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Solutions Chapter 14

Solutions Chapter 14. Why does a raw egg swell or shrink when placed in different solutions?. PLAY MOVIE. Some Definitions. A solution is a HOMOGENEOUS mixture of 2 or more substances in a single phase. One constituent is usually regarded as the SOLVENT and the others as SOLUTES.

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Solutions Chapter 14

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  1. SolutionsChapter 14 Why does a raw egg swell or shrink when placed in different solutions? PLAY MOVIE

  2. Some Definitions A solution is a HOMOGENEOUS mixture of 2 or more substances in a single phase. One constituent is usually regarded as the SOLVENT and the others as SOLUTES.

  3. Definitions Solutions can be classified as saturated or unsaturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. PLAY MOVIE

  4. Definitions Solutions can be classified as unsaturated or saturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. SUPERSATURATED SOLUTIONS contain more than is possible and are unstable. PLAY MOVIE

  5. Dissolving An Ionic Solid See Active Figure 14.9

  6. Concentration Units An IDEAL SOLUTION is one where the properties depend only on the concentration of solute. Need concentration units to tell us the number of solute particles per solvent particle. The unit molarity(review) does not do this!

  7. Concentration Units MOLE FRACTION, X For a mixture of A, B, and C MOLALITY, m WEIGHT % = grams solute per 100 g solution

  8. Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate mol fraction, molality, and weight % of glycol.

  9. Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate X, m, and % of glycol. 250. g H2O = 13.9 mol X glycol = 0.0672

  10. Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate X, m, and % of glycol. Calculate molality Calculate weight %

  11. Energetics of the Solution Process See Energetics of Dissolution of KF simulation

  12. Energetics of the Solution Process If the enthalpy of formation of the solution is more negative than that of the solvent and solute, the enthalpy of solution is negative. The solution process is exothermic! PLAY MOVIE

  13. Supersaturated Sodium Acetate • One application of a supersaturated solution is the sodium acetate “heat pack.” • Sodium acetate has an ENDOthermic heat of solution. PLAY MOVIE

  14. Supersaturated Sodium Acetate Sodium acetate has an ENDOthermic heat of solution. NaCH3CO2 (s) + heat Na+(aq) + CH3CO2-(aq) Therefore, formation of solid sodium acetate from its ions is EXOTHERMIC. Na+(aq) + CH3CO2-(aq)  NaCH3CO2 (s) + heat

  15. Dissolving Gases & Henry’s Law PLAY MOVIE Gas solubility (mol/L) = kH·Pgas kH for O2 = 1.66 x 10-6 M/mmHg (1.3 x 10-3 mol/kg * bar) When Pgas drops, solubility drops.

  16. Raoult’s Law An ideal solution obeys this law. Because solvent vapor pressure & the relative # of solvent molecules are proportional: Psolvent = Xsolvent · Posolvent Because mole fraction of solvent, XA, is always less than 1, then PA is always less than PoA. The vapor pressure of solvent over a solution is always LOWERED!

  17. Raoult’s Law Assume the solution containing 62.1 g of glycol in 250. g of water is ideal. What is the vapor pressure of water over the solution at 30 oC? (The VP of pure H2O is 31.8 mm Hg; see App. E.) Solution Xglycol = 0.0672 and so Xwater = ? Because Xglycol + Xwater = 1 Xwater = 1.000 - 0.0672 = 0.9328 Pwater = Xwater· Powater = (0.9328)(31.8 mm Hg) Pwater = 29.7 mm Hg

  18. 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 • Osmosis is possible (osmotic pressure) 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.

  19. Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution.

  20. Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution. PLAY MOVIE PLAY MOVIE

  21. Understanding Colligative Properties VP of H2O over a solution depends on the number of H2O molecules per solute molecule. Psolventproportional to Xsolvent Psolvent = Xsolvent · Posolvent VP of solvent over solution = (Mol frac solvent)•(VP pure solvent) RAOULT’S LAW

  22. Changes in Freezing and Boiling Points of Solvent See Figure 14.13

  23. Vapor Pressure Lowering See Figure 14.13

  24. The boiling point of a solution is higher than that of the pure solvent. PLAY MOVIE

  25. Elevation of Boiling Point Elevation in BP = ∆TBP = KBP·m (where KBP is characteristic of solvent from table 14.3 p.633)

