1. www.cengage.com/chemistry/cracolice Mark S. Cracolice Edward I. Peters Mark S. Cracolice The University of Montana Chapter 16 Solutions

2. Solution Solution is a homogenous mixture. A solution can exist in any of the three states: Solid solution: steel, brass, bronze Liquid solution: alcohol in water, sugar water. Gaseous solution: air, any mixture of gases.

3. Solution Terminology Solute The substance present in a relatively small amount in a solution; the solid or gas when a substance in that state is dissolved in a liquid to make a solution. Solvent The substance present in a relatively large amount in a solution; the liquid when a solid or gas is dissolved to make a solution.

4. Solution Terminology Concentrated Solution has a relatively large quantity of a specific solute per unit amount of solution. Dilute Solution has a relatively small amount of a specific solute per unit amount of solution.

5. Solution Terminology Saturated solution A solution that can exist in equilibrium with undissolved solute is a saturated solution. Solubility The concentration of the saturated solution is called the solubility of the solute. Unsaturated A solution is unsaturated if its concentration is smaller than the solubility.

6. Solution Terminology Supersaturated A solution is said to be supersaturated if its concentration is greater than the solubility. Supersaturated solution is not stable, a slight physical disturbance can start crystallization

7. The Formation of a Solution The water molecule is polar. A polar molecule is one with an asymmetrical distribution of charge, resulting in positive and negative poles.

8. The Formation of a NaCl Solution

9. The formation of a sodium chloride solution from sodium chloride crystals is made possible by the interaction of water molecules with sodium and chloride ions. NaCl (s) ↔ Na + (aq) + Cl - (aq) The sodium and chloride ions in solution are surrounded by polar water molecules, and are said to be hydrated.

10. The Formation of a Solution The rate of dissolving depends on 1 The surface area. A finely divided solid dissolves more rapidly. 2 The diffusion of the solute from the surface. Stirring increases the rate. 3 Temperature. The higher the temperature, the higher the rate.

11. The Formation of a Solution Crystallization rate The rate per unit of surface area increases as the solution concentration at the surface increases. Equilibrium in a Saturated Solution When the rates of dissolving and crystallization are the same the solution is saturated.

12. The Formation of a Saturated Solution When crystallization rate is equal to the dissolving rate the solution is saturated.

13. Factors That Determine Solubility Solubility depends on three factors: Intermolecular forces Gas pressure (for a gas) Temperature

14. Factors That Determine Solubility Intermolecular Forces Generally, if the forces between molecules A are about the same as the forces between molecules B, A and B will probably dissolve in each other. The rule of solubility is like dissolves like. A nonpolar solvent can dissolve a nonpolar solute. A polar solvent can dissolve a polar or ionic solute

15. Factors That Determine Solubility Ethylene glycol HO-CH 2 –CH 2 –OH, which is polar and can form hydrogen bond, is soluble in water. Motor oil is not soluble in water because hydrocarbon molecules are nonpolar.

16. Factors That Determine Solubility Partial Pressure of Solute Gas Over Liquid Solution Pressure has little effect on the solubility of solids or liquids but has a pronounced effect on the solubility of gases. For an ideal solution, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas over the surface of the liquid.

17. Factors That Determine Solubility Solubility of carbon dioxide decreases when partial pressure of carbon dioxide drops

18. Factors That Determine Solubility Temperature The solubility of most solids increases with rising temperature (but there are notable exceptions). The solubilities of gases in liquids are generally lower at higher temperatures.

19. Factors That Determine Solubility

21. Percentage by Mass Percentage by mass can be used as conversion factor between mass of solution and mass of solute.

22. Percentage by Mass When 125 grams of a solution is evaporated to dryness, 42.3 grams of solute was recovered. What was the percentage of the solute?

