1. Solutions

2. Solution Definitions Solution:Solution: homogeneous mixture, evenly mixed at the particle level (like salt water). Solvent: Solvent: present in greater amount that can dissolve the solute Solute: Solute: substance being dissolved

3. Classes of Solutions aqueous solution: water = “the universal solvent” solvent = water amalgam:solvent = Hg e.g., dental amalgam tincture:solvent = alcohol e.g., tincture of iodine (for cuts) organic solution:solvent contains carbon e.g., gasoline, benzene, toluene, hexane

4. Solution Definitions alloy: solvent: the substance that dissolves the solute watersalt a solid solution of metals -- e.g., bronze = Cu + Sn; brass = Cu + Zn “will dissolve in” refers to two gases or two liquids that form a solution; more specific than the term “soluble” miscible: soluble:

5. Solubility Solubility - Solubility - maximum grams of solute that will dissolve in 100 g of solvent at a given temperature varies with temperature

6. Solubility SATURATED SOLUTION no more solute dissolves UNSATURATED SOLUTION more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form increasing concentration

7. Temp. ( o C) Solubility (g/100 g H 2 O) KNO 3 (s) KCl (s) HCl (g) SOLUBILITY CURVE Solubility  how much solute dissolves in a given amt. of solvent at a given temp. below unsaturated:solution could hold more solute; below line on saturated:solution has “just right” amt. of solute; on line supersaturated:solution has “too much” solute dissolved in it; above the line

8. Solubility Table LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World, 1996, page 517 0 10 20 30 40 50 60 70 80 90 100 Temperature (°C) Solubility vs. Temperature for Solids Solubility (grams of solute/100 g H 2 O) KI KCl 20 10 30 40 50 60 70 80 90 110 120 130 140 100 NaNO 3 KNO 3 HClNH 4 Cl NH 3 NaCl KClO 3 SO 2 shows the dependence of solubility on temperature gases solids

9. per 100 g H 2 O Classify as unsaturated, saturated, or supersaturated. 80 g NaNO 3 @ 30 o Cunsaturated 60 g KCl @ 60 o Csaturated 50 g NH 3 @ 10 o Cunsaturated 70 g NH 4 Cl @ 70 o Csupersaturated So sat. pt. @ 40 o C for 500 g H 2 O = 5 x 66 g = 330 g 120 g < 330 gunsaturated saturation point @ 40 o C for 100 g H 2 O = 66 g KNO 3 Per 500 g H 2 O, 120 g KNO 3 @ 40 o C

10. Solubility  Solids are more soluble at... high temperatures.  Gases are more soluble at... low temperatures

11. (A) Per 100 g H 2 O, 100 g Unsaturated; all solute NaNO 3 @ 50 o C. dissolves; clear solution. (B) Cool solution (A) very Supersaturated; extra slowly to 10 o C. solute remains in solution; still clear. Describe each situation below. (C) Quench solution (A) in Saturated; extra solute an ice bath to 10 o C. (20 g) can’t remain in solution, becomes visible.

12. Non-Solution Definitions insoluble:“will NOT dissolve in” e.g., sand and water immiscible:refers to two gases or two liquids that will NOT form a solution e.g., water and oil suspension:appears uniform while being stirred, but settles over time

13. Water HOT Solubility A B Before Water COLD Add 1 drop of red food coloring Miscible – “mixable” two gases or two liquids that mix evenly Experiment 1: Water HOT AFTER Water COLD A B

14. Solubility Water Oil T 30 sec AFTER Before Add oil to water and shake Immiscible – “does not mix” two liquids or two gases that DO NOT MIX Experiment 2: T 0 sec

15. Solutions What the solute and the solvent are determines: –whether a substance will dissolve. –how much will dissolve.

16. As size, rate As T, rate 3. mixing Factors Affecting the Rate of Dissolution 1. temperature 2. particle size 4. nature of solvent or solute More mixing, rate

17. Solid Solubility Timberlake, Chemistry 7 th Edition, page 297 KI NaNO 3 KNO 3 Na 3 PO 4 NaCl Temperature ( o C) Solubility (g solute / 100 g H 2 O) 200 180 160 140 120 100 80 60 40 20 0 406080100

18. Gas Solubility CH 4 O2O2 CO He Temperature ( o C) Solubility (mM) 2.0 1.0 01020304050 Higher Temperature …Gas is LESS Soluble

19. Particle Size and Mixing In order to dissolve - the solvent molecules must come in contact with the solute. Stirring moves fresh solvent next to the solute. The solvent touches the surface of the solute. Smaller pieces increase the amount of surface of the solute.

