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1 © 2006 Brooks/Cole - Thomson Solutions Why does a raw egg swell or shrink when placed in different solutions?

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Presentation on theme: "1 © 2006 Brooks/Cole - Thomson Solutions Why does a raw egg swell or shrink when placed in different solutions?"— Presentation transcript:

1 1 © 2006 Brooks/Cole - Thomson Solutions Why does a raw egg swell or shrink when placed in different solutions?

2 2 © 2006 Brooks/Cole - Thomson 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 3 © 2006 Brooks/Cole - Thomson Solutions can be classified as saturated or unsaturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. Definitions SUPERSATURATED SOLUTIONS contain more than is possible and are unstable.

4 4 © 2006 Brooks/Cole - Thomson Dissolving An Ionic Solid Active Figure 14.9

5 5 © 2006 Brooks/Cole - Thomson Energetics of the Solution Process Figure 14.8

6 6 © 2006 Brooks/Cole - Thomson 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 7 © 2006 Brooks/Cole - Thomson 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 8 © 2006 Brooks/Cole - Thomson 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 9 © 2006 Brooks/Cole - Thomson Colligative Properties On adding a solute to a solvent, the properties 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 10 © 2006 Brooks/Cole - Thomson 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” does not do this! Concentration Units

11 11 © 2006 Brooks/Cole - Thomson Concentration Units MOLE FRACTION, X For a mixture of A, B, and C WEIGHT % = grams solute per 100 g solution (also ppm, ppb, which is like %, which is parts per hundred) MOLALITY, m

12 12 © 2006 Brooks/Cole - Thomson 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 ethylene glycol.

13 13 © 2006 Brooks/Cole - Thomson 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. X glycol = Calculate mole fraction

14 14 © 2006 Brooks/Cole - Thomson 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. Calculate weight %

15 15 © 2006 Brooks/Cole - Thomson Dissolving Gases & Henry’s Law S g Gas solubility (mol/L) = k H P gas k H for O 2 = 1.66 x M/mmHg When P gas drops, solubility drops.

16 16 © 2006 Brooks/Cole - Thomson Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution.

17 17 © 2006 Brooks/Cole - Thomson P solvent = X solvent P o solvent Understanding Colligative Properties Vapor pressure 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 VP of solvent over solution = (Mol frac solvent)(VP pure solvent) RAOULT’S LAW RAOULT’S LAW

18 18 © 2006 Brooks/Cole - Thomson P A = X A P o A Raoult’s Law An ideal solution is one that obeys Raoult’s law. 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 !

19 19 © 2006 Brooks/Cole - Thomson Vapor Pressure Lowering Figure 14.14

20 20 © 2006 Brooks/Cole - Thomson 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) 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

21 21 © 2006 Brooks/Cole - Thomson Elevation of Boiling Point Elevation in BP = ∆T BP = K BP m (where K BP is characteristic of solvent) The boiling point of a solution is higher than that of the pure solvent.

22 22 © 2006 Brooks/Cole - Thomson 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 ∆T BP = o C/molal (4.00 molal) ∆T BP = o C/molal (4.00 molal) ∆T BP = o C BP = o C

23 23 © 2006 Brooks/Cole - Thomson 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 Pure water Ethylene glycol/water solution

24 24 © 2006 Brooks/Cole - Thomson Lowering the Freezing Point Water with and without antifreezeWhen a solution freezes, the solid phase is pure water. The solution becomes more concentrated.

25 25 © 2006 Brooks/Cole - Thomson How much NaCl must be dissolved in 4.00 kg of water to lower FP to o C?. Solution Calc. required molality (*This one is a little different because the solute is ionic, not molecular.) ∆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

26 26 © 2006 Brooks/Cole - Thomson 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 concentration of all dissolved particles. Recall that 1 mol NaCl(aq)--> 1 mol Na + (aq) + 1 mol Cl - (aq) Freezing Point Depression

27 27 © 2006 Brooks/Cole - Thomson 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. Each mol of NaCl contributes 2 mol of dissolved particles. 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

28 28 © 2006 Brooks/Cole - Thomson Boiling Point Elevation and Freezing Point Depression ∆T = Kmi ∆T = Kmi 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

29 29 © 2006 Brooks/Cole - Thomson Osmosis Dissolving the shell in vinegar Egg in corn syrupEgg in pure water

30 30 © 2006 Brooks/Cole - Thomson 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

31 31 © 2006 Brooks/Cole - Thomson Process of Osmosis

32 32 © 2006 Brooks/Cole - Thomson Osmosis at the Particulate Level Figure 14.17

33 33 © 2006 Brooks/Cole - Thomson Osmosis at the Particulate Level Figure 14.17

34 34 © 2006 Brooks/Cole - Thomson Osmosis 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.

35 35 © 2006 Brooks/Cole - Thomson Osmosis and Living Cells

36 36 © 2006 Brooks/Cole - Thomson Reverse Osmosis Water Desalination Water desalination plant in Tampa

37 37 © 2006 Brooks/Cole - Thomson 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) ∏ = 10.0 mmHg (1 atm / 760 mmHg) = atm (b)Calculate molarity

38 38 © 2006 Brooks/Cole - Thomson 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. Solution (b)Calc. molarity from ∏ = iMRT


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