2. 1 Solutions Why does a raw egg swell or shrink when placed in different solutions? Chemistry I – Chapters 15 & 16

3. 2 Some Definitions A solution is a HOMOGENEOUS mixture of 2 or more substances in a single phase. One constituent is usually regarded as the SOLVENT and the others as SOLUTES.

4. 3 Parts of a Solution SOLUTE – the part of a solution that is being dissolved (usually the lesser amount) SOLVENT – the part of a solution that dissolves the solute (usually the greater amount) Solute + Solvent = Solution SoluteSolventExample solid Alloys (brass, steel) solidliquidSalt water gassolidAir bubbles in ice cubes liquid “suicides” (mixed drinks) gasliquidSoft drinks gas Air

5. 4 Definitions Solutions can be classified as saturated or unsaturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. An unsaturated solution contains less than the maximum amount of solute that can dissolve at a particular temperature

6. 5 Example: Saturated and Unsaturated Fats Unsaturated fats have at least one double bond between carbon atoms; monounsaturated means there is one double bond, polysaturated means there are more than one double bond. Thus, there are some bonds that can be broken, chemically changed, and used for a variety of purposes. These are REQUIRED to carry out many functions in the body. Fish oils (fats) are usually unsaturated. Game animals (chicken, deer) are usually less saturated, but not as much as fish. Olive and canola oil are monounsaturated. Saturated fats are called saturated because all of the bonds between the carbon atoms in a fat are single bonds. Thus, all the bonds on the carbon are occupied or “saturated” with hydrogen. These are stable and hard to decompose. The body can only use these for energy, and so the excess is stored. Thus, these should be avoided in diets. These are usually obtained from sheep and cattle fats. Butter and coconut oil are mostly saturated fats.

7. 6 Definitions SUPERSATURATED SOLUTIONS contain more solute than is possible to be dissolved Supersaturated solutions are unstable. The supersaturation is only temporary, and usually accomplished in one of two ways: 1.Warm the solvent so that it will dissolve more, then cool the solution 2.Evaporate some of the solvent carefully so that the solute does not solidify and come out of solution.

8. 7 Supersaturated Sodium Acetate Supersaturated Sodium Acetate One application of a supersaturated solution is the sodium acetate “heat pack.”One application of a supersaturated solution is the sodium acetate “heat pack.”

9. 8 C. Johannesson B. Solvation NONPOLAR POLAR “Like Dissolves Like”

10. 9 C. Johannesson B. Solvation  Soap/Detergent –polar “head” with long nonpolar “tail” –dissolves nonpolar grease in polar water

11. 10 C. Johannesson

12. 11 C. Johannesson C. Solubility Solubility CurveSolubility Curve –shows the dependence of solubility on temperature

13. 12 C. Johannesson C. Solubility Solubility CurveSolubility Curve –shows the dependence of solubility on temperature –The line represents- saturation –Below the line- unsaturated –Above the line- Supersaturated

14. 13 Some questions to answer What mass of solute will dissolve in 100g of water at the following temperatures. Which is most soluble KNO3 at 70C NaCl at 100C NH4CL at 90C

15. 14 C. Johannesson C. Solubility Solids are more soluble at...Solids are more soluble at... –high temperatures.  Gases are more soluble at... low temperatures & high pressures (Henry’s Law). EX: nitrogen narcosis, the “bends,” soda  Gases are more soluble at... low temperatures & high pressures (Henry’s Law). EX: nitrogen narcosis, the “bends,” soda

16. 15 Henry’s Law At a given temperature the solubility of a gas in a liquid (S) is directly proportional to the pressure of the gas above the liquid (P). In other words if the pressure increases the solubility increases S 1 = S 2 P 1 P 2

17. 16 IONIC COMPOUNDS Compounds in Aqueous Solution Many reactions involve ionic compounds, especially reactions in water — aqueous solutions. KMnO 4 in water K + (aq) + MnO 4 - (aq)

18. 17 How do we know ions are present in aqueous solutions? The solutions conduct electricity! They are called ELECTROLYTES HCl, MgCl 2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions. Aqueous Solutions

19. 18 Aqueous Solutions Some compounds dissolve in water but do not conduct electricity. They are called nonelectrolytes. Examples include: sugarethanol ethylene glycol Examples include: sugarethanol ethylene glycol

20. 19

21. 20 It’s Time to Play Everyone’s Favorite Game Show… Electrolyte or Nonelectrolyte! Chem videos

22. 21 Electrolytes in the Body  Carry messages to and from the brain as electrical signals  Maintain cellular function with the correct concentrations electrolytes Make your own 50-70 g sugar One liter of warm water Pinch of salt 200ml of sugar free fruit squash Mix, cool and drink

23. 22 Concentration of Solute The amount of solute in a solution is given by its concentration The amount of solute in a solution is given by its concentration. Molarity (M) = moles solute liters of solution

24. 23 1.0 L of water was used to make 1.0 L of solution. Notice the water left over.

25. 24 Steps to make a solution from a SOLID Step 1: Weigh out the amount of solid needed Step 2: place the weighed solid into a VOLUMETRIC FLASK Step 3: while mixing, dissolve, add solvent to bring the level up to Mark on the flask neck. 