20. Molecular Polarity Nonpolar molecules: electrons are shared equally –Molecule has no charge e.g. Ethane

21. O 2- H+H+ H+H+  ++ Polar Molecules Electrons are not equally shared Molecule has slight charge “Like dissolves like” (Polarity)

22. Solubility Rules 1.Most nitrate (NO 3 1- ) salts are soluble. 2.Most salts of Na +, K +, and NH 4 + are soluble. 3.Most chloride salts are soluble. Notable exceptions are AgCl, PbCl 2, and Hg 2 Cl 2. 4.Most sulfate salts are soluble. Notable exceptions are BaSO 4, PbSO 4, and CaSO 4. 5.Most hydroxide compounds are only slightly soluble.* The important exceptions are NaOH and KOH, Ba(OH) 2 and Ca(OH) 2 are only moderately soluble. 6.Most sulfide (S 2- ), carbonate (CO 3 2- ), and phos- phate (PO 4 3- ) salts are only slightly soluble. *The terms insoluble and slightly soluble really mean the same thing. Such a tiny amount dissolves that it is not possible to detect it with the naked eye. General Rules for Solubility of Ionic Compounds (Salts) in Water at 25 o C Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 218

23. Oil and Water Don’t Mix Oil is nonpolar Water is polar “Like dissolves like” polar + polar = solution nonpolar + nonpolar = solution polar + nonpolar = suspension (won ’ t mix evenly) Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 470

24. Solvation  Soap / Detergent polar “head” with long nonpolar “tail” dissolves nonpolar grease in polar water micelle

25. Solubility of Common Compounds NO 3 - salts Na +, K +, NH 4 + salts Soluble compounds Cl -, Br -, I - salts S 2-, CO 3 2-, PO 4 3- salts OH 1- salts SO 4 2- salts Except for those containing Ag +, Hg 2 2+, Pb 2+ Except for those containing Ba 2+, Pb 2+, Ca 2+ Insoluble compounds Except for those containing Na +, K +, Ca 2+ Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 218

26. Making Molar Solutions …from liquids (More accurately, from stock solutions)

27. Concentration…a measure of solute-to-solvent ratio concentrated vs. dilute “lots of solute” “not much solute” “watery” Add water to dilute a solution; boil water off to concentrate it.

28. Concentration “The amount of solute in a solution” mol L M A. mass % = mass of solute mass of sol’n B. parts per million (ppm)  also, ppb and ppt – commonly used for minerals or contaminants in water supplies C. molarity (M) = moles of solute L of sol’n – used most often in this class D. molality (m) = moles of solute kg of solvent M = mol L % by mass – medicated creams % by volume – rubbing alcohol

29. V = 1000 mL n = 2 moles Concentration = 2 molar V = 1000 mL n = 4 moles [ ] = 4 molar V = 5000 mL n = 20 moles [ ] = 4 molar Molarity (M) = # of moles volume (L) V = 250 mL n = 8 moles [ ] = 32 molar

30. Glassware

31. precise; expensive; good for making solutions Range: Glassware – Precision and Cost beakervs.volumetric flask When filled to 1000 mL line, how much liquid is present? beaker 5% of 1000 mL = 50 mL volumetric flask 1000 mL + 0.30 mL 950 mL – 1050 mL 999.70 mL– 1000.30 mL imprecise; cheap Range:

32. How to mix a Standard Solution Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 480 Wash bottle Volume marker (calibration mark) Weighed amount of solute

33. How to mix solid chemicals How many grams of sodium hydroxide would we need to make 100. mL of 3.0 M NaOH? M = mol L 3 M = ? mol 1 L How much will this weigh? 1 Na @ 23g/mol + 1O @ 16g/mol + 1 H @ 1 g/mol MM NaOH = 40g/mol 40.0 g NaOH 1 mol NaOH X g NaOH = 3.0 mol NaOH = To mix this: add 120 g NaOH into 1L volumetric flask with~750 mL cold H 2 O. Mix, allow to return to room temperature – bring volume to 1 L. 120 g NaOH

