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Chapter 12—Solutions and Properties

Chapter 12—Solutions and Properties. Chapter 12—Solutions (and their properties). Solutions have several terms to remember or else problem solving is nearly impossible Solute—material present in limited amount Solvent—material present is greater amount solution is the solvent + solute(s).

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Chapter 12—Solutions and Properties

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  1. Chapter 12—Solutions and Properties • Chapter 12—Solutions (and their properties). • Solutions have several terms to remember or else problem solving is nearly impossible • Solute—material present in limited amount • Solvent—material present is greater amount • solution is the solvent + solute(s). • First things first…need to define what a solution IS, starting with a ‘homogeneous mixture’ where solute and solvent are mixed IN THE SAME PHASE (see Chapter 3 for EXACT definitions. • Some real world ‘solutions’ some of which may not be obvious

  2. Aqueous Solutions of Ionic Compounds • The forces causing an ionic solid to dissolve in water are ion–dipole forces, the attraction of water dipoles for cations and anions. • The attractions of water dipoles for ions pulls the ions out of the crystalline lattice and into aqueous solution. • The extent to which an ionic solid dissolves in water is determined largely by the competition between: • interionic attractions that hold ions in a crystal, and • ion–dipole attractions that pull ions into solution.

  3. Ion–Dipole Forces in Dissolution Negative ends of dipoles attracted to cations. Positive ends of dipoles attracted to anions.

  4. Predict whether each combination is likely to be a solution or a heterogeneous mixture: (a) methanol, CH3OH, and water, HOH (b) pentane, CH3(CH2)3CH3, and octane, CH3(CH2)6CH3 (c) sodium chloride, NaCl, and carbon tetrachloride, CCl4 (d) 1-decanol, CH3(CH2)8CH2OH, and water, HOH • –intermolecular forces are nearly identical…like-like (solution) • Pentane/octane…straight chain hydrocarbons…(solution) • Non-polar solvent (CCl4), ionic solid (NaCl)…ahhh…not so much • Decanol (deca—10; -ol means OH) and water…some similar

  5. Some Solubility Terms • Liquids that mix in all proportions are called miscible. • The methanol/water question on previous panel • When there is a dynamic equilibrium between an undissolved solute and a solution, the solution is saturated. • Even NaCl isn’t infinitely soluble in water… • The concentration of the solute in a saturated solution is the solubility of the solute. • A solution which contains less solute than can be held at equilibrium is unsaturated.

  6. Formation of a Saturated Solution (3 step) Eventually, the rates of dissolving and of crystallization are equal; no more solute appears to dissolve. Solid begins to dissolve. Longer standing does not change the amount of dissolved solute. As solid dissolves, some dissolved solute begins to crystallize.

  7. Solubility and Temperature • Most ionic compounds have aqueous solubilities that increase significantly with increasing temperature. • MORE ENERGY (temp) to break ionic bonds… • A few have solubilities that change little with temperature. • A very few have solubilities that decrease with increasing temperature. • If solubility increases with temperature, a hot, saturated solution may be cooled (carefully!) without precipitation of the excess solute. This creates a supersaturated solution. • Supersaturated solutions ordinarily are unstable …

  8. Solutions have units…lots of units • Familiar with some of them…molarity (moles of solute per liter of solution). • Bad news…there are a LOT of concentration units. • Good news? They’re really easy to remember • Mole fraction…moles of solute/moles of (solute+solvent) • Weight %...(mass of component/ mass of solution) * 100 • we’ll come back to that multiplication factor • Keep in mind that these “solutions” can have several components (salt water has Na+, Cl-, and H2O)

  9. Couple of quick examples… • Suppose you have a 10.0 g glucose solution that dissolves in 100.0 g of H2O. What is the wt. %. • Simple…10.0 g / (100.0 g + 10.0 g) = 9.09 wt. % glucose • A sin of sorts…didn’t convert to moles…did you? • Other mass comparisons? ppm? ppb? SAME CALC • Different conversion though…but not too difficult • mass component / mass total solution * 100 wt % • * 106 (ppm…106 is 1 million) • * 109 (ppb…same as above…but a billion)

