Chapter14

Chapter14 - Solutions and their Behavior QuickTime and a...

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Unformatted text preview: Solutions and their Behavior QuickTime and a TIFF (Uncompressed) decompressor are needed to see this picture. + QuickTime and a TIFF (Uncompressed) decompressor are needed to see this picture. QuickTime and a QuickTime and this decompressor TIFF (Uncompressed) picture.decompressor TIFF see a are needed to(Uncompressed) are needed to see this picture. Solutions Why does a raw egg swell or shrink when placed in different solutions? Some Definitions A solution is a mixture of 2 or more substances in a single phase. One constituent is usually regarded as the solvent and the others as solutes. Definitions Solutions can be saturated or unsaturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. SUPERSATURATED SOLUTIONS contain more than is possible and are unstable. Dissolving An Ionic Solid Energetics of the Solution Energetics of the Solution 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! Supersaturated - Sodium Acetate One application of a supersaturated solution is the sodium acetate "heat pack." Sodium acetate has an endothermic heat of solution. 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 Colligative Properties 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. 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" does not do this! Concentration Units Mole Fraction, X For a mixture of A, B, and C Mol fraction A = ________mol A_____ Mol A + Mol B + Mol C MOLALITY, m = mol solute kg solvent WEIGHT % = grams solute per 100 g solution 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. Calculating Concentrations Calculate # moles then mole percent 250. g H2O = 13.9 mol x glycol 1.00mol = = 0.0672mol% 1.00mol + 13.9mol Calculate molality 1.00mol molality = = 4.00molal 0.250kg S Calculate weight % S 62.1g %glycol = 100% = 19.9% 62.1g + 250g 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. Understanding Colligative Properties To understand colligative properties, study the liquid-vapor equilibrium for a solution. Understanding Colligative Properties VP of H2O over a solution depends on the number of H2O molecules per solute molecule. Psolvent = Xsolvent P=osolvent VP of solvent over solution = (Mol frac solvent) (VP pure solvent) RAOULT'S LAW The mole fraction of solvent, XA, is always less than 1 PA is always less than PoA The VP of solvent over a solution is always lowered! 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) 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 Raoult's Law For a 2-comp. 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. Elevation of Boiling Point Elevation in BP = TBP = KBPm (where KBP is characteristic of solvent) VP Pure solvent 1 atm P VP solvent after adding solute BP solution T BP pure solvent 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. 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 Freezing Point Depression Calculate the FP of a 4.00 molal glycol/water solution. KFP = -1.86 oC/molal Solution TFP = KFP m = (-1.86 oC/molal)(4.00 m) TFP = -7.44 oC Recall that TBP = +2.05 C for this solution. Freezing Point Depression How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 oC?. Solution Calculate required molality TFP = KFP m -10.00 oC = (-1.86 oC/molal) Conc Conc = 5.38 molal This means we need 5.38 mol of total dissolved particles per kg of solvent. Freezing Point Depression How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 oC?. Solution 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 BP Elevation and FP Depression T = Kmi A generally useful equation i = van't Hoff factor = number of particles produced per formula unit. Compound glycol NaCl CaCl2 Theoretical Value of i 1 2 3 Osmosis Semipermeable membrane allows only the movement of solvent molecules. Driving force is entropy. Osmotic pressure, = cRT (c is conc. in mol/L) Osmosis at the Particle Level Reverse Osmosis H2O Desalination 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. (a)Calc. in atmospheres (b) S Calc. concentration 1atm S = 10mmHg = 0.0132atm 760mmHg 0.0132atm C= = 5.39 10- 4 mol /L L atm (0.0821 )(298K) K mol (c) S 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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