26. 25 PROBLEM: Dissolve 5.00 g of NiCl 2 6 H 2 O in enough water to make 250 mL of solution. Calculate the Molarity. Step 1: Calculate moles of NiCl 2 6H 2 O Step 2: Calculate Molarity NiCl 2 6 H 2 O [NiCl 2 6 H 2 O ] = 0.0841 M

27. 26 C. Johannesson D. Preparing Solutions Copyright © 1995-1996 NT Curriculum Project, UW-Madison (above: “Filling the volumetric flask”)

28. 27 Step 1: Change mL to L. 250 mL * 1L/1000mL = 0.250 L Step 2: Calculate. Moles = (0.0500 mol/L) (0.250 L) = 0.0125 moles Step 3: Convert moles to grams. (0.0125 mol)(90.00 g/mol) = 1.13 g USING MOLARITY moles = MV What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. mL of a 0.0500 M solution?

29. 28 Learning Check How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution? 1)12 g 2)48 g 3) 300 g

30. 29 Solution M = moles of solute Liters of solution M * V = moles 3.0 mol/L * 0.400 L = 1.2 mol NaOH 1.2 mole NaOH x 40.0 g NaOH 1 mole NaOH = 48 g NaOH

31. 30 Molarity from Solubility curve Determine the Molarity of a saturated NaCl solution at 25C Go back to solubility curve

32. 31 An IDEAL SOLUTION is one where the properties depend only on the concentration of solute. Need conc. units to tell us the number of solute particles per solvent particle. The unit “molarity” does not do this! Concentration Units

33. 32 Two Other Concentration Units grams solute X100 grams solution MOLALITY, m % by mass = % by mass m of solution= mol solute kilograms solvent

34. 33 Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H 2 O. Calculate molality and % by mass of ethylene glycol.

35. 34 Calculating Concentrations Calculate molality Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H 2 O. Calculate m & % of ethylene glycol (by mass). Calculate mass %

36. 35 Learning Check A solution contains 15 g Na 2 CO 3 and 235 g of H 2 O? What is the mass % of the solution? 1) 15% Na 2 CO 3 2) 6.4% Na 2 CO 3 3) 6.0% Na 2 CO 3

37. 36 Solution mass solute = 15 g Na 2 CO 3 mass solution= 15 g + 235 g = 250 g %(by mass) = 15 g Na 2 CO 3 x 100 250 g solution = 6.0% Na 2 CO 3 solution

38. 37 Using mass % How many grams of NaCl are needed to prepare 250 g of a 10.0% (by mass) NaCl solution? 250 g NaCl soln x 10.0 g NaCl = 25 g NaCl 100 g NaCl soln

39. 38 Try this molality problem 25.0 g of NaCl is dissolved in 5000. mL of water. Find the molality (m) of the resulting solution. m = mol solute / kg solvent 25 g NaCl 1 mol NaCl 58.5 g NaCl = 0.427 mol NaCl Since the density of water is 1 g/mL, 5000 mL = 5000 g, which is 5 kg 0.427 mol NaCl 5 kg water = 0.0854 m salt water

40. 39 Mole Fraction ( ✗ ) N is the number of moles of each substance solute and solvent ✔ all mole fractions add up to = 1

41. 40 Problems If I add 1.65 L of water to 112 grams of sodium nitrate a)What is the molality of NaNO 3 in this solution? b)What is the percent by mass of sodium nitrate in this solution? c)What is the mole fraction of water in this solution?

42. 41 If I add 1.65 L of water to 112 grams of sodium Nitrate… a)What is the molality of NaNO3 in this solution? 0.82 m b)What is the percent by mass of sodium acetate in this solution? 6.36% c)What is the mole fraction of water in this solution? 0.985

43. 42 Preparing Solutions Two types 1) Weigh out a solid solute and dissolve in a given quantity of solvent. 1) Weigh out a solid solute and dissolve in a given quantity of solvent. 2) Dilute a concentrated solution to give one that is less concentrated.2) Dilute a concentrated solution to give one that is less concentrated.

44. 43 SAFETY Safety: “Do as you oughtta, add acid to watta!”

45. 44 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M Dilute the solution!

46. 45 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Conclusion: add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH.

47. 46 A shortcut A shortcut M 1 V 1 = M 2 V 2 M 1 V 1 = M 2 V 2 Moles = Moles Moles = Moles Preparing Solutions by Dilution

48. 47 Making a solution from a STOCK solution link

49. 48 You try this dilution problem You have a stock bottle of hydrochloric acid, which is 12.1 M. You need 400 mL of 0.10 M HCl. How much of the acid and how much water will you need? M 1 V 1 = M 2 V 2 M 1 = 12.1 M V 1 = ??? L M 2 = 0.10 M V 2 = 400 mL  0.400 L V 1 = 0.0033 L (or 3.3 mL HCl) Then add enough water so that the total volume is 400 mL. It should be ABOUT 396.7 mL (400 – 3.3), but it will be off slightly due to the density of the HCl not being 1.00 g/mL

50. 49 Colligative Properties On adding a solute to a solvent, the properties of the PURE solvent are modified. Vapor pressure decreasesVapor pressure decreases Freezing point decreasesFreezing point decreases Boiling point increasesBoiling point increases Osmosis is possible (osmotic pressure)Osmosis is possible (osmotic pressure) These changes are called COLLIGATIVE PROPERTIES. They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles.