23. Molarity Molarity, M Moles of solute per liter of solution: Molarity can be used as conversion factor between liters of solution and moles of solute. Volume of solution x molarity = number of moles

24. Molarity Example: Calculate the molarity of a solution made by dissolving 15 g of NaOH in water and diluting to 1.00 x 10 2 mL Solution: First calculate the number of moles NaOH

25. Molarity Next calculate the volume of solution in liters 1.00 x 10 2 Calculate the answer using the definition

26. Molarity Example: How many grams of silver nitrate must be dissolved to prepare 5.00 x10 2 mL of 0.150 M AgNO 3 5.00 x10 2 mL x ( 1 L/ 1000 mL) x(0.150 mol AgNO 3 /L) x(169.9 g AgNO 3 /mol AgNO 3 ) = 12.7 g AgNO 3

27. Molarity Example Find the volume of a 1.40 M solution that contains 0.287 mole of ammonia. Solution 0.287 mole NH 3 x ( 1L/ 1.40 mole NH 3 ) = 0.205 L

28. Molarity To prepare a solution of a specified molarity: 1. Weigh the appropriate amount of solute. 2. Add less than the total volume of solvent. 3. Mix to completely dissolve the solute. 4. Add additional solvent until the total solution volume is appropriate.

29. Preparation of 250 mL Solution KMnO 4 0.0100M 250 mL x (1L/1000 mL) x (0.00100 mol KMnO 4 /1L) x (158.04 g KMnO 4 / 1 mol KMnO 4 ) = 0.395 g KMnO 4

30. Molality Molality, m The number of moles of solute dissolved in one kilogram of solvent: Molarity (mol/L) is temperature dependent; molality is temperature independent.

31. Normality: Equivalent Equivalent, eq One equivalent of acid is the quantity that yields one mole of hydrogen ions in a chemical reaction. Once equivalent of base is the quantity that reacts with one mole of hydrogen ions.

32. Normality: Equivalent eq /mol H 3 PO 4 + NaOH → NaH 2 PO 4 + H 2 O1 H 3 PO 4 + 2 NaOH → Na 2 HPO 4 + 2 H 2 O2 H 3 PO 4 + 3 NaOH → Na 3 PO 4 + 3 H 2 O3 Number of equivalents of acid = Number of equivalents of base

33. Normality: Equivalent Mass Equivalent mass: The number of grams per equivalent. H 3 PO 4 + NaOH → NaH 2 PO 4 + H 2 O97.99 g/eq H 3 PO 4 + 2 NaOH → Na 2 HPO 4 + 2 H 2 O97.99 g/2 eq H 3 PO 4 + 3 NaOH → Na 3 PO 4 + 3 H 2 O97.99 g/3 eq Equivalent mass of H 3 PO 4 for the three reactions are respectively: 97.99 g/eq, 49.00 g/eq, and 32.66 g/eq.

34. Normality Normality, N The number of equivalents, eq, per liter of solution:

35. Normality Example: What is the normality of a solution made by dissolving 2.50 grams of sodium hydroxide in water to make 5.00 × 10 2 mL of solution? Solution: By definition, Therefore, we need to find equivalents of solute and liters of solution.

36. Normality What is the normality of a solution made by dissolving 2.50 grams of sodium hydroxide in water to make 5.00 × 10 2 mL of solution? Equivalents of solute: G IVEN: 2.5 g NaOHW ANTED: eq NaOH

37. Normality What is the normality of a solution made by dissolving 2.50 grams of sodium hydroxide in water to make 5.00 × 10 2 mL of solution? G IVEN: 5.00 × 10 2 mLW ANTED: L

38. Normality What is the normality of a solution made by dissolving 2.50 grams of sodium hydroxide in water to make 5.00 × 10 2 mL of solution? We now have both parts of the normality fraction, 0.0625 eq NaOH and 0.500 L solution:

39. Solution Concentration: Summary

40. Dilution of Concentrated Solutions Concentrated solutions are diluted by adding more solvent. The amount of solute remains the same before and after a dilution. M c x V c = number of moles = M d x V d M c x V c = M d x V d

41. Dilution of Concentrated Solutions Example: How many milliliters of concentrated hydrochloric acid, which is 11.6 M, should be used to prepare 5.50 liters of 0.500 M HCl ? V c x M c = V d x M d V c x 11.6 M = 5.50 L x 0.500 M V c = 5.50 L x 0.500 M /11.6 M = 0.237 L = 237 mL

42. Dilution of Concentrated Solutions If the volume of solution is increased 10 times, the concentration of diluted solution is 10 times smaller.

