Solutions Chapter 14 Why does a raw egg swell or shrink when placed in different solutions? Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

2 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Solutions can be classified as unsaturated or saturated. DefinitionsDefinitions

4 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Solutions can be classified as unsaturated or saturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. DefinitionsDefinitions

5 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Solutions can be classified as unsaturated or saturated. A saturated solution contains the maximum quantity of solute that dissovles at that temperature. SUPERSATURATED SOLUTIONS contain more than is possible and are unstable. DefinitionsDefinitions

6 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Energetics of the Solution Process If the enthalpy of formation of the solution is more negative that that of the solvent and solute, the enthalpy of solution is negative. The solution process is exothermic!

7 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Supersaturated Sodium Acetate Supersaturated Sodium Acetate One application of a supersaturated solution is the sodium acetate “heat pack.”One application of a supersaturated solution is the sodium acetate “heat pack.” Sodium acetate has an ENDOthermic heat of solution.Sodium acetate has an ENDOthermic heat of solution.

8 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Supersaturated Sodium Acetate Sodium acetate has an ENDOthermic heat of solution. NaCH 3 CO 2 (s) + heat ----> Na + (aq) + CH 3 CO 2 - (aq) Therefore, formation of solid sodium acetate from its ions is EXOTHERMIC. Na + (aq) + CH 3 CO 2 - (aq) ---> NaCH 3 CO 2 (s) + heat

9 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Colligative Properties On adding a solute to a solvent, the props. of the solvent are modified. Vapor pressure decreasesVapor pressure decreases Melting point decreasesMelting point decreases Boiling point increasesBoiling point increases Osmosis is possible (osmotic pressure)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. On adding a solute to a solvent, the props. of the solvent are modified. Vapor pressure decreasesVapor pressure decreases Melting point decreasesMelting point decreases Boiling point increasesBoiling point increases Osmosis is possible (osmotic pressure)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.

10 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved An IDEAL SOLUTION is one where the properties depend only on the concentration of solute. Need conc. units to tell us the number of solute particles per solvent particle. The unit “molarity” does not do this! An IDEAL SOLUTION is one where the properties depend only on the concentration of solute. Need conc. units to tell us the number of solute particles per solvent particle. The unit “molarity” does not do this! Concentration Units

11 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved MOLE FRACTION, X For a mixture of A, B, and C MOLE FRACTION, X For a mixture of A, B, and C Concentration Units

12 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Concentration Units MOLE FRACTION, X For a mixture of A, B, and C MOLALITY, m MOLE FRACTION, X For a mixture of A, B, and C MOLALITY, m

13 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Concentration Units MOLE FRACTION, X For a mixture of A, B, and C MOLALITY, m WEIGHT % = grams solute per 100 g solution MOLE FRACTION, X For a mixture of A, B, and C MOLALITY, m WEIGHT % = grams solute per 100 g solution

14 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H 2 O. Calculate mol fraction, molality, and weight % of glycol.

15 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculating Concentrations 250. g H 2 O = 13.9 mol Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H 2 O. Calculate X, m, and % of glycol.

16 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculating Concentrations 250. g H 2 O = 13.9 mol X glycol = g H 2 O = 13.9 mol X glycol = Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H 2 O. Calculate X, m, and % of glycol.

17 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculating Concentrations Calculate molality Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H 2 O. Calculate X, m, and % of glycol.

18 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculating Concentrations Calculate molality Calculate weight % Calculate molality Calculate weight % Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H 2 O. Calculate X, m, and % of glycol.

19 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Dissolving Gases & Henry’s Law Gas solubility (M) = k H P gas k H for O 2 = 1.66 x M/mmHg When P gas drops, solubility drops.

20 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Lake Nyos, Cameroon Courtesy of George Kling, page

21 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution.

22 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Understanding Colligative Properties To understand colligative properties, study the LIQUID- VAPOR EQUILIBRIUM for a solution.

23 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Understanding Colligative Properties VP of H 2 O over a solution depends on the number of H 2 O molecules per solute molecule. P solvent proportional to X solvent P solvent proportional to X solvent OR OR P solvent = X solvent P o solvent P solvent = X solvent P o solvent VP of solvent over solution = (Mol frac solvent)(VP pure solvent) RAOULT’S LAW RAOULT’S LAW VP of H 2 O over a solution depends on the number of H 2 O molecules per solute molecule. P solvent proportional to X solvent P solvent proportional to X solvent OR OR P solvent = X solvent P o solvent P solvent = X solvent P o solvent VP of solvent over solution = (Mol frac solvent)(VP pure solvent) RAOULT’S LAW RAOULT’S LAW

24 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Raoult’s Law An ideal solution is one that obeys Raoult’s law. P A = X A P o A P A = X A P o A Because mole fraction of solvent, X A, is always less than 1, then P A is always less than P o A. The vapor pressure of solvent over a solution is always LOWERED ! An ideal solution is one that obeys Raoult’s law. P A = X A P o A P A = X A P o A Because mole fraction of solvent, X A, is always less than 1, then P A is always less than P o A. The vapor pressure of solvent over a solution is always LOWERED !

