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Chapters 17 and 18

Chapters 17 and 18. Water & Aqueous Solutions. Aqueous Solutions. 1) solute: substance being dissolved 2) solvent: substance doing the dissolving 3) solution: solute + solvent Ex) NaCl + H 2 O salt water (solute) (solvent) (solution).

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Chapters 17 and 18

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  1. Chapters 17 and 18 Water & Aqueous Solutions

  2. Aqueous Solutions 1) solute: substance being dissolved 2) solvent: substance doing the dissolving 3) solution: solute + solvent Ex) NaCl + H2O salt water (solute) (solvent) (solution) 4) aqueous solution: water is the solvent 5) solvation: process by which a solute is dissolved

  3. Ex) Solvation with Water The positive end of water molecules are: -attracted to a negative ion, or the -negative end of a polar molecule Water molecules then surround the particle keeping it from rejoining other like particles The same process works for the negative end of water molecules -polar solvents dissolve only ionic or polar cmpds -nonpolar solvents dissolve only nonpolar cmpds “like dissolves like”

  4. Solvation of NaCl

  5. 6) saturated: a solution containing the maximum amount of solute 7)solubility: the amount of solute added that will produce a saturated solution (usually per 100 g of solvent) 8) insoluble: solute will not dissolve in a solvent 9) unsaturated: a solution containing less than the maximum amount of solute 10) supersaturated: a solution containing more solute than it can theoretically hold.

  6. 11) miscible: two liquids that will dissolve in each other 12) immiscible: not miscible, liquids won’t mix 13) concentrated: a large amount of solute for a given amount of solvent 14) dilute: a small amount of solute 15) dissociation: separation of solute particles from each other

  7. Checking for Understanding If you have a glass of Gatorade and it has sugar at the bottom of the glass. What vocabulary word would best describe it. If you had a glass of Kool Aid and you can dissolve more sugar in it. What two vocabulary words could you use to describe it.

  8. What part of the water molecule would Be attracted to a negative chlorine ion. What part of the water molecule would be Attracted to a + Na ion? Which 2 vocab words can describe when oil is mixed with water?

  9. Factors Affecting Solubility 1) Solid dissolved in a liquid: generally, solubility increases with temperature 2) Gas dissolved in a liquid: solubility decreases with temperature Rate of Solution Formation 1) Particle size: powdered solutes will dissolve faster because of a greater surface area exposed to solvent molecules 2) Agitation: stirring increases more frequent solute contact with solvent

  10. Pressure and solubility of a gas As the pressure of the gas above a liquid is increased so does the solubility of the gas in the liquid. Oppositely, if you lower the pressure of the gas above the liquid the solubility of that gas decreases.

  11. Lake Nyos in Cameroon was the site of a natural disaster, in which a concentrated solution of carbon dioxide in water suddenly released enough carbon dioxide gas to suffocate 1700 people. The carbon dioxide concentration was high due to the high pressure deep in the lake: High pressure increases the solubility of carbon dioxide in water.

  12. Checking for understanding. If two beakers are filled with 100 mL of water One of the beakers is at 100 degrees C and the Other is at 25 degrees. Which beaker can dissolve more sugar?

  13. Checking for understanding. If two beakers are filled with 100 mL of water One of the beakers is at 100 degrees C and the Other is at 25 degrees. Which beaker can dissolve more carbon dioxide?

  14. Checking for understanding. If two bottles of Dr. Pepper are left in the sun, or In the refrigerator. One of the bottles is capped And the other does not. Which soda will contain More carbon dioxide? Why?

  15. Solution Problems

  16. Molarity Is a measurement of concentration. 1) Molarity: moles solute per liters of solution moles solute liters solution M =

  17. Ex) What is the molarity of a 2.5 L solution containing 4.62 g of CsBr? convert grams to moles: molar mass: 133 + 80 = 213 g/mole (Cs) (Br) mol 213 g 4.62 g CsBr = 0.0217 mol CsBr moles solute liters soln 0.0217 mole 2.5 Liters M = = =8.7 x 10-3 M

  18. Ex) How many grams of HNO3 would you need to make 1.5 liters of an 8M solution? moles solute liters soln Moles solute 1.5 liters M = 8 = Moles solute = 12 But I need grams molar mass HNO3 = 1 + 14 + (3x16) = 63 g/mol 12 moles HNO3 63 grams 1 moles HNO3 = 756 g HNO3

  19. Ex) How many grams of CuSO4 are needed to prepare 2500 mL of a 0.55 M solution CuSO4. Remember our volume must be in Liters. 2500 mL = 2.5 liters moles solute liters solution moles solute = (M) (liters) M = moles solute = (.55 ) (2.5 L) Now I need to change The moles to grams 1.38 moles CuSO4 1.38 moles CuSO4 159.6 g =219 g CuSO4 1 mol CuSO4