51. 50 Change in Freezing Point The freezing point of a solution is LOWER than that of the pure solvent Pure water Ethylene glycol/water solution

52. 51 Change in Freezing Point Common Applications of Freezing Point Depression Propylene glycol Ethylene glycol – deadly to small animals

53. 52 Common Applications of Freezing Point Depression Which would you use for the streets of Bloomington to lower the freezing point of ice and why? Would the temperature make any difference in your decision? a)sand, SiO 2 b)Rock salt, NaCl c)Ice Melt, CaCl 2 Change in Freezing Point

54. 53 Change in Boiling Point Common Applications of Boiling Point Elevation

55. 54 Boiling point elevation The solute gets in the way of the molecules escaping to become vapor..

56. 55 Boiling Point Elevation and Freezing Point Depression ∆T = Kmi ∆T = Kmi i = van’t Hoff factor = number of particles produced per molecule/formula unit. For covalent compounds, i = 1. For ionic compounds, i = the number of ions present (both + and -) CompoundTheoretical Value of i glycol1 NaCl2 CaCl 2 3 Ca 3 (PO 4 ) 2 5

57. 56 Boiling Point Elevation and Freezing Point Depression ∆T = Kmi ∆T = Kmi SubstanceKbKb benzene2.53 camphor5.95 carbon tetrachloride5.03 ethyl ether2.02 water0.52 m = molality K = molal freezing point/boiling point constant SubstanceKfKf benzene5.12 camphor40. carbon tetrachloride30. ethyl ether1.79 water1.86

58. 57 Change in Boiling Point Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the boiling point of the solution? K b = 0.52 o C/molal for water (see K b table). Solution∆T BP = K b m i 1.Calculate solution molality = 4.00 m 2.∆T BP = K b m i ∆T BP = 0.52 o C/molal (4.00 molal) (1) ∆T BP = 0.52 o C/molal (4.00 molal) (1) ∆T BP = 2.08 o C BP = 100 + 2.08 = 102.08 o C (water normally boils at 100)

59. 58 Calculate the Freezing Point of a 4.00 molal glycol/water solution. K f = 1.86 o C/molal (See K f table) Solution ∆T FP = K f m i = (1.86 o C/molal)(4.00 m)(1) = (1.86 o C/molal)(4.00 m)(1) ∆T FP = 7.44 FP = 0 – 7.44 = -7.44 o C (because water normally freezes at 0) Freezing Point Depression

60. 59 At what temperature will a 5.4 molal solution of NaCl freeze? Solution ∆T FP = K f m i ∆T FP = (1.86 o C/molal) 5.4 m 2 ∆T FP = (1.86 o C/molal) 5.4 m 2 ∆T FP = 20.1 o C ∆T FP = 20.1 o C FP = 0 – 20.1 = -20.1 o C FP = 0 – 20.1 = -20.1 o C Freezing Point Depression

61. 60 Colligative properties tutorial and quiz linkColligative properties tutorial and quiz

62. 61 Molecular mass determination A Solution of 7.50 g of a nonvolatile compound in 22.60 g of water boils at 100.78°C at 760 mm Hg. What is the molecular mass of the solute? (assume i =1) ΔT b = K b X m X i  m = ΔT K b The K b for water is 0.512 °C/molal m = 0.78°C 0.512 °C/molal m =1.52 the units of Molaity are moles solute/Kg solvent

63. 62 Now calculate moles of solute in solution 1.5 m X 22.6 g of water X 1kg/1000g =0.034 mol solute Finally use the number of moles of solute and its mass to determine the molecular mass of the solute. MM of solute = mass of solute/moles of solute =7.50 g/0.0344mol= 2.2X10 2 g/mol

64. 63 Problem The freezing point for water is lowered to - 0.390 C when 3.90 grams of a molecular solute is dissolved in 475 g of water. Calculate the molar mass of the solute.

65. 64 osmosis

66. 65 Setup for titrating an acid with a base

67. 66 TitrationTitration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. 3.Indicator shows when exact stoichiometric reaction has occurred. (Acid = Base) This is called NEUTRALIZATION. This is called NEUTRALIZATION.

68. 67 35.62 mL of NaOH is neutralized with 25.2 mL of 0.0998 M HCl by titration to an equivalence point. What is the concentration of the NaOH? LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately determine its concentration.

69. 68 35.62 mL of NaOH is neutralized with 25.2 mL of 0.0998 M HCl by titration to an equivalence point. What is the concentration of the NaOH? M a V a = M b V b M a V a = M b V b (0.0998 M) (25.2 mL) = 0.0706 M (35.62 mL) M x V = moles Since the moles of the acid and base are equal and do not change: M x V (acid) = M x V (base)