  10. Summary of concentration units • moLALITY (must mean a MOLE unit…yes?). Just a different solvent unit. Moles of solute (same as before) divided by the kg of solvent. • Unique applications of this unit…we’ll get more into that in a bit

  11. Now with molality…etc…sample prob? • Find mass % in each of the following… • 4.12 g NaOH in 100.0 g water • 5.00 mL ethanol (d = 0.789 g/mL) in 50.0 g water • 1.50 mL glycerol (d = 1.324 g/mL) in 22.25 mL water (d=.998) • Express the following in the indicated unit • 1.0 ug benzene/L H2O (in ppb) • 0.0035% NaCl by mass, as ppm NaCl • 2.4 ppm F-, as molarity of fluoride ion • How many moles of ethylene glycol are present in 2.30 L of 6.27 m (small m) glycol solution (d = 1.035)

  12. Should finish Chapter 11 this week • Finished Friday with a couple of sample problems • Today—finish up the sample calcs • Phase diagram…and how they apply to… • Colligative Properties

  13. Expanded ppm—Mol problem • Recall that ppm  [g solute / g solution] * 106 • So if given ppm, and you need to convert…first divide by 106 • You are then left with units of grams solute / g solution

  14. Quick sample problem of molality • Yep, that’s the “small m” concentration unit • Good thing about molality is that this is pretty much it as far as this unit is concerned • Rest of course will focus on molarity • Find molality of a sucrose solution (m.m. = 342.3) formed when 1.45 g sucrose is dissolved in 30.0 mL of water. • Convert sucrose to mol • Convert volume to mass, then to kg

  15. Pressure increase = [increase] • The [gases] increases with increased pressure. • Can of Coke is probably best example (open)

  16. Colligative Properties • Colligative Properties: Depend only on the number of solute particles in solution. These affect properties of the solvent. • There are four main colligative properties: • Vapor pressure lowering • Freezing point depression • Boiling point elevation • Osmotic pressure

  17. Solution Vapor Pressures • So the past few days of summer have been humid—which means there is water vapor in the air • For water at 30 °C (vp = 31.8 mmHg) • So what happens when pure water (at 30 °C) becomes a salt solution? • The vapor pressure drops! Not as much water vapor in the air!! • V.p of water drops to say…25 mm Hg)… what does this really mean? • First…it’s time to head back to Chap 10 (phase diagram)

  18. Phase Diagrams A—C, liquid-vapor equilibrium. A phase diagram is a graphical representation of the conditions of temperature and pressure under which a substance exists as a solid, liquid, a gas, or some combination of these in equilibrium. A—D, solid-liquid equilibrium. A—B, solid-vapor equilibrium. Triple point

  19. Vapor Pressure Lowering by a Nonvolatile Solute … the vapor pressure from the pure solvent. Raoult’s Law: the vapor pressure from a solution (nonvolatile solute) is lower than … Result: the boiling point of the solution increases by DTb.

  20. Cook food a bit faster? • Boiling-Point Elevation (∆Tb): The boiling point of the solution (Tb) minus the boiling point of the pure solvent (T°b):∆Tb = Tb – T°b • ∆Tb is proportional to concentration:∆Tb = Kb m Kb= molal boiling-point elevation constant.

  21. So…salt doesn’t help… • Left off with our attempt at inducing heart disease. We fail. • Food for thought (76,, 81, 84-87, 92-99, 101, 106, 107, 108, 112, 116, 133, 135, 136) • Now let’s look at a situation closer to reality. Will the snow EVER melt?

  22. Boiling-Point Elevation and Freezing-Point Depression

  23. Why the snow melts • Freezing-Point Depression (∆Tf):The freezing point of the pure solvent (T°f) minus the freezing point of the solution (Tf). ∆Tf = T°f – Tf • ∆Tf is proportional to concentration:∆Tf = Kf mKf = molal freezing-point depression constant.