  26. Change in Boiling Point Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the BP of the solution? KBP = +0.512 oC/molal for water (see Table 14.3). Solution 1. Calculate solution molality = 4.00 m 2. ∆TBP = KBP· m ∆TBP = +0.512 oC/molal (4.00 molal) ∆TBP = +2.05 oC BP = 102.05 oC

  27. Change in Freezing Point Ethylene glycol/water solution Pure water The freezing point of a solution is LOWERthan that of the pure solvent. FP depression = ∆TFP = KFP·m PLAY MOVIE PLAY MOVIE

  28. Lowering the Freezing Point Water with and without antifreeze When a solution freezes, the solid phase is pure water. The solution becomes more concentrated.

  29. Freezing Point Depression Calculate the FP of a 4.00 molal glycol/water solution. KFP = -1.86 oC/molal (Table 14.3) Solution ∆TFP = KFP· m = (-1.86 oC/molal)(4.00 m) ∆TFP = -7.44 oC Recall that ∆TBP = +2.05 ˚C for this solution.

  30. Freezing Point Depression How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 oC? Solution Calc. required molality ∆TFP = KFP· m -10.00 oC = (-1.86 oC/molal)(Conc) Conc = 5.38 molal

  31. Freezing Point Depression How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 oC?. Solution Concreq’d = 5.38 molal This means we need 5.38 mol of dissolved particles per kg of solvent. Recognize that m represents the total concentration of all dissolved particles. Recall that 1 mol NaCl(aq)  1 mol Na+(aq) + 1 mol Cl-(aq)

  32. Freezing Point Depression How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 oC?. Solution Concreq’d = 5.38 molal We need 5.38 mol of dissolved particles per kg of solvent. NaCl(aq)  Na+(aq) + Cl-(aq) To get 5.38 mol/kg of particles we need 5.38 mol / 2 = 2.69 mol NaCl / kg 2.69 mol NaCl / kg  157 g NaCl / kg (157 g NaCl / kg)(4.00 kg) = 629 g NaCl

  33. Boiling Point Elevation and Freezing Point Depression ∆T = K·m·i A generally useful equation i = van’t Hoff factor = number of particles produced per formula unit. Compound Theoretical Value of i glycol 1 NaCl 2 CaCl2 3

  34. Osmosis PLAY MOVIE Dissolving the shell in vinegar Egg in pure water Egg in corn syrup

  35. Osmosis The semipermeable membrane allows only the movement of solvent molecules. Solvent molecules move from pure solvent to solution in an attempt to make both have the same concentration of solute. Driving force is entropy

  36. Process of Osmosis

  37. Osmotic pressure Osmotic Pressure, ∏ Equilibrium is reached when pressure — the OSMOTIC PRESSURE, ∏ — produced by extra solution counterbalances pressure of solvent molecules moving thru the membrane. ∏ = cRT (c is conc. in mol/L) (R = 0.082057 L*atm/K*mol)

  38. Osmosis PLAY MOVIE PLAY MOVIE

  39. Osmosis at the Particulate Level See Figure 14.17

  40. Osmosis • Osmosis of solvent from one solution to another can continue until the solutions are ISOTONIC— they have the same concentration.

  41. Osmosis and Living Cells

  42. Reverse OsmosisWater Desalination Water desalination plant in Tampa

  43. Osmosis Calculating a Molar Mass Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. ∏ measured to be 10.0 mm Hg at 25 ˚C. Calc. molar mass of hemoglobin. Solution (a) Calc. ∏ in atmospheres ∏ = (10.0 mmHg)(1 atm / 760 mmHg) = 0.0132 atm (b) Calc. concentration

  44. Osmosis Calculating a Molar Mass • Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. ∏ measured to be 10.0 mm Hg at 25 ˚C. Calc. molar mass of hemoglobin. • Solution • Calc. concentration from ∏ = cRT • Conc. =0.0132 atm / (0.0821) (298 K) • L*atm/K*mol Conc = 5.39 x 10-4 mol/L (c) Calc. molar mass Molar mass = 35.0 g / 5.39 x 10-4 mol/L Molar mass = 65,100 g/mol

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