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Solutions. Chemical Stewardship Be responsible in how you dispose of and use chemicals. Chemical pollution can travel far – and harm organisms. Frog with.

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Presentation on theme: "Solutions. Chemical Stewardship Be responsible in how you dispose of and use chemicals. Chemical pollution can travel far – and harm organisms. Frog with."— Presentation transcript:

1 Solutions

2 Chemical Stewardship Be responsible in how you dispose of and use chemicals. Chemical pollution can travel far – and harm organisms. Frog with three legs – it has mutated from chemical exposure.

3 Solutions What the solute and the solvent are determines –whether a substance will dissolve. –how much will dissolve. A substance dissolves faster if it is stirred or shaken. –The particles are made smaller. –The temperature is increased. Why?

4 Solution = Solute + Solvent Solute - gets dissolved Solvent- does the dissolving –A–Aqueous (water) –T–Tincture (alcohol) –A–Amalgam (mercury) –O–Organic Polar Non-polar Dental filling Nightmare on White Street Chem Matters, December 1996

5 Solution Definitions solution: alloy: solvent: the substance that dissolves the solute watersalt a homogeneous mixture -- evenly mixed at the particle level -- e.g., salt water 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 “soluble” -- e.g., food coloring and water miscible: soluble:

6 Types of Solutions SoluteSolventSolution Gaseous Solutions gas liquid gas air (nitrogen, oxygen, argon gases) humid air (water vapor in air) Liquid Solutions gas liquid solid liquid carbonated drinks (CO 2 in water) vinegar (CH 3 COOH in water) salt water (NaCl in water) Solid Solutions liquid solid dental amalgam (Hg in Ag) sterling silver (Cu in Ag) Charles H.Corwin, Introductory Chemistry 2005, page 369

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

8 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

9 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

10 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

11 Muddy Water: Dissolved Solids Muddy Water T 1 min T 5 min AFTER Water Before Add soil to water, shake well, and allow to settle Experiment 3: Dissolved solids can be calculated as a percentage: v/v (volume/volume) w/v(weight/volume) w/w(weight/weight) 5 mL solid / 95 mL water 5% v/v soil in water 5 mL / 100 mL = 5%

12 Centrifugation Spin sample very rapidly: denser materials go to bottom (outside) Separate blood into serum and plasma –Serum (clear) –Plasma (contains red blood cells ‘RBCs’) Check for anemia (lack of iron) Blood RBC’s Serum A AFTER Before B C

13 Making solutions 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.

14 O 2- H+H+ H+H+ H2OH2O  ++ ++ Water Molecule Water is a POLAR molecule

15 Copyright © 2007 Pearson Benjamin Cummings. All rights reserved. Water molecules “stick” together to create surface tension to support light weight objects.

16 Water Molecule What is a polar molecule? How does the polarity of water effect this molecule? O H H Hydrogen bond  

17 Hydrogen bonds occur between two polar molecules, or between different polar regions of one large macro- molecule. One “relatively” negative region is attracted to a second “relatively” positive region. O H H H N H H Electronegative atoms Hydrogen bond

18 Interstitial Spaces Water dissolvedsolid Oil red food coloring Layer Non-polar Polar "immiscible"

19 Na + Cl - NaCl solid salt NaCl (aq) = Na + = Cl - Dissolving of solid NaCl Animation by Raymond Chang All rights reserved.