34. 40.0 g NaOH 1 mol NaOH How many moles of solute are required to make 1.35 L of 2.50 M solution? mol = M L B. What mass of magnesium phosphate is this? A. What mass of sodium hydroxide is this? mol L M 3.38 mol = 2.50 M (1.35 L) = X g NaOH = 3.38 mol NaOH = 135 g NaOH 262.9 g Mg 3 (PO 4 ) 2 1 mol Mg 3 (PO 4 ) 2 X g Mg 3 (PO 4 ) 2 = 3.38 mol Mg 3 (PO 4 ) 2 = 889 g Mg 3 (PO 4 ) 2

35. 0.342 mol 5.65 L Find molarity if 58.6 g of barium hydroxide are in 5.65 L solution. 171.3 g Ba(OH) 2 1 mol Ba(OH) 2 Step 2). What is the molarity of a 5.65 L solution containing 0.342 mol solute? Step 1). How many moles barium hydroxide is this? 0.0605 M Ba(OH) 2 = X mol Ba(OH) 2 = 58.6 g Ba(OH) 2 = 0.342 mol Ba(OH) 2 M = mol L M =

36. You have 10.8 g potassium nitrate. How many mL of solution will make this a 0.14 M solution? convert to mL

37. Molality mass of solvent only 1 kg water = 1 L water

38. Molality  Find the molality of a solution containing 75 g of MgCl 2 in 250 mL of water. 75 g MgCl 2 1 mol MgCl 2 95.21 g MgCl 2 = 3.2 m MgCl 2 0.25 kg water

39. Molality  How many grams of NaCl are required to make a 1.54 m solution using 0.500 kg of water? 0.500 kg water1.54 mol NaCl 1 kg water = 45.0 g NaCl 58.44 g NaCl 1 mol NaCl

40. 500 mL volumetric flask Preparing Solutions 500 mL of 1.54M NaCl 500 mL water 45.0 g NaCl –mass 45.0 g of NaCl –add water until total volume is 500 mL –mass 45.0 g of NaCl –add 0.500 kg of water (volume will be just over 500 total mL) 500 mL mark 1.54m NaCl in 0.500 kg of water molality molarity

41. Dilution Preparation of a desired solution by adding water to a concentrate. Moles of solute remain the same.

42. Dilution What volume of 15.8M HNO 3 is required to make 250 mL of a 6.0M solution? GIVEN: M 1 = 15.8M V 1 = ? M 2 = 6.0M V 2 = 250 mL WORK: M 1 V 1 = M 2 V 2 (15.8M) V 1 = (6.0M)(250mL) V 1 = 95 mL of 15.8M HNO 3

43. occurs when neutral combinations of particles separate into ions while in aqueous solution. sodium chloride sodium hydroxide hydrochloric acid sulfuric acid acetic acid Dissociation: NaCl  Na 1+ + Cl 1– NaOH  Na 1+ + OH 1– HCl  H 1+ + Cl 1– H 2 SO 4  2 H 1+ + SO 4 2– CH 3 COOH  CH 3 COO 1– + H 1+

44. NaCl Na 1+ + Cl 1– CH 3 COOH CH 3 COO 1– + H 1+ Weak electrolytes exhibit little dissociation. “Strong” or “weak” is a property of the substance. We can’t change one into the other. Strong electrolytes exhibit nearly 100% dissociation. NOT in water:10000 0 in aq. solution: 1 999 999 NOT in water: 1000 0 0 in aq. solution: 980 20 20

45. electrolytes: solutes that dissociate in solution -- conduct electric current because of free-moving ions e.g., acids, bases, most ionic compounds -- are crucial for many cellular processes -- obtained in a healthy diet -- For sustained exercise or a bout of the flu, sports drinks ensure adequate electrolytes. nonelectrolytes: solutes that DO NOT dissociate -- DO NOT conduct electric current (not enough ions) e.g., any type of sugar

46. …normal boiling point (NBP)…higher BP FREEZING PT. DEPRESSION BOILING PT. ELEVATION Colligative Properties  depend on concentration of a solution Compared to solvent‘s: a solution w/that solvent has a: …normal freezing point (NFP)…lower FP

47. 1. salting roads in winter FP BP water0 o C (NFP) 100 o C (NBP) 2. antifreeze (AF) /coolant FP BP water0 o C (NFP) 100 o C (NBP) water + a little AF–10 o C110 o C 50% water + 50% AF–35 o C130 o C water + a little salt water + more salt –11 o C 103 o C –18 o C105 o C Applications (NOTE: Data are fictitious.)

