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Solution (Homogeneous Mixture) Evenly mixed, uniform particles

Dental filling. Solution (Homogeneous Mixture) Evenly mixed, uniform particles. Solute - gets dissolved Solvent - does the dissolving Aqueous ( water) Tincture ( alcohol ) Amalgam ( mercury ) Organic (carbon based) Polar Non-polar. Types of Solutions.

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Solution (Homogeneous Mixture) Evenly mixed, uniform particles

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  1. Dental filling Solution (Homogeneous Mixture)Evenly mixed, uniform particles • Solute - gets dissolved • Solvent - does the dissolving • Aqueous(water) • Tincture(alcohol) • Amalgam(mercury) • Organic (carbon based) • Polar • Non-polar

  2. Types of Solutions Charles H.Corwin, Introductory Chemistry 2005, page 369

  3. Source of electric power Pure water Pure water does not conduct an electric current Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 215

  4. Source of electric power Free ions present in water Ionic Solutions conduct a Current Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 215

  5. NaCl(aq) Na+ + Cl- HF(aq) H+ + F- Electrolytes Electrolytes - solutions that carry an electric current strong electrolyte weak electrolyte nonelectrolyte Timberlake, Chemistry 7th Edition, page 290

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

  7. Dissolving of NaCl Timberlake, Chemistry 7th Edition, page 287

  8. Oil and Water Don’t Mix • Oil is nonpolar • Water is polar “Like dissolves like” Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 470

  9. gases solids Solubility vs. Temperature Solubility Curves 140 KI 130 120 NaNO3 110 100 KNO3 90 80 HCl NH4Cl • shows the dependence • of solubility on temperature 70 Solubility (grams of solute/100 g H2O) KCl 60 NH3 50 40 30 NaCl KClO3 20 10 SO2 0 10 20 30 40 50 60 70 80 90 100 temp (0C) LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517

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

  11. Classify as unsaturated, saturated, or supersaturated. 80 g NaNO3 @ 30oC 40 g KCl @ 60oC 50 g NH3 @ 10oC 70 g NH4Cl @ 70oC per 100 g H2O Per 500 g H2O, 120 g KNO3 @ 40oC saturation point @ 40oC for 100 g H2O = 55 g KNO3 So sat. pt. @ 40oC for 500 g H2O = 5 x 55 g = 303 g 120 g < 303 g

  12. Classify as unsaturated, saturated, or supersaturated. per 100 g H2O 80 g NaNO3 @ 30oC unsaturated 40 g KCl @ 60oC saturated 50 g NH3 @ 10oC unsaturated 70 g NH4Cl @ 70oC supersaturated Per 500 g H2O, 120 g KNO3 @ 40oC saturation point @ 40oC for 100 g H2O = 55 g KNO3 So sat. pt. @ 40oC for 500 g H2O = 5 x 55 g = 303 g 120 g < 303 g unsaturated

  13. Factors Affecting Solubility As To , rate (except gases) 1. temperature As size , rate 2. particle size More mixing, rate 3. mixing 4. nature of solvent or solute Polar vs. non-polar vs. ionic

  14. Gas Solubility CO2 2.0 O2 Higher Temperature …Gas is LESS Soluble CO Solubility (mM) 1.0 He 0 10 20 30 40 50 Temperature (oC)

  15. 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.

  16. mol D. molality (m) = moles of solute kg of solvent M L Concentration“The amount of solute in a solution” A. mass % = mass of solute X 100% (% solution) mass of solution B. parts per million (ppm) also, ppb, ppt, pph – commonly used for minerals or ppm = mass of solute X 1,000,000 contaminants in water supplies mass of solvent C. molarity (M) = moles of solute L of solution –used most often in this class E. mole ratio (or mole fraction) = moles (n) solute (or solvent) moles (n) solution

  17. One molar solution.

  18. Molarity • Find the molarity of a solution containing 75 g of MgCl2 in 250 ml of solution. 75 g MgCl2 1 mol MgCl2 95.21 g MgCl2 0.25 L solution = 3.2M MgCl2

