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Ch 14 Solutions

Ch 14 Solutions. Forming Solutions (You should know most of this from ws ) Terms: Solution – homogeneous mixture – alike throughout a. ex: soda, air, brass b. Solvent - substance doing the dissolving c. Solute – substance being dissolved 2. Aqueous soln – water is solvent.

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Ch 14 Solutions

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  1. Ch 14 Solutions

  2. Forming Solutions (You should know most of this from ws) Terms: • Solution – homogeneous mixture – alike throughout a. ex: soda, air, brass b. Solvent - substance doing the dissolving c. Solute – substance being dissolved 2. Aqueous soln – water is solvent

  3. Types of solns

  4. 14.1 Solubility 1. defn – ability to dissolve 2. when ionic solids dissolve they become electrolytes a. conduct electricity b. due to ions moving freely 3. molecular substance are not electrolytes  molecules stay together in soln

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

  6. Experiment 2: • Solubility - immiscible Add oil to water and shake AFTER Before Immiscible – “does not mix” two liquids or two gases that DO NOT MIX Oil Water Water

  7. 4. Effects of pressure and solubility a. for gases pressure effects solubility b. Henry’s Law – the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas on the surface of the liquid - ex CO2 in soda c. effervescence – the rapid of escape of a gas from a liquid in which it is dissolved (fizzing in pop)

  8. 200 200 180 180 160 160 140 140 120 120 100 100 80 80 60 60 40 40 20 20 0 0 Solubility • Temperature makes a dif KI KNO3 Solubility (g solute / 100 g H2O) Solubility (g solute / 100 g H2O) NaNO3 Na3PO4 NaCl 20 40 60 80 100 Temperature (oC)

  9. Gas Solubility • dec CH4 2.0 O2 Higher Temperature …Gas is LESS Soluble CO Solubility (mM) 1.0 He 0 10 20 30 40 50 Temperature (oC)

  10. Solubility and temp CsCl 180 NaNO3 170 160 NaC2H3O2 150 140 130 RbCl 120 LiCl 110 Solubility in grams per 100 g of water 100 90 80 NH4Cl 70 60 KCl 50 NaCl 40 30 Li2SO4 20 10 SO2 0 90 80 100 10 60 70 30 40 50 20 0 Temperature (oC)

  11. Calculations using Solubility curves Using previous slide 1. 100 g H2O is saturated with KClO3 at 100°C. How much KClO3 will come out of solution if the temperature is decreased to 48°C? 57 g - 18 g = 39 g How much would precipitate under the same circumstances if 350 g water were used (57 g - 18 g) x 3.5 = 137 g

  12. Ionic solns dissolve to form ions  conduct electricity

  13. Another visual for salt dissolving:

  14. 5. Process of dissolving a. solute – solute break (takes energy) b. solvent-solvent break (takes energy) c. solute-solvent form (releases energy) 6. Likes dissolve likes a. polar molecules will dissolve polar molecules b. nonpolar will dissolve nonpolar c. that is why polar water will not dissolve nonpolar oil, so oil floats on water ? Why is a good thing that oil floats on water? (think ocean)

  15. + H H O - - + - + NaCl dissolving Na+ + + hydrated ions - + - Cl-

  16. 14.2 Soln composition: An Intro 1. saturated – soln containing as much solute as it can dissolve at that temp (full) 2. unsaturated – can dissolve more (not full) 3. supersatd – contains more solute than can be satdsoln (over full) 4. concentrated soln – lots of solute 5. dilute soln – little solute

  17. Which beaker has a more concentrated soln?

  18. Which beaker has a more concentrated soln?

  19. They are the same! • X has 1 liter of soln with 2 particles in it so 2:1 • And Y has 2 liters of soln with 4 particles in it so also 2:1

  20. C. Factors Affecting the Rate of Dissolving 1. surface area (crush) 2. stirring 3. temperature

  21. Bellringer 3/25 If 25 g of salt can dissolve easily in water, and you have added 10g of salt… • This solution is considered to be ____. (due to amount of salt dissolved) • Salt is considered to be ______ because it dissolves in water. • Which substance is the solute? • Which substance is the solvent?