20 Dissolving of NaCl Timberlake, Chemistry 7 th Edition, page 287 HH O Na Cl hydrated ions

21 Dissolving of Salt in Water NaCl(s) + H 2 O  Na + (aq) + Cl - (aq) Cl - ions Na + ions Water molecules

22 Solubility vs. Temperature 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)

23 Gas Solubility CH 4 O2O2 CO He Temperature ( o C) Solubility (mM) Higher Temperature …Gas is LESS Soluble

24 Electrolytes Timberlake, Chemistry 7 th Edition, page 290 Electrolytes Electrolytes - solutions that carry an electric current NaCl(aq) Na + + Cl - HF(aq) H + + F - strong electrolyteweak electrolytenonelectrolyte

25 Effect of Salinity on Cells Timberlake, Chemistry 7 th Edition, page 312 isotonic solution no change hypotonic solution hemolysis crenation hypertonic solution

26 Copyright © 2007 Pearson Benjamin Cummings. All rights reserved. (a) Cells in dilute salt solution (b) Cells in distilled water(c) Cells in concentrated salt solution IsotonicHypotonicHypertonic

27 Concentration = # of fish volume (L) Concentration = V = 1000 mL n = 2 fish Concentration = 2 “fishar” V = 1000 mL n = 4 fish [ ] = 4 “fishar” V = 5000 mL n = 20 fish [ ] = 4 “fishar” 1 fish 1 (L) Concentration = 1 “fishar”

28 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 Concentration = # of moles volume (L) V = 250 mL n = 8 moles [ ] = 32 molar

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

30 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.

31 Making a Dilute Solution Timberlake, Chemistry 7 th Edition, page 344 initial solution remove sample diluted solution same number of moles of solute in a larger volume mix moles of solute

32 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

33 precise; expensive 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 mL 950 mL – 1050 mL mL– mL imprecise; cheap Range:

34 Markings on Glassware TC 20 o C “to contain at a temperature of 20 o C” TD “to deliver” T s “time in seconds” mL + 5%Range = 500 mL + 25 mL 475 – 525 mL Beaker Graduated Cylinder Volumetric Flask 500 mL mLRange = – mL 500 mL + 5 mLRange = 500 mL + 5 mL 495 – 505 mL

35 How to mix solid chemicals Lets mix chemicals for the upcoming soap lab. We will need 1000 mL of 3 M NaOH per class. How much sodium hydroxide will I need, for five classes, for this lab? M = mol L 3 M = ? mol 1 L ? = 3 mol NaOH/class How much will this weigh? 1 23g/mol + 16g/mol g/mol MM NaOH = 40g/mol 40.0 g NaOH 1 mol NaOH X g NaOH = 15.0 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. FOR EACH CLASS: x 5 classes 15 mol NaOH 600 g NaOH

36 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

37 Process of Making a Standard Solution from Liquids Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 483

38 Identify each volume to two decimal places (values tell you how much you have expelled) mL mL5.00 mL Reading a pipette

39 Dilution of Solutions Solution Guide Formula Weight Specific Gravity Molarity Reagent Percent To Prepare 1 Liter of one molar Solution Acetic Acid Glacial (CH 3 COOH) %57.3 mL Ammonium Hydroxide (NH 4 OH) %69.0 mL Formic Acid (HCOOH) %42.5 mL Hydrochloric Acid (HCl) %82.5 mL Hydrofluoric Acid (HF) %34.5 mL Nitric Acid (HNO 3 ) %63.0 mL Perchloric Acid 60% (HClO 4 ) %110 mL Perchloric Acid 70% (HClO 4 ) %85.5 mL Phosphoric Acid (H 3 PO4) %67.5 mL Potassium Hydroxide (KOH) %57.3 mL Sodium Hydroxide (NaOH) %85.5 mL Sulfuric Acid (H 2 SO 4 ) %51.5 mL M Conc. V Conc. = M Dilute V Dilute

40 C = concentrate D = dilute Dilutions of Solutions  Dilution Equation: Concentrated H 3 PO 4 is 14.8 M. What volume of concentrate is required to make L of M H 3 PO 4 ? V C = L = 845 mL Acids (and sometimes bases) are purchased in concentrated form (“concentrate”) and are easily diluted to any desired concentration. **Safety Tip: When diluting, add acid or base to water.**

41 safety glasses Be sure to wear your safety glasses! 1. Measure out L of concentrated H 3 PO In separate container, obtain ~20 L of cold H 2 O. 3. In fume hood, slowly pour [H 3 PO 4 ] into cold H 2 O. 4. Add enough H 2 O until L of solution is obtained. How would you mix the above solution?

42 Yes; we’re OK. You have 75 mL of conc. HF (28.9 M); you need 15.0 L of M HF. Do you have enough to do the experiment? > mol HAVE1.50 mol NEED M C V C = M D V D 28.9 M (0.075 L) = M (15.0 L)

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

44 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

45 Preparing Solutions How to prepare 500 mL of 1.54 M NaCl solution –mass 45.0 g of NaCl –add water until total volume is 500 mL –Shake to dissolve salt 500 mL volumetric flask 500 mL mark 45.0 g NaCl solute

46 500 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 kg of water 500 mL mark 1.54m NaCl in kg of water molality molarity

47 Spec-20 Instrument Sample holder cover Amplifier control knob Light control knob Wavelength control knob

48 Spec-20 Absorbance (wavelength) Absorbance 100 % Absorbance 0 % Transmittance 0 % Absorbance 100 % Transmittance Insert Photograph Spec-20 Sample-holder Meter Zero adjust and power control Wavelength scale Wavelength control Light control

49 Schematic representation of a spectrophotometer Detector Sample-holder Meter Zero adjust and power control Wavelength scale Wavelength control Light control Monochromator Light Source (1) Sample cell (3) (2) (4) (5) IoIo I Meter A spectrophotometer is an instrument that measures the fraction (I/I o ) of an incident beam of light (I o ) which is transmitted (I) by a sample at a particular wavelength. For a given substance, the amount of light absorbed depends on: a) the concentration; b) the cell or path length; c) the wavelength of light; and d) the solvent.

50 Wavelength (nm) Absorbance Absorbance of Chlorophyll Frequency (Hz) Wavelength (nm) cosmic rays gamma rays x-raysultra- violet infra- red radio (microwave) radartele- vision radiopower transmission VioletBlueGreen YellowOrangeRed UV Near Infrared 400 nm 500 nm 600 nm 700 nm

51 Calibration Curve Concentration Absorbance 100 % Absorbance 0 % Transmittance 0 % Absorbance 100 % Transmittance out of linear range (too concentrated) x 2 (fixed wavelength) ? Dilute sample with water 50:50. Run sample, read concentration.

52 1 mol CaF 2 2 atoms F 18 g 6.02 x molecules 1 g1000 mL What mass of CaF 2 must be added to 1,000 L of water so that fluoride atoms are present at a conc. of 1.5 ppm? = 3.34 x m’cules H 2 O X m’cule H 2 O = 1000 L 1 L1 mL 1 mol6.02 x10 23 m’cule 1 mol 1.5 atom F X atoms F 1,000,000 m’cule H 2 O 3.34 x m’cule H 2 O = 1 molecule CaF 2 X = 5.01 x atoms F times =2.505 x molecules CaF 2 1 mol CaF 2 X g CaF 2 = x molecules =3.25 g CaF g CaF 2

53 40.0 g NaOH 1 mol NaOH How many moles solute are required to make 1.35 L of 2.50 M solution? mol = M L B. What mass magnesium phosphate is this? A. What mass 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 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

54 0.342 mol 5.65 L Find molarity if 58.6 g barium hydroxide are in 5.65 L solution g Ba(OH) 2 1 mol Ba(OH) 2 Step 2). What is the molarity of a 5.65 L solution containing mol solute? Step 1). How many moles barium hydroxide is this? M Ba(OH) 2 = X mol Ba(OH) 2 = 58.6 g Ba(OH) 2 = mol Ba(OH) 2 M = mol L M =

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

56 occurs when neutral combinations of particles separate into ions while in aqueous solution. sodium chloride sodium hydroxide hydrochloric acid sulfuric acid acetic acid Dissociation In general, acids yield hydrogen ions in aqueous solution; bases yield hydroxide ions. 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+ (H 1+ ) (OH 1– ) ? ?

57 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: in aq. solution: NOT in water: in aq. solution:

58 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

59 …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

60 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.)

61 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

62 Effect of Pressure on Boiling Point Boiling Point of Water at Various Locations Location Feet above sea level P atm (kPa) Boiling Point (  C) Top of Mt. Everest, Tibet29, Top of Mt. Denali, Alaska20, Top of Mt. Whitney, California14, Leadville, Colorado10, Top of Mt. Washington, N.H.6, Boulder, Colorado5, Madison, Wisconsin New York City, New York Death Valley, California

63 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 Br 1– ……i = 3

64 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)

65 (NONELECTROLYTE) 168 g glucose (C 6 H 12 O 6 ) 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. i = 1  T b = K b m i = (0.373) (1) = 0.19 o C BP = ( ) o C = o C  T f = K f m i = –1.86 (0.373) (1) = –0.69 o C FP = (0 + –0.69) o C = –0.69 o C

66 168 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 = (0.316) (2) = 0.32 o C BP = ( ) o C = 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–

67 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 ?

68 Step 1) Identify the species present in the combined solution, and determine what reaction occurs. Step 2) Write the balanced net ionic equation for the reaction. Step 3) Calculate the moles of reactants. Step 4) Determine which reactant is limiting. Step 5) Calculate the moles of product or products, as required. Step 6) Convert to grams or other units, as required. Stoichiometry steps for reactions in solution Stoichiometry for Reactions in Solution

69 461 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 = L of 4.0 M KI

70 = mol CuSO 4 How many mL of a 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 mol CuSO M CuSO 4 = L L 1000 mL 1 L = 346 mL

71 63.55 g Cu 1 mol Cu Stoichiometry Problems How many grams of Cu are required to react with 1.5 L of 0.10M AgNO 3 ? 1.5 L.10 mol AgNO 3 1 L = 4.8 g Cu Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 1 mol Cu 2 mol AgNO 3 ? g 1.5L 0.10M Courtesy Christy Johannesson

72 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 g ? L 0.90 L 2.5M Courtesy Christy Johannesson

73 Limiting Reactants 79.1 g Zn 1 mol Zn g Zn = 27.1 L H 2 1 mol H 2 1 mol Zn 22.4 L H 2 1 mol H 2 Zn + 2HCl  ZnCl 2 + H g ? L 0.90 L 2.5M Courtesy Christy Johannesson

74 Limiting Reactants 22.4 L H 2 1 mol H L 2.5 mol HCl 1 L = 25 L H 2 1 mol H 2 2 mol HCl Zn + 2HCl  ZnCl 2 + H g ? L 0.90 L 2.5M Courtesy Christy Johannesson

75 Limiting Reactants Zn: 27.1 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

76 A Hydrocarbon Typical petroleum product Non-polar CH 2 CH 3 CH 2 CH 3 C 18 H 38

77 Oil and Water Don’t Mix Oil is nonpolar Water is polar “Like dissolves like” Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 470

78 Hydrogenation Timberlake, Chemistry 7 th Edition, page H 2 vegetable oils shortening stick margarine tub (soft) margarine heat, nickel catalyst

79 Molecular Polarity H–C–H H H HH O nonpolar molecules:-- e – are shared equally -- tend to be symmetric e.g., fats and oils polar molecules:-- e – NOT shared equally e.g., water Like dissolves like “Like dissolves like.” polar + polar = solution nonpolar + nonpolar = solution polar + nonpolar = suspension (won’t mix evenly)

80 Using Solubility Principles C=C Cl Chemicals used by body obey solubility principles. -- fat-soluble vitamins:e.g., vits. A, D Dry cleaning employs nonpolar liquids. -- polar liquids damage wool, silk -- also, dry clean for stubborn stains (ink, rust, grease) -- (tetra) perchloroethylene is in common use -- water-soluble vitamins: e.g., vit. C

81 MODEL OF A SOAP MOLECULE NONPOLAR HYDROCARBON TAIL POLAR HEAD Na 1+ emulsifying agent (emulsifier): -- molecules w/both a polar AND a nonpolar end -- allows polar and nonpolar substances to mix detergentlecithineggse.g., soap

82 Saponification glyceryl tripalmitate (tripalmitin) glycerol 3 sodium palmitate (soap) CH 2 – O – C – (CH 2 ) 14 CH 3 CH – O – C – (CH 2 ) 14 CH 3 CH 2 – O – C – (CH 2 ) 14 CH 3 O O O CH 2 – OH CH – OH CH 2 – OH O 3 Na + - OC – (CH 2 ) 14 CH 3 3 NaOH + + sodium hydroxide Process of making soap from animal fat or vegetable oil using a base.

83 A Phospholipid polar head nonpolar tails (a) chemical structure of a phospholipid (b) simplified way to draw a phospholipid polar head nonpolar tails Timberlake, Chemistry 7 th Edition, page 576

84 Lipid Bilayers Micelle Bilayer Liposome Individual units are cylindrical (cross-section of head equals that of side chain) Individual units are wedge-shaped (cross-section of head greater than that of side chain) Aqueous cavity

85 A Model of a Cell Membrane Timberlake, Chemistry 7 th Edition, page 587 Polar Nonpolar Cholesterol Proteins Phospholipid bilayer

86 Cleaning Action of Soap Micelle Timberlake, Chemistry 7 th Edition, page 573

87 SOAP vs.DETERGENT -- made from animal and-- made from petroleum vegetable fats-- works better in hard water Hard water contains minerals w /ions like Ca 2+, Mg 2+, and Fe 3+ that replace Na 1+ at polar end of soap molecule. Soap is changed into an insoluble precipitate (i.e., soap scum). micelle: a liquid droplet covered w /soap or detergent molecules

88 Solvation NONPOLAR POLAR “Like Dissolves Like”

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

90 Lava Lamp It is… a philosophy the primordial ooze once ruled our world the moment an art form a classic progressive prehistoric post-modern here to stay Polar mixture Water Polyethylene glycol Nonpolar mixture Chlorinated paraffin Paraffin from kerosene Heat transfer coil Bulb gives heat and light HHHHHHHHHHHHHHHHHHHH H-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-H HHHHHHHHHHHHHHHHHHHH ClHHHH HHHHH HHHH HHH H-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-H HHHClHHHH HHHHH HHH H HHHHHHHH H-O-C-C-O-C-C-O-C-C-O-C-C-O-H HHH H HHHH H H O

91 Dialysis A semi-permeable membrane allows small particles to pass through while blocking larger particles. Dialysis is used to clean blood when people suffer kidney failure.

92 Solution, Suspension, Colloid Timberlake, Chemistry 7 th Edition, page 309

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

94 Solubility Table LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World, 1996, page Solubility vs. Temperature for Solids Solubility (grams of solute/100 g H 2 O) KI KCl NaNO 3 KNO 3 HClNH 4 Cl NH 3 NaCl KClO 3 SO 2 shows the dependence of solubility on temperature gases solids

95 Solubility Solubility maximum grams of solute that will dissolve in 100 g of solvent at a given temperature varies with temperature based on a saturated solution