  19. Molality • Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water. 75 g MgCl2 1 mol MgCl2 95.21 g MgCl2 0.25 kg water = 3.2m MgCl2

  20. Molality • How many grams of NaCl are req’d to make a 1.54m solution using 0.500 kg of water? 0.500 kg water 1.54 mol NaCl 1 kg water 58.44 g NaCl 1 mol NaCl = 45.0 g NaCl

  21. 500 mL of 1.54M NaCl Preparing Solutions molarity molality 1.54m NaCl in 0.500 kg of water • 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 500 mLwater 500 mL volumetric flask 500 mL mark 45.0 gNaCl

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

  23. Dilution • Preparation of a desired solution by adding water to a concentrate (i.e stock solution). • Moles of solute remain the same.

  24. Dilution Acids (and sometimes bases) are purchased in concentrated form (“concentrate”) and are easily diluted to any desired concentration. Dilutions of Solutions  **Safety Tip: When diluting, add acid or base to water.** C = concentrate D = dilute Dilution Equation: Concentrated H3PO4 is 14.8 M. What volume of concentrate is required to make 25.00 L of 0.500 M H3PO4? VC = 0.845 L = 845 mL

  25. M M mol M = L mol mol V V P P __ __ __ __ Molarity and Stoichiometry mol = M x L M x L M x L 1 2 1 2 Pb(NO3)2(aq) + KI (aq)  PbI2(s) + KNO3(aq) What volume of 4.0 M KI solution is required to yield 89 g PbI2 with excess lead nitrate?

  26. mol L M = mol M 0.39 mol KI 4.0 M KI L = = 1 Pb(NO3)2(aq) + 2 KI (aq) 1 PbI2(s) + 2 KNO3(aq) 89 g ? L 4.0 M What volume of 4.0 M KI solution is required to yield 89 g PbI2 with excess lead nitrate? Strategy: (1) Find mol KI needed to yield 89 g PbI2. (2) Based on (1), find volume of 4.0 M KI solution. 1 mol PbI2 2 mol KI X mol KI = 89 g PbI2 = 0.39 mol KI 461 g PbI2 1 mol PbI2 = 0.098 L of 4.0 M KI

  27. mol M mol L M = L = 0.173 mol CuSO4 0.500 M CuSO4 How many mL of a 0.500 M CuSO4 solution will react with excess Al to produce 11.0 g Cu? Al3+ SO42– __CuSO4(aq) + __Al (s)  __Cu(s) + __Al2(SO4)3(aq) CuSO4(aq) + Al (s)  Cu(s) + Al2(SO4)3(aq) 3 2 3 1 x mol 11 g 1 mol Cu 3 mol CuSO4 X mol CuSO4 = 11 g Cu = 0.173 mol CuSO4 63.5 g Cu 3 mol Cu = 0.346 L 1000 mL 0.346 L = 346 mL 1 L

  28. How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3? Cu + 2AgNO3 2Ag + Cu(NO3)2 1.5L 0.10M ? g .10 mol AgNO3 1 L 1 mol Cu 2 mol AgNO3 63.55 g Cu 1 mol Cu 1.5 L = 4.8 g Cu Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  29. 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  ZnCl2 + H2 ? L 79.1 g 0.90 L 2.5M Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  30. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 ? L 79.1 g 0.90 L 2.5M 79.1 g Zn 1 mol H2 1 mol Zn 1 mol Zn 65.39 g Zn 22.4 L H2 1 mol H2 = 27.1 L H2 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  31. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 ? L 79.1 g 0.90 L 2.5M 0.90 L 2.5 mol HCl 1 L 1 mol H2 2 mol HCl 22.4 L H2 1 mol H2 = 25 L H2 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  32. left over zinc Limiting Reactants Zn: 27.1 L H2 HCl: 25 L H2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25 L H2 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  33. Colligative Properties 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 FREEZING PT. DEPRESSION …normal boiling point (NBP) …higher BP BOILING PT. ELEVATION VAPOR PRESSURE DEPRESSION …normal vapor pressure (NVP) …lower VP OSMOTIC PRESSURE ELEVATION …normal osmotic pressure (NOP) …higher OP