25 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 o C? (The VP of pure H 2 O is 31.8 mm Hg; see App. E.) Solution X glycol = and so X water = ? Because X glycol + X water = 1 X water = = P water = X water P o water = (0.9382)(31.8 mm Hg) P water = 29.7 mm Hg 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 o C? (The VP of pure H 2 O is 31.8 mm Hg; see App. E.) Solution X glycol = and so X water = ? Because X glycol + X water = 1 X water = = P water = X water P o water = (0.9382)(31.8 mm Hg) P water = 29.7 mm Hg

26 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Raoult’s Law For a 2-component system where A is the solvent and B is the solute P A = VP lowering = X B P o A  P A = VP lowering = X B P o A VP lowering is proportional to mol frac solute! For very dilute solutions, P A = Kmolality B where K is a proportionality constant. For very dilute solutions,  P A = Kmolality B where K is a proportionality constant. This helps explain changes in melting and boiling points. For a 2-component system where A is the solvent and B is the solute P A = VP lowering = X B P o A  P A = VP lowering = X B P o A VP lowering is proportional to mol frac solute! For very dilute solutions, P A = Kmolality B where K is a proportionality constant. For very dilute solutions,  P A = Kmolality B where K is a proportionality constant. This helps explain changes in melting and boiling points.

27 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Changes in Freezing and Boiling Points of Solvent See Figure 14.13

28 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The boiling point of a solution is higher than that of the pure solvent.

29 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Elevation of Boiling Point Elevation in BP = t BP = K BP m Elevation in BP =  t BP = K BP m (where K BP is characteristic of solvent)

30 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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? K BP = o C/molal for water (see Table 14.3). Solution 1.Calculate solution molality = 4.00 m 2.t BP = K BP m 2.  t BP = K BP m t BP = o C/molal (4.00 molal)  t BP = o C/molal (4.00 molal) t BP = o C  t BP = o C BP = o C

31 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Change in Freezing Point The freezing point of a solution is LOWER than that of the pure solvent. FP depression = t FP = K FP m FP depression =  t FP = K FP m Pure water Ethylene glycol/water solution

32 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Freezing Point Depression Consider equilibrium at melting point Liquid solvent Solid solvent Rate at which molecules go from S to L depends only on the nature of the solid.Rate at which molecules go from S to L depends only on the nature of the solid. BUT — rate for L ---> S depends on how much is dissolved. This rate is SLOWED for the same reason VP is lowered.BUT — rate for L ---> S depends on how much is dissolved. This rate is SLOWED for the same reason VP is lowered. Therefore, to bring S ---> L and L ---> S rates into equilibrium for a solution, T must be lowered.Therefore, to bring S ---> L and L ---> S rates into equilibrium for a solution, T must be lowered. Thus, FP for solution < FP for solvent FP depression = t FP = K FP m FP depression =  t FP = K FP m

33 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculate the FP of a 4.00 molal glycol/water solution. K FP = o C/molal (Table 14.4) Solution t FP = K FP m  t FP = K FP m = (-1.86 o C/molal)(4.00 m) = (-1.86 o C/molal)(4.00 m) t FP = o C  t FP = o C Freezing Point Depression

34 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved How much NaCl must be dissolved in 4.00 kg of water to lower FP to o C?. Solution Calc. required molality t FP = K FP m  t FP = K FP m o C = (-1.86 o C/molal) Conc o C = (-1.86 o C/molal) Conc Conc = 5.38 molal Conc = 5.38 molal Freezing Point Depression

35 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved How much NaCl must be dissolved in 4.00 kg of water to lower FP to o C?. Solution Conc req’d = 5.38 molal This means we need 5.38 mol of dissolved particles per kg of solvent. Recognize that m represents the total conc. of all dissolved particles. Recall that 1 mol NaCl(aq) --> 1 mol Na + (aq) + 1 mol Cl - (aq) Freezing Point Depression

36 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved How much NaCl must be dissolved in 4.00 kg of water to lower FP to o C?. Solution Conc req’d = 5.38 molal We need 5.38 mol of dissolved particles per kg of solvent. NaCl(aq) --> Na + (aq) + Cl - (aq) 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 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 Freezing Point Depression

37 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Boiling Point Elevation and Freezing Point Depression t = K m i  t = K m i A generally useful equation i = van’t Hoff factor = number of particles produced per formula unit. CompoundTheoretical Value of i glycol1 NaCl2 CaCl 2 3

38 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved OsmosisOsmosis

39 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved OsmosisOsmosis The semipermeable membrane should allow only the movement of solvent molecules. Therefore, solvent molecules move from pure solvent to solution.

40 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved OsmosisOsmosis The semipermeable membrane should allow only the movement of solvent molecules. Therefore, solvent molecules move from pure solvent to solution.

41 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved OsmosisOsmosis Equilibrium is reached when pressure produced by extra solution — the OSMOTIC PRESSURE, the OSMOTIC PRESSURE,  = cRT (where c is conc. in mol/L)  = cRT (where c is conc. in mol/L) counterbalances pressure of solvent molecules moving thru the membrane.

42 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved OsmosisOsmosis

43 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved OsmosisOsmosis Osmosis of solvent from one solution to another can continue until the solutions are ISOTONIC — they have the same concentration.Osmosis of solvent from one solution to another can continue until the solutions are ISOTONIC — they have the same concentration. Osmotic pressure in living systems: FIGURE 14.16Osmotic pressure in living systems: FIGURE 14.16

44 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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. 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 (a) Calc.  in atmospheres = 10.0 mmHg (1 atm / 760 mmHg)  = 10.0 mmHg (1 atm / 760 mmHg) = atm (b)Calc. concentration

45 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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. 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 (b)Calc. concentration from = cRT (b)Calc. concentration from  = cRT

46 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Osmosis Calculating a Molar Mass Conc = 5.39 x mol/L Conc = 5.39 x mol/L (c)Calc. molar mass Molar mass = 35.0 g / 5.39 x mol/L Molar mass = 35.0 g / 5.39 x mol/L Molar mass = 65,100 g/mol Molar mass = 65,100 g/mol 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. 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 (b)Calc. concentration from = cRT (b)Calc. concentration from  = cRT