96 Boiling Points of Covalent Hydrides Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page Period of X (H 2 X) Boiling point ( o C) H2OH2O H2SH2S H 2 Se H 2 Te 2345 Li 3 He 2 C6C6 N7N7 O8O8 F9F9 Ne 10 Na 11 B5B5 Be 4 H1H1 Al 13 Si 14 P 15 S 16 Cl 17 Ar 18 K 19 Ca 20 Sc 21 Ti 22 V 23 Cr 24 Mn 25 Fe 26 Co 27 Ni 28 Cu 29 Zn 30 Ga 31 Ge 32 As 33 Se 34 Br 35 Kr 36 Rb 37 Sr 38 Y 39 Zr 40 Nb 41 Mo 42 Tc 43 Ru 44 Rh 45 Pd 46 Ag 47 Cd 48 In 49 Sn 50 Sb 51 Te 52 I 53 Xe 54 Cs 55 Ba 56 Hf 72 Ta 73 W 74 Re 75 Os 76 Ir 77 Pt 78 Au 79 Hg 80 Tl 81 Pb 82 Bi 83 Po 84 At 85 Rn 86 Fr 87 Ra 88 Rf 104 Db 105 Sg 106 Bh 107 Hs 108 Mt 109 Mg   Hydrogen Bonding

97 Boiling Points of Covalent Hydrides Molecular mass Boiling point ( o C) H2OH2O H2SH2S H 2 Se H 2 Te CH 4 SiH 4 GeH 4 SnH 4

98 Group16 Hydrogen Compounds CompoundMolar Mass Melting Point ( o C) Boiling Point ( o C) H fusion (cal/mol) H vapor (cal/mol) H2OH2O H2SH2S H 2 Se H 2 Te

99 ? K+K+ K+K+ K+K+ NO 3 - Precipitation Reaction Between AgNO 3 and KCl AgNO 3 (aq) + KCl(aq)unknown white solid Ag + + NO K + + Cl - in silver nitrate solution in potassium chloride solution unknown white solid Ag + + NO K + + Cl - Ag + + Cl - + K + + NO 3 - product AgNO 3 (aq) + KCl(aq)  KNO 3 (aq) + AgCl(s) AgCl AgNO 3 (aq) + KCl(aq) AgCl(s) + KNO 3 (aq) AgCl precipitate Ag + Cl -

100 Clogged Pipes – Hard Water limestone ‘hard’ water carbonic acid Step 1: Acid rain is formed Step 2: Acid rain dissolves limestone Water softener H 2 O + CO 2 H 2 CO 3 H 2 CO 3 + CaCO 3 Ca(HCO 3 ) 2 H 2 CO 3 + MgCO 3 Mg(HCO 3 ) 2

101 Pipes develop Scales Step 3: Hard water is heated and deposits scales Ca(HCO 3 ) 2 CaCO 3 (s) + H 2 O + CO 2 Mg(HCO 3 ) 2 MgCO 3 (s) + H 2 O + CO 2 scales on pipes   1 gram = 1 / 7000th pound 300 gallons / day with 10 grain water supply

102 No More Hard Water Scale No More Ugly Hard Water Spotting Protects Plumbing and Appliances Saves Money on Cleaning Products

103 Water Purification Hard Water Cation Exchanger Anion Exchanger Deionized Water (a) (b)(c) H+H+ H+H+ H+H+ H+H+ H+H+ H+H+ H+H+ H+H+ OH - (a)The cations in hard water are exchanged for H +. (b)The anions in hard water are exchanged for OH -. (c)The H + and OH - combine to give H 2 O. Hard water is softened by exchanging Na + for Ca 2+, Mg 2+, and Fe 3+. Corwin, Introductory Chemistry  2005, page 361 Na + Ca 2+ Mg 2+ Fe 3+

104 Which ions are removed from hard water to produce soft water? Hard water is softened by exchanging Na + for Ca 2+, Mg 2+, and Fe 3+. Na + (aq) + H (resin) Na (resin) + H + (aq) Cl - (aq) + (resin) OH (resin) Cl + OH - (aq) Notice that the ion exchange resin produces both hydrogen ions and hydroxide ions which can readily combine to give water. H + (aq) + OH - (aq) H 2 O (aq) The net result is that the resin removes all ions from water passing through the deionizing system. Corwin, Introductory Chemistry  2005, page 361

105 semipermeable membrane pressure fresh water Reverse Osmosis salt water


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