48. 3. law enforcement white powder starts melting at… finishes melting at… penalty, if convicted A109 o C175 o Ccomm. service B150 o C180 o C2 years C194 o C196 o C20 years

49. Calculations for Colligative Properties The change in FP or BP is found using…  T x = K x m i  T x = change in T o (below NFP or above NBP) K x = constant depending on… (A) solvent (B) freezing or boiling m = molality of solute = mol solute / kg solvent i = integer that accounts for any solute dissociation any sugar (all nonelectrolytes)……………...i = 1 table salt, NaCl  Na 1+ + Cl 1– ………………i = 2 barium bromide, BaBr 2  Ba 2+ + 2 Br 1– ……i = 3

50. Freezing Point DepressionBoiling Point Elevation  T f = K f m i  T b = K b m i Then use these in conjunction with the NFP and NBP to find the FP and BP of the mixture. (K b = ebullioscopic constant, which is 0.51 K kg/mol for the boiling point of water) (K f = cryoscopic constant, which is 1.86 K kg/mol for the freezing point of water)

51. (NONELECTROLYTE) 168 g glucose, C 6 H 12 O 6, (a sugar, nonelectrolyte) are mixed w /2.50 kg H 2 O. Find BP and FP of mixture. For water, K b = 0.512, K f = –1.86. i = 1  T b = K b m i = 0.512 (0.373) (1) = 0.19 o C BP = (100 + 0.19) o C = 100.19 o C  T f = K f m i = –1.86 (0.373) (1) = –0.694 o C FP = (0 + –0.69) o C = –0.694 o C

52. 168 g cesium bromide are mixed w /2.50 kg H 2 O. Find BP and FP of mixture. For H 2 O, K b = 0.512, K f = –1.86.  T b = K b m i = 0.512 (0.316) (2) = 0.32 o C BP = (100 + 0.32) o C = 100.32 o C  T f = K f m i = –1.86 (0.316) (2) = –1.18 o C FP = (0 + –1.18) o C = –1.18 o C Cs 1+ Br 1– i = 2 CsBr  Cs 1+ + Br 1–

53. 1 Pb(NO 3 ) 2 (aq) + KI (aq)  PbI 2 (s) + KNO 3 (aq) __ 212 Molarity and Stoichiometry M M V V P P mol M L M = mol L mol = M L What volume of 4.0 M KI solution is required to yield 89 g PbI 2 ?

54. 461 g PbI 2 Strategy: 1 Pb(NO 3 ) 2 (aq) + 2 KI (aq)  1 PbI 2 (s) + 2 KNO 3 (aq) X mol KI = 89 g PbI 2 1 mol PbI 2 2 mol KI = 0.39 mol KI (1) Find mol KI needed to yield 89 g PbI 2. (2) Based on (1), find volume of 4.0 M KI solution. What volume of 4.0 M KI solution is required to yield 89 g PbI 2 ? 89 g ? L 4.0 M M = mol L L = mol M = 0.39 mol KI 4.0 M KI = 0.098 L of 4.0 M KI

55. = 0.173 mol CuSO 4 How many mL of a 0.500 M CuSO 4 solution will react w /excess Al to produce 11.0 g Cu? __CuSO 4 (aq) + __Al (s)  Al 3+ SO 4 2– CuSO 4 (aq) + Al (s)  Cu(s) + Al 2 (SO 4 ) 3 (aq)3231 x mol11 g __Cu(s) + __Al 2 (SO 4 ) 3 (aq) X mol CuSO 4 = 11 g Cu 1 mol Cu 63.5 g Cu 3 mol CuSO 4 3 mol Cu M = mol L L = mol M 0.173 mol CuSO 4 0.500 M CuSO 4 = 0.346 L 0.346 L 1000 mL 1 L = 346 mL

56. Limiting Reactants 79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP? Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? L 0.90 L 2.5M Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

57. Limiting Reactants 79.1 g Zn 1 mol Zn 65.39 g Zn = 27 L H 2 1 mol H 2 1 mol Zn 22.4 L H 2 1 mol H 2 Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? L 0.90 L 2.5M Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

58. Limiting Reactants 22.4 L H 2 1 mol H 2 0.90 L 2.5 mol HCl 1 L = 25 L H 2 1 mol H 2 2 mol HCl Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? L 0.90 L 2.5M Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

59. Limiting Reactants Zn: 27 L H 2 HCl: 25 L H 2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25 L H 2 left over zinc Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem