solute particles. We can also predict whether a solid will dissolve in a liquid by applying the like dissolves like rule.

11. 11 Chapter 14 © 2011 Pearson Education, Inc. Like Dissolves Like for Solids Ionic compounds, like sodium chloride, are soluble in polar solvents and insoluble in nonpolar solvents. Polar compounds, like table sugar (C 12 H 22 O 11 ), are soluble in polar solvents and insoluble in nonpolar solvents. Nonpolar compounds, like naphthalene (C 10 H 8 ), are soluble in nonpolar solvents and insoluble in polar solvents.

12. 12 Chapter 14 © 2011 Pearson Education, Inc. Chemistry Connection: Colloids Why is the flashlight beam visible in only one container? The solution, at the right of this slide, is a colloid. A colloid is a solution with large solute particles (ranging from 1 to 100 nm). The solute particles in a colloid are large enough to scatter light via a phenomenon known as the Tyndall effect.

13. 13 Chapter 14 © 2011 Pearson Education, Inc. The Dissolving Process When a soluble crystal is placed into a solvent, it begins to dissolve. When a sugar crystal is placed in water, the water molecules attack the crystal and begin pulling part of it away and into solution. The sugar molecules are held within a cluster of water molecules called a solvent cage.

14. 14 Chapter 14 © 2011 Pearson Education, Inc. Dissolving of Ionic Compounds When a sodium chloride crystal is placed in water, the water molecules attack the edge of the crystal. In an ionic compound, the water molecules pull individual ions off of the crystal. The anions are surrounded by the positively charged hydrogens on water. The cations are surrounded by the negatively charged oxygen on water.

15. 15 Chapter 14 © 2011 Pearson Education, Inc. Rate of Dissolving There are three ways we can speed up the rate of dissolving for a solid compound: 1.Heating the solution This increases the kinetic energy of the solvent, and the solute is attacked faster by the solvent molecules. 2.Stirring the solution This increases the interaction between solvent and solute molecules. 3.Grinding the solid solute There is more surface area for the solvent to attack.

16. 16 Chapter 14 © 2011 Pearson Education, Inc. Solubility of Solids and Temperature The solubility of a compound is the maximum amount of solute that can dissolve in 100 g of water at a given temperature. In general, a compound becomes more soluble as the temperature increases.

17. 17 Chapter 14 © 2011 Pearson Education, Inc. Saturated Solutions A solution containing exactly the maximum amount of solute at a given temperature is a saturated solution. A solution that contains less than the maximum amount of solute is an unsaturated solution. Under certain conditions, it is possible to exceed the maximum solubility of a compound. A solution with greater than the maximum amount of solute is a supersaturated solution.

18. 18 Chapter 14 © 2011 Pearson Education, Inc. Supersaturated Solutions At 55  C, the solubility of NaC 2 H 3 O 2 is 100 g per 100 g water. If a saturated solution at 55  C is cooled to 20  C, the solution is supersaturated. Supersaturated solutions are unstable. The excess solute can readily be precipitated.

19. 19 Chapter 14 © 2011 Pearson Education, Inc. Supersaturation A single crystal of sodium acetate added to a supersaturated solution of sodium acetate in water causes the excess solute to rapidly crystallize from the solution.

20. 20 Chapter 14 © 2011 Pearson Education, Inc. Concentration of Solutions The concentration of a solution tells us how much solute is dissolved in a given quantity of solution. We often hear imprecise terms such as a “dilute solution” or a “concentrated solution.” There are two precise ways to express the concentration of a solution: 1.Mass/mass percent 2.Molarity

21. 21 Chapter 14 © 2011 Pearson Education, Inc. Mass Percent Concentration Mass percent concentration compares the mass of solute to the mass of solvent. The mass/mass percent (m/m %) concentration is the mass of solute dissolved in 100 g of solution. mass of solute mass of solution x 100% = m/m % g solute g solute + g solvent x 100% = m/m %

22. 22 Chapter 14 © 2011 Pearson Education, Inc. Calculating Mass/Mass Percent A student prepares a solution from 5.00 g NaCl dissolved in 97.0 g of water. What is the concentration in m/m %? 5.50 g NaCl 5.00 g NaCl + 97.0 g H 2 O x 100% = m/m % 5.00 g NaCl 102 g solution x 100% = 4.90 %

23. 23 Chapter 14 © 2011 Pearson Education, Inc. Mass Percent Unit Factors We can write several unit factors based on the concentration 4.90 m/m % NaCl: 4.90 g NaCl 100 g solution 4.90 g NaCl 100 g solution 4.90 g NaCl 95.1 g water 4.90 g NaCl 95.1 g water 100 g solution 95.1 g water 100 g solution

24. 24 Chapter 14 © 2011 Pearson Education, Inc. Mass Percent Calculation What mass of a 5.00 m/m % solution of dextrose contains 25.0 grams of dextrose? We want grams solution; we have grams dextrose. = 500 g solution 100 g solution 5.00 g dextrose 25.0 g dextrose x

25. 25 Chapter 14 © 2011 Pearson Education, Inc. Molar Concentration The molar concentration, or molarity (M), is the number of moles of solute per liter of solution, and is expressed as moles/liter. Molarity is the most commonly used unit of concentration. moles of solute liters of solution = M

26. 26 Chapter 14 © 2011 Pearson Education, Inc. Calculating Molarity What is the molarity of a solution containing 24.0 g of NaOH in 0.100 L of solution? We also need to convert grams NaOH to moles NaOH (M = 40.00 g/mol). = 6.00 M NaOH x 24.0 g NaOH 0.100 L solution 1 mol NaOH 40.00 g NaOH

27. 27 Chapter 14 © 2011 Pearson Education, Inc. Molarity Unit Factors We can write several unit factors based on the concentration 6.00 M NaOH: 6.00 mol NaOH 1 L solution 6.00 mol NaOH 1 L solution 6.00 mol NaOH 1000 mL solution 6.00 mol NaOH 1000 mL solution

28. 28 Chapter 14 © 2011 Pearson Education, Inc. Molar Concentration Problem How many grams of K 2 Cr 2 O 7 are in 250.0 mL of 0.100 M K 2 Cr 2 O 7 ? We want mass K 2 Cr 2 O 7 ; we have mL solution. = 7.36 g K 2 Cr 2 O 7 0.100 mol K 2 Cr 2 O 7 1000 mL solution 250.0 mL solution x x 294.2 g K 2 Cr 2 O 7 1 mol K 2 Cr 2 O 7

29. 29 Chapter 14 © 2011 Pearson Education, Inc. Molar Concentration Problem, Continued What volume of 12.0 M HCl contains 9.15 g of HCl solute (M = 36.46 g/mol)? We want volume; we have grams HCl. = 20.9 mL solution 1 mol HCl 36.46 g HCl 9.15 g HCl x x 1000 mL solution 12.0 mol HCl

30. 30 Chapter 14 © 2011 Pearson Education, Inc. Critical Thinking: Water Fluoridation Cities often add fluoride to drinking water. Tooth enamel is made mostly of the mineral hydroxyapatite, Ca 10 (PO 4 ) 6 (OH) 2. Fluoride prevents tooth decay by converting some of the hydroxyapatite to Ca 10 (PO 4 ) 6 F 2, which is more resistant to acid. Typically, fluoridation levels are less than 1 mg/L.

31. 31 Chapter 14 © 2011 Pearson Education, Inc. Dilution of a Solution Rather than prepare a solution by dissolving a solid in water, we can prepare a solution by diluting a more concentrated solution. When performing a dilution, the amount of solute does not change, only the amount of solvent. The equation we use is: M 1 x V 1 = M 2 x V 2. –M 1 and V 1 are the initial molarity and volume, and M 2 and V 2 are the new molarity and volume.

32. 32 Chapter 14 © 2011 Pearson Education, Inc. Dilution Problem What volume of 6.0 M NaOH needs to be diluted to prepare 5.00 L if 0.10 M NaOH? We want final volume and we have our final volume and concentration. M 1 x V 1 = M 2 x V 2 (6.0 M) x V 1 = (0.10 M) x (5.00 L) V 1 == 0.083 L (0.10 M) x (5.00 L) 6.0 M

33. 33 Chapter 14 © 2011 Pearson Education, Inc. Solution Stoichiometry In Chapter 10, we performed mole calculations involving chemical equations: stoichiometry problems. We can also apply stoichiometry calculations to solutions. molarity known  moles known  moles unknown  mass unknown solution concentration balanced equation molar mass

34. 34 Chapter 14 © 2011 Pearson Education, Inc. Solution Stoichiometry Problem What mass of silver bromide is produced from the reaction of 37.5 mL of 0.100 M aluminum bromide with excess silver nitrate solution? AlBr 3 (aq) + 3 AgNO 3 (aq) → 3 AgBr(s) + Al(NO 3 ) 3 (aq) We want g AgBr; we have volume of AlBr 3. = 2.11 g AgBr 37.5 mL soln x 3 mol AgBr 1 mol AlBr 3 0.100 mol AlBr 3 1000 mL soln x 1 mol AgBr 187.77 g AgBr x

35. 35 Chapter 14 © 2011 Pearson Education, Inc. Chapter Summary Gas solubility decreases as the temperature increases. Gas solubility increases as the pressure increases. When determining whether a substance will be soluble in a given solvent, apply the like dissolves like rule. –Polar molecules dissolve in polar solvents. –Nonpolar molecules dissolve in nonpolar solvents.

36. 36 Chapter 14 © 2011 Pearson Education, Inc. Chapter Summary, Continued Three factors can increase the rate of dissolving for a solute: 1.Heating the solution 2.Stirring the solution 3.Grinding the solid solute In general, the solubility of a solid solute increases as the temperature increases. A saturated solution contains the maximum amount of solute at a given temperature.

37. 37 Chapter 14 © 2011 Pearson Education, Inc. Chapter Summary, Continued The mass/mass percent concentration is the mass of solute per 100 grams of solution. The molarity of a solution is the moles of solute per liter of solution. moles of solute liters of solution = M mass of solute mass of solution x 100% = m/m %

38. 38 Chapter 14 © 2011 Pearson Education, Inc. Chapter Summary, Continued You can make a solution by diluting a more concentrated solution. M 1 x V 1 = M 2 x V 2 We can apply stoichiometry to reactions involving solutions using the molarity as a unit factor to convert between moles and volume.