43. Solution Stoichiometry How many grams of lead(II) iodide will precipitate when excess potassium iodide solution is added to 50.0 mL of 0.811 M lead(II) nitrate. Pb(NO 3 ) 2 (aq) + 2 KI(aq)  PbI 2 (s) + 2 KNO 3 (aq) Given: 50.0 mL of 0.811 M lead(II) nitrate Wanted: gram of lead(II) iodide. Need to convert volume of solution to moles of solute

44. Solution Stoichiometry How many grams of lead(II) iodide will precipitate when excess potassium iodide solution is added to 50.0 mL of 0.811 M lead(II) nitrate. Pb(NO 3 ) 2 (aq) + 2 KI(aq)  PbI 2 (s) + 2 KNO 3 (aq) mL of Pb(NO 3 ) 2  liters of Pb(NO 3 ) 2  moles of Pb(NO 3 ) 2  moles of PbI 2  gram of PbI 2 50.0 mL of Pb(NO 3 ) 2 x (L of Pb(NO 3 ) 2 / 1000 mLof Pb(NO 3 ) 2 ) x (0.811 mol of Pb(NO 3 ) 2 /L Pb(NO 3 ) 2 ) x (1 mol PbI 2 /1 mol Pb(NO 3 ) 2 ) x (461.0 g PbI 2 /mol PbI 2 ) = 18.7 g PbI 2

45. Titration Using Molarity Titration: the very careful addition of one solution to another by a burette.

46. Titration Using Molarity Standardize Determination of the concentration of a solution to be used in a titration by titrating it against a primary standard. Primary Standard A soluble solid of reasonable cost that is very stable and pure, preferably with a high molar mass, that can be weighed accurately for use in a titration.

47. Titration Using Molarity 1.18 g oxalic acid, H 2 C 2 O 4. 2H 2 O (126.07 g/mol), are dissolved in water and the solution is titrated with a solution of NaOH of unknown concentration. 28.3 ml NaOH are required to neutralize the acid. Calculate the molarily of the NaOH solution. H 2 C 2 O 4 + 2 NaOH → Na 2 C 2 O 4 + 2 H 2 O Given 1.18 g H 2 C 2 O 4. 2H 2 O, 28.3 mL (0.0283L) NaOH Wanted mol/L NaOH g H 2 C 2 O 4. 2H 2 O  mol H 2 C 2 O 4. 2H 2 O  mol H 2 C 2 O 4   mol NaOH  molarity NaOH

48. Titration Using Molarity 1.18 g oxalic acid, H 2 C 2 O 4. 2H 2 O (126.07 g/mol), are dissolved in water and the solution is titrated with a solution of NaOH of unknown concentration. 28.3 ml NaOH are required to neutralize the acid. Calculate the molarily of the NaOH solution. H 2 C 2 O 4 + 2 NaOH → Na 2 C 2 O 4 + 2 H 2 O 1.18 g H 2 C 2 O 4. 2H 2 O x (1 mol H 2 C 2 O 4. 2H 2 O/ 126.07g H 2 C 2 O 4. 2H 2 O ) x (1 mol H 2 C 2 O 4. / 1 mol H 2 C 2 O 4. 2H 2 O) x ( 2 mol NaOH/1 mol H 2 C 2 O 4.) x ( 1/ 0.0283L NaOH) = 0.661 mol/L

49. Titration Using Normality The number of equivalents of all species in a reaction is the same. For an acid–base reaction, equivalents of acid = equivalents of base

50. Titration Using Normality A 25.0 mL sample of an electroplating solution is analyzed for its sulfuric acid concentration. It takes 46.8 mL of the 0.661 N NaOH to neutralize the sample, find the normality of the acid. N a x V a = N b x V b N a x 25.0 mL = 46.8 mL x 0.661 N N a = (46.8 mL / 25.0 mL ) x 0.661 N = 1.24 N H 2 SO 4

51. Colligative Properties of Solutions A pure solvent has distinct physical properties. Introducing a solute into the solvent affects these properties. In dilute solutions of certain solutes, the change in some of these properties is proportional to the molal concentration of the solute particles. Colligative Property Solution property that is determined only by the number of solute particles dissolved in a fixed quantity of solvent and not by the identity of the solute particles.