  20. 2) Dilutions M1V1 = M2V2 initial soln molarity & volume final soln molarity & volume Ex) Calculate the amount of solution needed to prepare 400 mL of a 1.5 M solution of NaOH from a 2.8 M stock solution. M2V2 M1 (1.5) (400) 2.8 V1 = = = 214 mL

  21. 3) Percent Solutions volume solute volume solution % by Volume: x 100 Ex) What is the percent by volume of a solution made by mixing 22 mL of alcohol with 125 mL of water? 22 mL 22mL + 125 mL x 100 = % by Volume = 15% (v/v)

  22. mass solute(g) volume solution (mL) % by mass / vol: x 100 Ex) What is the percent by mass/ vol of a solution made by mixing 0.892 g of KCl in 54.6 mL of solution? % by mass / vol = 0.892 g 54.6 mL x 100 = 1.6% (m/v) do: Dilutions & % Soln WS B

  23. 4) Molality: moles solute per kilograms of solvent moles solute kg solvent m = Ex) What is the molality of a solution containing 7.78 g of urea, (NH2)2CO, in 203 g of water? molar mass: 28 + 4 + 12 + 16 = 60 g/mol mol 60 g 7.78 g urea = 0.13 mol 0.13 mol 0.203 kg molality = = 0.64 m

  24. 5) Mole Fraction moles solute moles solution mole fraction = moles solute moles solute + moles solvent =

  25. Ex) A solution is made by adding 200.4 g of ethanol (C2H5OH) to 143.9 g of water. Calculate the mole fraction of ethanol. - convert each component to moles molar mass ethanol: 24 + 6.0 + 16 = 46 g/mol molar mass water: 16 + 2 = 18 g/mol 200.4 g ethanol mol 46 g 4.4 mol ethanol = 143.9 g water mol 18 g 8.0 mol water =

  26. mole fraction of ethanol moles solute moles solute + moles solvent = 4.4 mol ethanol 4.4 mol ethanol + 8.0 mol water = = 0.35

  27. Colligative Properties On adding a solute to a solvent, the properties of the solvent are modified. • Vapor pressure decreases • Melting point decreases • Boiling point increases 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.

  28. Change in Freezing Point Ethylene glycol/water solution Pure water The freezing point of a solution is LOWERthan that of the pure solvent

  29. Change in Boiling Point Common Applications of Boiling Point Elevation

  30. Boiling Point Elevation and Freezing Point Depression ∆T = K•m•i i = number of particles produced per molecule/formula unit. For covalent compounds, i = 1. For ionic compounds, i = the number of ions present (both + and -) Compound Theoretical Value of i glycol 1 NaCl 2 CaCl2 3 Ca3(PO4)2 5

  31. Boiling Point Elevation and Freezing Point Depression ∆T = K•m•i m = molality K = molal freezing point/boiling point constant

  32. Boiling Point Elevation – DTb – Addition of non-volatile solute, reducing vapor pressure • Higher temperature needed to elevate vapor pressure to atmospheric pressure in order to boil • DTb - difference between boiling point of solution and boiling point of puresolvent • DTb = Kbm, where Kb is molal boiling point elevation constant (different for each solvent) and m is molality

  33. 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? Kb = 0.52 oC/molal for water (see Kb table). Solution ∆TBP = Kb • m • i 1. Calculate solution molality = 4.00 m 2. ∆TBP = Kb • m • i ∆TBP = 0.52 oC/molal (4.00 molal) (1) ∆TBP = 2.08 oC BP = 100 + 2.08 = 102.08 oC (water normally boils at 100)

  34. Practice Problem: • Water with sugar added to it will boil at a higher temperature than pure water. By how much will the boiling point of water be elevated if 100. g of sucrose, C12H22O11, is added to 500. g of water? • (Kb = 0.52 oC/m)

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

  36. Freezing Point Depression Calculate the Freezing Point of a 4.00 molal glycol/water solution. Kf = 1.86 oC/molal (See Kf table) Solution ∆TFP = Kf • m • i = (1.86 oC/molal)(4.00 m)(1) ∆TFP = 7.44 FP = 0 – 7.44 = -7.44 oC(because water normally freezes at 0)

  37. Freezing Point Depression – ability of dissolved solute to lower freezing point of its solution • Freezing point of substance is temperature at which vapor pressure of solid and liquid phases are same • DTf = Kfm, where Kf is molal freezing point depression constant (different for each solvent) and m is molality

  38. Freezing Point Depression At what temperature will a 5.4 molal solution of NaCl freeze? Solution ∆TFP = Kf • m • i ∆TFP = (1.86 oC/molal) • 5.4 m • 2 ∆TFP = 20.1 oC FP = 0 – 20.1 = -20.1 oC

  39. Practice problem: • While antifreeze protects a car from freezing, it also protects it from overheating. Calculate the freezing point depression of a solution of 100. g of ethylene glycol, C2H6O2, in 0.500 kg of water. (Kf = 1.86 oC/m)

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