Presentation on theme: "Why does a raw egg swell or shrink when placed in different solutions?"— Presentation transcript:
1 Why does a raw egg swell or shrink when placed in different solutions?
2 Some DefinitionsA 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.SUPERSATURATED SOLUTIONS contain more than is possible and are unstable.
6 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 Supersaturated Sodium Acetate One application of a supersaturated solution is the sodium acetate “heat pack.”Sodium acetate has an ENDOthermic heat of solution.
8 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
9 Colligative Properties On adding a solute to a solvent, the properties of the solvent are modified.Vapor pressure decreasesMelting point decreasesBoiling point increasesOsmosis 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 Concentration UnitsAn 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!
11 Concentration Units MOLE FRACTION, X MOLALITY, m For a mixture of A, B, and CMOLALITY, mWEIGHT % = grams solute per 100 g solution(also ppm, ppb, which is like %, which is partsper hundred)
12 Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate mol fraction, molality, and weight % of ethylene glycol.
13 Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate X, m, and % of glycol.Calculate mole fraction250. g H2O = 13.9 molX glycol =
14 Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate X, m, and % of glycol.Calculate molalityCalculate weight %
15 Dissolving Gases & Henry’s Law Sg Gas solubility (mol/L) = kH • PgaskH for O2 = x 10-6 M/mmHgWhen Pgas drops, solubility drops.
16 Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution.
17 Understanding Colligative Properties Vapor pressure of H2O over a solution depends on the number of H2O molecules per solute molecule.Psolvent proportional to XsolventPsolvent = Xsolvent • PosolventVP of solvent over solution = (Mol frac solvent)•(VP pure solvent)RAOULT’S LAW
18 Raoult’s Law PA = XA • PoA An ideal solution is one that obeys Raoult’s law.PA = XA • PoABecause 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!
20 Raoult’s Law Pwater = 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 oC? (The VP of pure H2O is 31.8 mm Hg)SolutionXglycol = and so Xwater = ?Because Xglycol + Xwater = 1Xwater = =Pwater = Xwater • Powater = (0.9382)(31.8 mm Hg)Pwater = 29.7 mm Hg
21 Elevation of Boiling Point Elevation in BP = ∆TBP = KBP•m(where KBP is characteristic of solvent)The boiling point of a solution is higher than that of the pure solvent.
22 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 = oC/molal for water (see Table 14.3).Solution1. Calculate solution molality = 4.00 m2. ∆TBP = KBP • m∆TBP = oC/molal (4.00 molal)∆TBP = oCBP = oC
23 Change in Freezing Point Ethylene glycol/watersolutionPure waterThe freezing point of a solution is LOWER than that of the pure solvent.FP depression = ∆TFP = KFP•m
24 Lowering the Freezing Point Water with and without antifreezeWhen a solution freezes, the solid phase is pure water. The solution becomes more concentrated.
25 Freezing Point Depression How much NaCl must be dissolved in kg of water to lower FP to oC?.SolutionCalc. required molality (*This one is a little different because the solute is ionic, not molecular.)∆TFP = KFP • moC = (-1.86 oC/molal) • ConcConc = molal
26 Freezing Point Depression How much NaCl must be dissolved in 4.00 kg of water to lower FP to oC?.SolutionConc req’d = 5.38 molalThis 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)
27 Freezing Point Depression How much NaCl must be dissolved in 4.00 kg of water to lower FP to oC?.SolutionConc req’d = 5.38 molalWe 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)To get 5.38 mol/kg of particles we need5.38 mol / 2 = mol NaCl / kg2.69 mol NaCl / kg ---> 157 g NaCl / kg(157 g NaCl / kg)•(4.00 kg) = 629 g NaCl
28 Boiling Point Elevation and Freezing Point Depression ∆T = K•m•iA generally useful equationi = van’t Hoff factor = number of particles produced per formula unit.Compound Theoretical Value of iglycol 1NaCl 2CaCl2 3
29 Osmosis Dissolving the shell in vinegar Egg in pure water Egg in corn syrup
30 Osmosis Driving force is entropy 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 Reverse Osmosis Water Desalination Water desalination plant in Tampa
37 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∏ = mmHg • (1 atm / 760 mmHg)= atm(b) Calculate molarity
38 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. molarity from ∏ = iMRTConc = 5.39 x 10-4 mol/L(c) Calc. molar massMolar mass = 35.0 g / 5.39 x 10-4 mol/LMolar mass = 65,100 g/mol