Chemistry and Chemical Reactivity 6th Edition 1 John C. Kotz Paul M. Treichel Gabriela C. Weaver CHAPTER 14 Solutions and Their Behavior Lectures written by John Kotz ©2006 2006 Brooks/Cole Thomson © Brooks/Cole - Thomson 2 Solutions Chapter 14 Why does a raw egg swell or shrink when placed in different solutions? © 2006 Brooks/Cole - Thomson Some Definitions A solution is a One constituent is usually HOMOGENEOUS regarded as the SOLVENT mixture of 2 or more and the others as substances in a single SOLUTES. phase. © 2006 Brooks/Cole - Thomson 3 Definitions Solutions can be classified as saturated or unsaturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. © 2006 Brooks/Cole - Thomson 4 Definitions 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. © 2006 Brooks/Cole - Thomson 5 6 Dissolving An Ionic Solid Active Figure 14.9 © 2006 Brooks/Cole - Thomson Energetics of the Solution Process Figure 14.8 © 2006 Brooks/Cole - Thomson 7 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! © 2006 Brooks/Cole - Thomson 8 9 Supersaturated Sodium Acetate • One application of a supersaturated solution is the sodium acetate “heat pack.” • Sodium acetate has an ENDOthermic heat of solution. © 2006 Brooks/Cole - Thomson 10 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 © 2006 Brooks/Cole - Thomson Colligative Properties On adding a solute to a solvent, the props. 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. © 2006 Brooks/Cole - Thomson 11 Concentration Units 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! © 2006 Brooks/Cole - Thomson 12 Concentration Units MOLE FRACTION, X For a mixture of A, B, and C mol A X A mol fraction A = mol A + mol B + mol C MOLALITY, m mol solute m of solute = kilograms solvent WEIGHT % = grams solute per 100 g solution © 2006 Brooks/Cole - Thomson 13 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. © 2006 Brooks/Cole - Thomson 14 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 1.00 mol glycol X glycol = 1.00 mol glycol + 13.9 mol H2O X glycol = 0.0672 © 2006 Brooks/Cole - Thomson 15 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 1.00 mol glycol conc (molality) = 4.00 molal 0.250 kg H2O Calculate weight % 62.1 g %glycol = x 100% = 19.9% 62.1 g + 250. g © 2006 Brooks/Cole - Thomson 16 Dissolving Gases & Henry’s Law Gas solubility (mol/L) = kH • Pgas kH for O2 = 1.66 x 10-6 M/mmHg When Pgas drops, solubility drops. © 2006 Brooks/Cole - Thomson 17 Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution. © 2006 Brooks/Cole - Thomson 18 Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution. © 2006 Brooks/Cole - Thomson 19 Understanding Colligative Properties VP of H2O over a solution depends on the number of H2O molecules per solute molecule. Psolvent proportional to Xsolvent Psolvent = Xsolvent • Posolvent VP of solvent over solution = (Mol frac solvent)•(VP pure solvent) RAOULT’S LAW © 2006 Brooks/Cole - Thomson 20 Raoult’s Law An ideal solution is one that obeys Raoult’s law. PA = XA • PoA 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! © 2006 Brooks/Cole - Thomson 21 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 = (0.9382)(31.8 mm Hg) Pwater = 29.7 mm Hg © 2006 Brooks/Cole - Thomson 22 Raoult’s Law For a 2-component system where A is the solvent and B is the solute ∆PA = VP lowering = XBPoA VP lowering is proportional to mol frac solute! For very dilute solutions, ∆PA = K•molalityB where K is a proportionality constant. This helps explain changes in melting and boiling points. © 2006 Brooks/Cole - Thomson 23 Changes in Freezing and Boiling Points of Solvent VP Pure solvent 1 atm VP solvent after adding solute P BP solution BP pure solvent © 2006 Brooks/Cole - Thomson T See Figure 14.14 24 25 Vapor Pressure Lowering Figure 14.14 © 2006 Brooks/Cole - Thomson The boiling point of a solution is higher than that of the pure solvent. © 2006 Brooks/Cole - Thomson 26 Elevation of Boiling Point Elevation in BP = ∆TBP = KBP•m (where KBP is characteristic of solvent) VP Pure solvent 1 atm VP solvent after adding solute P BP solution BP pure solvent © 2006 Brooks/Cole - Thomson T 27 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 © 2006 Brooks/Cole - Thomson 28 Change in Freezing Point Pure water Ethylene glycol/water solution The freezing point of a solution is LOWER than that of the pure solvent. FP depression = ∆TFP = KFP•m © 2006 Brooks/Cole - Thomson 29 30 Lowering the Freezing Point Water with and without antifreeze © 2006 Brooks/Cole - Thomson When a solution freezes, the solid phase is pure water. The solution becomes more concentrated. Freezing Point Depression Calculate the FP of a 4.00 molal glycol/water solution. KFP = -1.86 oC/molal (Table 14.4) Solution ∆TFP = KFP • m = (-1.86 oC/molal)(4.00 m) ∆TFP = -7.44 oC Recall that ∆TBP = +2.05 ˚C for this solution. © 2006 Brooks/Cole - Thomson 31 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 © 2006 Brooks/Cole - Thomson 32 Freezing Point Depression How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 oC?. 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) © 2006 Brooks/Cole - Thomson 33 Freezing Point Depression How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 oC?. 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) 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 © 2006 Brooks/Cole - Thomson 34 35 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 © 2006 Brooks/Cole - Thomson 36 Osmosis Dissolving the shell in vinegar © 2006 Brooks/Cole - Thomson Egg in pure water Egg in corn syrup 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 © 2006 Brooks/Cole - Thomson 37 38 Process of Osmosis © 2006 Brooks/Cole - Thomson Osmosic Pressure, ∏ Equilibrium is reached when pressure — the OSMOTIC PRESSURE, ∏ — produced by extra Osmotic pressure solution counterbalances pressure of solvent molecules moving thru the membrane. ∏ = cRT (c is conc. in mol/L) © 2006 Brooks/Cole - Thomson 39 Osmosis © 2006 Brooks/Cole - Thomson 40 Osmosis at the Particulate Level Figure 14.17 © 2006 Brooks/Cole - Thomson 41 Osmosis • 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 14.18 © 2006 Brooks/Cole - Thomson 42 43 Osmosis and Living Cells © 2006 Brooks/Cole - Thomson Reverse Osmosis Water Desalination Water desalination plant in Tampa © 2006 Brooks/Cole - Thomson 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 (a) Calc. ∏ in atmospheres ∏ = 10.0 mmHg • (1 atm / 760 mmHg) = 0.0132 atm (b) Calc. concentration © 2006 Brooks/Cole - Thomson 45 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 (b) Calc. concentration from ∏ = cRT 0.0132 atm Conc = (0.0821 L • atm/K • mol)(298K) (c) Conc = 5.39 x 10-4 mol/L Calc. molar mass Molar mass = 35.0 g / 5.39 x 10-4 mol/L Molar mass = 65,100 g/mol © 2006 Brooks/Cole - Thomson 46