  34. Calculations for Colligative Properties The change in FP or BP is found using… DTx = Kx m i DTx = change in To (below NFP or above NBP) Kx = constant depends on… (A) solvent (B) freezing or boiling Water: KB = 0.512 0C kg solvent / mol solute KF = 1.86 0C kg solvent / mol solute m = molality of solute = mol solute / kg solvent i = integer (van’t Hoff factor that accounts for any solute particles) any sugar (all nonelectrolytes)……………...i = 1 table salt, NaCl  Na1+ + Cl1–………………i = 2 barium bromide, BaBr2  Ba2+ + 2 Br1–……i = 3

  35. Colligative Properties (Ex.) 168 g glucose (C6H12O6) are mixed w/2.50 kg H2O. Find BP and FP of mixture. For H2O, Kb = 0.512, Kf = –1.86. i = 1 (NONELECTROLYTE) DTb = Kb m i = 0.512 (0.373) (1) = 0.19oC BP = (100 + 0.19)oC = 100.19oC DTf = Kf m i = –1.86 (0.373) (1) = –0.69oC FP = (0 + –0.69)oC = –0.69oC

  36. Colligative Properties (Ex.) 168 g cesium bromide are mixed w/2.50 kg H2O. Find BP and FP of mixture. For H2O, Kb = 0.512, Kf = –1.86. Cs1+ Br1– i = 2 CsBr Cs1+ + Br1– DTb = Kb m i = 0.512 (0.316) (2) = 0.32oC BP = (100 + 0.32)oC = 100.32oC DTf = Kf m i = –1.86 (0.316) (2) = –1.18oC FP = (0 + –1.18)oC = –1.18oC

  37. Applications (Data are estimates) 1. salting roads in winter water + a little salt –11oC 103oC water + more salt –18oC 105oC 2. antifreeze (AF) /coolant

  38. Calculations for Colligative Properties The vapor pressure of a solution… Psolution = iXsolvent P0solvent Raoult’s Law: Psolution = Observed vapor pressure of solution Xsolvent = Mole fraction of the solvent P0solvent = Vapor Pressure of pure solvent i = integer that accounts for any solute dissociation π = iMRT The osmotic pressure of a solution… π = Osmotic pressure (atm) M = Molarity of solution i = integer that accounts for any solute dissociation R = Gas law constant T = Kelvin temperature

  39. semipermeable membrane Net Solvent Flow Dilute solution Net flow of solvent Net flow of solvent Concentrated solution Solute particle Solvent particle Ralph A. Burns, Fundamentals of Chemistry 1999, page 430

  40. Osmosis Water (solvent) Solution Semipermeable membrane Kelter, Carr, Scott, Chemistry A Wolrd of Choices 1999, page 267

  41. Pressure here Reverse Osmosis Water (solvent) Solution Semipermeable membrane Kelter, Carr, Scott, Chemistry A Wolrd of Choices 1999, page 267

  42. Colligative Properties (Ex.) Predict the vapor pressure of a solution prepared by mixing 35.0 g of solid Na2SO4 (molar mass = 142 g/mol) with 175 g water at 250C. The vapor pressure of pure water at 250C is 23.76 torr. 2Na1+ SO42– ie. 1 mole Na2SO4 = 3 moles ions nwater = 175 g x 1 mol/18.0 g = 9.72 mol H2O nNa2SO4 = 35.0 g x 1 mol/142 g x 3 = 0.738 mol Na2SO4 Xwater = 9.72 mol/(0.738 mol + 9.72 mol) = 0.929 Psolution = 0.929 x 23.76 = 22.1 torr

  43. Water Purification Cation Exchanger Anion Exchanger Deionized Water Hard Water H+ OH- H+ OH- H+ Mg2+ OH- Na+ H+ (b) (c) (a) OH- OH- Fe3+ H+ Ca2+ H+ OH- H+ OH- H+ OH- Hard water is softened by exchanging Na+ for Ca2+, Mg2+, and Fe3+. • The cations in hard water are exchanged for H+. • The anions in hard water are exchanged for OH-. • The H+ and OH- combine to give H2O. Corwin, Introductory Chemistry2005, page 361

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