Chapter 6: Solutions, Acids, And Bases Goss

Colligative Properties of SolutionsColligative properties = physical properties of solutions that depend on the # of particles dissolved, not the kind of particle.Colligative PropertiesLowering vapor pressureRaising boiling pointLowering freezing pointGenerating an osmotic pressureColligative PropertiesLowering vapor pressureRaising boiling pointLowering freezing pointGenerating an osmotic pressureBoiling Point Elevationa solution that contains a nonvolatile solute has a higher boiling pt than the pure solvent; the boiling pt elevation is proportional to the # of moles of solute dissolved in a given mass of solvent.

Boiling Point ElevationTb = kbmwhere:Tb = elevation of boiling ptm = molality of solutekb = the molal boiling pt elevation constant

kb values are constants; see table 15-4, p. 472 (honors text)

kb for water = 0.52 C/m Ex: What is the normal boiling pt of a 2.50 m glucose, C6H12O6, solution? normal implies 1 atm of pressureTb = kbmTb = (0.52 C/m)(2.50 m)Tb = 1.3 C

Tb = 100.0 C + 1.3 C = 101.3 C

Ex: How many grams of glucose, C6H12O6, would need to be dissolved in 535.5 g of water to produce a solution that boils at 101.5C?Tb = kbm1.5 C= (0.52 C/m)(m)m = 2.885

Freezing/Melting Point DepressionThe freezing point of a solution is always lower than that of the pure solvent.

Freezing/Melting Point DepressionTf = kfmwhere:Tf = lowering of freezing pointm = molality of solutekf = the freezing pt depression constant

kf for water = 1.86 C/m kf values are constants; see table 15-5, p. 474 (honors text)

Ex: Calculate the freezing pt of a 2.50 m glucose solution.Tf = kfmTf = (1.86 C/m)(2.50 m)Tf = 4.65 C

Tf = 0.00C - 4.65 C = -4.65C

Ex: When 15.0 g of ethyl alcohol, C2H5OH, is dissolved in 750 grams of formic acid, the freezing pt of the solution is 7.20C. The freezing pt of pure formic acid is 8.40C. Determine Kf for formic acid.Tf = kfm1.20 C= (kf)( 0.4348 m)kf = 2.8 C/m

Ex: An antifreeze solution is prepared containing 50.0 cm3 of ethylene glycol, C2H6O2, (d = 1.12 g/cm3), in 50.0 g water. Calculate the freezing point of this 50-50 mixture. Would this antifreeze protect a car in Chicago on a day when the temperature gets as low as 10 F? (-10 F = -23.3 C)Tf = kfmTf = (1.86C/m)(18.06 m)Tf = 33.6 CTf = 0 C 33.6 C = -33.6 C

YES!

Electrolytes and Colligative Properties Colligative properties depend on the # of particles present in solution. Because ionic solutes dissociate into ions, they have a greater effect on freezing pt and boiling pt than molecular solids of the same molal conc.

Electrolytes and Colligative PropertiesFor example, the freezing pt of water is lowered by 1.86C with the addition of any molecular solute at a concentration of 1 m.Such as C6H12O6, or any other covalent compound

However, a 1 m NaCl solution contains 2 molal conc. of IONS. Thus, the freezing pt depression for NaCl is 3.72Cdouble that of a molecular solute.NaCl Na+ + Cl- (2 particles)

Electrolytes - Boiling Point Elevation and Freezing Point DepressionThe relationships are given by the following equations:

Tf = kf mn or Tb = kbmn

Tf/b = f.p. depression/elevation of b.p.m = molality of solutekf/b = b.p. elevation/f.p depression constantn = # particles formed from the dissociation of each formula unit of the soluteEx: What is the freezing pt of:a) a 1.15 m sodium chloride solution?NaCl Na+ + Cl-n=2

Tf = kfmnTf = (1.86 C/m)(1.15 m)(2)Tf = 4.28 C

Tf = 0.00C - 4.28 C = -4.28C

Ex: What is the freezing pt of:b) a 1.15 m calcium chloride solution?CaCl2 Ca2+ + 2Cl-n=3

Tf = kfmnTf = (1.86 C/m)(1.15 m)(3)Tf = 6.42 C

Tf = 0.00C 6.42 C = -6.42C

Ex: What is the freezing pt of:c) a 1.15 m calcium phosphate solution?Ca3(PO4)2 3Ca2+ + 2PO43-n=5

Tf = kfmnTf = (1.86 C/m)(1.15 m)(5)Tf = 10.7 C

Tf = 0.0C 10.7 C = -10.7C

Determining Molecular Weights by Freezing Point Depression

Tf = 0.56CTf = kfm0.56C = (5.12C/m)(m)m = 0.1094

Ex: A 1.20 g sample of an unknown molecular compound is dissolved in 50.0 g of benzene. The solution freezes at 4.92C. Determine the molecular weight of the compound. The freezing pt of pure benzene is 5.48C and the Kf for benzene is 5.12C/m. Ex: A 37.0 g sample of a new covalent compound was dissolved in 200.0 g of water. The resulting solution froze at 5.58C. What is the molecular weight of the compound? Tf = 5.58CTf = kfm5.58C = (1.86C/m)(m)m = 3.00 m