52. Colligative Properties of Solutions Boiling Point Elevation The boiling point of a solution is higher than the boiling point of the pure solvent. Example: The normal boiling point of water is 100.0°C. The normal boiling point of a 1 m solution of sugar water is 100.5°C.

53. Colligative Properties of Solutions Freezing Point Depression The freezing point of a solution is lower than the freezing point of the pure solvent. Example: The normal freezing point of water is 0.0°C. The normal boiling point of a 1 m solution of sugar water is –1.9°C.

54. Colligative Properties of Solutions

55. Boiling point elevation and freezing point depression are colligative properties: They depend on the number of solute particles but not their identity. ∆T b = change in boiling temperature ∆T f = change in freezing temperature

56. Colligative Properties of Solutions ∆T b = K b × m∆T f = K f × m K b = molal boiling-point elevation constant K f = molal freezing-point depression constant

57. Colligative Properties of Solutions Molal Boiling Point Elevation Constant Values SubstanceBoiling Point (°C)K b (°C/m) Benzene 802.5 Carbon disulfide 462.4 Carbon tetrachloride 775.0 Water1000.52 Molal Freezing Point Depression Constant Values SubstanceFreezing Point (°C)K f (°C/m) Benzene 6 5.1 Carbon disulfide–112 3.8 Carbon tetrachloride – 2330 Water 0 1.86

58. Colligative Properties of Solutions What is the freezing point of a solution made by adding 12.0 g of urea, CO(NH 2 ) 2, to 2.50 × 10 2 grams of water?

59. Colligative Properties of Solutions ∆T f = K f x m = 1.86 °C/m x 0.800 m = 1.49 0 C T f = 0°C – 1.49°C = –1.49 °C What is the freezing point of a solution made by adding 12.0 g of urea, CO(NH 2 ) 2, to 2.50 × 10 2 grams of water?

60. Determination of molar mass Calculate the Molar Mass of a Solute from Freezing Point Depression or Boiling Point Elevation Data 1. Calculate molality from m = ∆T f /K f or m = ∆T b /K b. Express as mol solute/kg solvent. 2. Using molality as a conversion factor between moles of solute and kilograms of solvent, find the number of moles of solute. 3. Use the defining equation for molar mass, MM = g/mol, to calculate the molar mass of the solute.

61. Determination of molar mass from ∆T b Example. K b of benzene is 2.5 0 C/m. A solution of 15.2 g of unknown solute in 91.0 g benzene boils at a temperature 2.1 C higher than the boiling point of pure benzene. Calculate the molar mass of the solute. Given: ΔT b = 2.1 0 C, K b = 2.5 0 C/m Need: molar mass First, calculate molality m : ∆T b = K b × m

62. Determination of molar mass from ∆T b Example. K b of benzene is 2.5 0 C/m. A solution of 15.2 g of unknown solute in 91.0 g benzene boils at a temperature 2.1 C higher than the boiling point of pure benzene. Calculate the molar mass of the solute. ∆T b = K b × m m = ΔT b / K b = 2.1 0 C/(2.5 0 C/m) = 0.84 mol/kg Then calculate moles of solute and molar mass Moles of solute in 91.0 g of benzene = 0.84 mol/kg x ( 1kg/1000g) x 91.0 g = 0.077 mol Molar mass = g/ mol = 15.2 g/ 0.077 mol = 197.40 g/mol = 200 g/mol

63. Homework Homework: 33, 43, 45, 49, 61, 71, 85, 89, 93, 103, 111, 121, 125, 127, 132.