  24. Another wrinkle in the fabric • van’t Hoff Factor, i: This factor equals the number of ions produced from each molecule of a compound upon dissolving. • i = 1 for CH3OH i = 3 for CaCl2 • i = 2 for NaCl i = 5 for Ca3(PO4)2 • For compounds that dissociate on dissolving, use: ∆Tb = iKbm∆Tf = iKf m ∆P = ix2P°1

  25. Boiling-Point Elevation and Freezing-Point Depression Nonelectrolytes Electrolytes DTb = Kbm DTb = Kbm i DTf = Kfm DTf = Kfmi

  26. T = DH DS Boiling-Point Elevation and Freezing-Point Depression DG = DH - TDS At equilibrium, DG = 0.

  27. T = DH DS Boiling-Point Elevation and Freezing-Point Depression DG = DH - TDS At equilibrium, DG = 0.

  28. mol kg Boiling-Point Elevation and Freezing-Point Depression What is the freezing point (in °C) of a solution prepared by dissolving 7.40 g of MgCl2 in 110 g of water? The van’t Hoff factor for MgCl2 is i = 2.7. Calculate the moles of MgCl2: 7.40 g 1 mol x = 0.0777 mol 95.2 g Calculate the molality of the solution: 0.0777 mol 1000 g x = 0.71 110 g 1 kg

  29. °C kg mol mol kg Boiling-Point Elevation and Freezing-Point Depression Calculate the freezing point of the solution: DTf = Kfmi = 1.86 x 0.71 x 2.7 = 3.6 °C Tf = 0.0 °C - 3.6 °C = -3.6 °C

  30. Last bit of Chapter 11… • You look so sad…but NEVER FEAR!! I have the PERFECT CURE for the disappointment of today • We can learn about OSMOSIS!! The perfect study technique for our exam on Monday. Well perhaps not the BEST way to study . • We left off yesterday (before your hopes for a snow day were dashed) with a Freezing Point depression calculation… • Want to spice things up a bit?

  31. Example 11.14 What is the freezing point of an aqueous sucrose solution that has 25.0 g C12H22O11 per 100.0 g H2O? Example 11.15 Sorbitol is a sweet substance found in fruits and berries and sometimes used as a sugar substitute. An aqueous solution containing 1.00 g sorbitol in 100.0 g water is found to have a freezing point of –0.102 °C. Elemental analysis indicates that sorbitol consists of 39.56% C, 7.75% H, and 52.70% O by mass. What are the (a) molar mass and (b) molecular formula of sorbitol? So, where do we get this MM of sorbitol? Conjure it out of thin air? I’m OK with that… well...not so much

  32. Yeah, so about that osmosis thing • Everybody knows about the idea of sleeping with a book under your pillow… • Theory: the knowledge of the book transferred into your brain (by osmosis) as you sleep. • Sounds good, but after the ‘actual’ discussion, you might change your mind.

  33. Osmosis and Osmotic Pressure • Osmosis: The selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. • Osmotic pressure (π or ∏): The pressure required to stop osmosis. π = iMRT R = 0.08206 (Latm)/(molK)

  34. Osmosis and Osmotic Pressure

  35. Osmosis and Osmotic Pressure • What is the osmotic pressure (in atm) of a 0.884 M sucrose solution at 16°C?

  36. Uses of Colligative Properties • Desalination:

  37. Osmosis and Osmotic Pressure • Isotonic: Solutions have equal concentration of solute, and so equal osmotic pressure. • Hypertonic: Solution with higher concentration of solute. • Hypotonic: Solution with lower concentration of solute.

  38. Practical Applications of Osmosis Ordinarily a patient must be given intravenous fluids that are isotonic—have the same osmotic pressure as blood. External solution is hypertonic; produces osmotic pressure > πint. Net flow of water out of the cell. Red blood cell in isotonic solution remains the same size. External solution is hypotonic; produces osmotic pressure < πint. Net flow of water into the cell.

  39. Example 12.16 An aqueous solution is prepared by dissolving 1.50 g of hemocyanin, a protein obtained from crabs, in 0.250 L of water. The solution has an osmotic pressure of 0.00342 atm at 277 K. (a) What is the molar mass of hemocyanin? (b) What should the freezing point of the solution be?

  40. Sample Problem • How many grams of ethylene glycol antifreeze, CH2(OH)CH2(OH), must you dissolve in one liter of water to get a freezing point of –20.0°C. The molar mass of ethylene glycol is 62.01 g. For water, Kf = 1.86 (°C·kg)/mol. What will be the boiling point?

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