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13 Properties of Solutions Solutions to Exercises 13.89 Proteins form hydrophilic colloids because they carry charges on their surface (Figure 13.30). When electrolytes are added to a suspension of proteins, the dissolved ions form ion pairs with the protein surface charges, effectively neutralizing them. The protein's capacity for ion-dipole interactions with water is diminished and the colloid separates into a protein layer and a water layer. 13.90 (a) The nonpolar hydrophobic tails of soap particles (the hydrocarbon chain of stearate ions) establish attractive intermolecular dispersion forces with the nonpolar oil molecules, while the charged hydrophilic head of the soap particles interacts with to keep the oil molecules suspended. (This is the mechanism by which laundry detergents remove greasy dirt from clothes.) (b) Electrolytes from the acid neutralize surface charges of the suspended particles in milk, causing the colloid to coagulate. Additional Exercises 13.91 The outer periphery of the BHT molecule is mostly hydrocarbon-like groups, such as -CH₃. The one -OH group is rather buried inside, and probably does little to enhance solubility in water. Thus, BHT is more likely to be soluble in the nonpolar hydrocarbon hexane, than in polar water. 13.92 In this equilibrium system, molecules move from the surface of the solid into solution, while molecules in solution are deposited on the surface of the solid. As molecules leave the surface of the small particles of powder, the reverse process preferentially deposits other molecules on the surface of a single crystal. Eventually, all molecules that were present in the 50 g of powder are deposited on the surface of a 50 g crystal; this can only happen if the dissolution and deposition processes are ongoing. 13.93 Assume that the density of the solution is 1.00 g/mL. (a) 4 ppm = 4 kg mg O₂ = 4 soln 1 mol O₂ = 1.25 = 10⁻⁴ M (b) = = 1.25 L L-atm 10⁻³ mol = = atm 13.94 (a) (b) = = 10⁻³ mol 1.12 atm = 8.1 10⁻⁷ M atm 13.95 0.10% by mass means 0.10 g glucose/100 blood. (a) ppm glucose = g solution glucose X = 100 glucose blood = 1000 ppm glucose g g (b) m = mol glucose/kg solvent. Assume that the mixture of nonglucose components is the 'solvent'. 386