  22. 14.3 Mass percent concentration – measure of amount of solute in a given amount of solvent or soln • Mass percent 1. eqn: mass % = mass of solute (100) mass of soln so: grams of solute (100) grams of solute + grams of solvent

  23. 2. ex: Calculate the mass percent of 1.0 g of ethanol in 100 g of water. mass % = mass of solute (100) mass of soln = 1.0 g eth (100) 101 g of soln = 0.99% ethanol

  24. 14.4 Molarity (M) 1. defn – the number of moles of solute in one liter of soln 2. eqn: M = moles solute L soln 3. Solutions are made by adding the correct amount of solute in a volumetric flask, dissolving the solute, and then adding enough solvent to make the amount. (Demo)

  25. 4. EX: You have 3.50 L of soln that contains 90.0g of sodium chloride, NaCl. What is the molarity of the soln? M = mole/L soln 90.0g NaCl 1 mole NaCl 58.5 g NaCl = 1.54 moles NaCl M = 1.54 moles NaCl 3.50 L soln = 0.440 M NaCl

  26. 5. Give the conc of all the ions in 0.50M Co(NO3)2 • Co(NO3)2 (s) Co+2(aq) + 2NO3-1(aq) • Which is saying : 1mole Co(NO3)2 (s) 1 mole Co+2(aq) + 2moles NO3-1(aq) Assuming 1 liter of soln, moles = M 0.5 M Co+2 1.0 M NO3-1

  27. Molality (m) 1. defn – number of moles of solute per kg of solvent 2. used when solvent is important 3. eqn: m = moles solute kg solvent

  28. 14.5 Dilution - the process of adding solvent to a solution to lower the conc. of solute 1. Helps make weak acid from strong acid, etc 2. eqn: McVc = MdVd M = molarity V = volume c = concentrated (strong) d = dilute (weak)

  29. 4. Ex: How would you make a 500 ml of 1.5M solution HCl from 12M HCl? McVc = MdVd (12 M) Vc = (1.5M) (500 ml) Vc = 62.5 ml  dilute 62.5 ml of 12 M HCl to 500 ml with water

  30. 200 200 180 180 160 160 140 140 120 120 100 100 80 80 60 60 40 40 20 20 0 0 Solubility • Temperature makes a dif KI KNO3 Solubility (g solute / 100 g H2O) Solubility (g solute / 100 g H2O) NaNO3 Na3PO4 NaCl 20 40 60 80 100 Temperature (oC)

  31. Gas Solubility • dec CH4 2.0 O2 Higher Temperature …Gas is LESS Soluble CO Solubility (mM) 1.0 He 0 10 20 30 40 50 Temperature (oC)

  32. Solubility and temp CsCl 180 NaNO3 170 160 NaC2H3O2 150 140 130 RbCl 120 LiCl 110 Solubility in grams per 100 g of water 100 90 80 NH4Cl 70 60 KCl 50 NaCl 40 30 Li2SO4 20 10 SO2 0 90 80 100 10 60 70 30 40 50 20 0 Temperature (oC)

  33. Calculations using Solubility curves Using fig 2 (pg. 316) 5. 100 g H2O is saturated with KClO3 at 100°C. How much KClO3 will come out of solution if the temperature is decreased to 48°C? 57 g - 18 g = 39 g How much would precipitate under the same circumstances if 350 g water were used (57 g - 18 g) x 3.5 = 137 g

  34. 14.6 Stoichiometry and Molarity • Steps: • 1. Write balanced eqn. If ions – show them • 2. use Molarity to find moles (M = moles/L) • 3. determine Limiting reactant • 4. find the amt needed

  35. Cu + 2AgNO3 2Ag + Cu(NO3)2 • Ex: How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3? ? g 1.5L 0.10M .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

  36. 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 79.1 g 0.90 L 2.5M ? L Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  37. 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

  38. 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

  39. 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

  40. Colligative Properties - property that dependent only on the amount of solute  so the more solute, the bigger the dif. • Boiling Point Elevation • Adding a solute to water (or any liquid) will raise the boiling point • Water at 1 atm boils at 100°C, but 1M NaClsoln (salt water) will boil at 104°C!

  41. FYI – just an explanation Why does the boiling point of a solution increase? - Forming a bubble in a solution - Solute particles block some of the water molecules trying to enter the bubble. • Need higher pressure to maintain the bubble.

  42. Comparing bubbles  So it takes more energy for right bubble to “boil” due to solutes

  43. 3. Thinking questions: • So why do you add salt to water before you boil it to cook noodles? • Why do you have antifreeze in car radiator instead of pure water?

  44. B. Freezing point depression 1. when a solute is added to a solvent, the freezing point is lowered 2. At 1 atm water freezes at 0°C, but a 1M NaClsoln (salt water) freezes at -1° 3. Questions? a. why is salt added to roads in winter?  when ice dissolves it makes salt soln, so water (snow) will freeze at lower temp b. why does soda freeze faster than H2O?

  45. 3. Questions? a. why is salt added to roads in winter?  when ice dissolves it makes salt soln, so water (snow) will freeze at lower temp b. why does soda freeze slower than H2O?

  46. Videos: • Supersatd: • http://www.youtube.com/watch?v=HnSg2cl09PI&feature=fvw • Water to ice instantly • http://www.youtube.com/watch?v=R_dnB-3YxLk&feature=related • Soln intro • http://www.youtube.com/watch?v=GaLYcHkgzU4

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