1. Physical Properties of Solutions Chapter 13

2. Factors Affecting Solubility Glucose (which has hydrogen bonding) is very soluble in water Cyclohexane (which only has dispersion forces) is not Fig 13.12 Structure and solubility

3. Pressure Effect on Gases in Solution Solubility of liquids and solids does not change appreciably with pressure Solubility of a gas in a liquid is directly proportional to its pressure Fig 13.14 Effect of pressure on gas solubility

4. Henry’s Law S g = kP g where S g ≡ solubility of the gas k ≡ the Henry’s Law constant for that gas in that solvent P g ≡ partial pressure of the gas above the liquid Fig 13.15 Solubility decreases as pressure decreases

5. Temperature Effect on Solids and Liquids Generally, the solubility of solid solutes in liquid solvents increases with increasing temperature Fig 13.17 Solubilities of several ion compounds as a function of temperature

6. The opposite is true of gases: Carbonated soft drinks are more “bubbly” if stored in the refrigerator Warm lakes have less O 2 dissolved in them than cool lakes Temperature Effect on Gases Fig 13.18 Variation of gas solubility with temperature

7. Concentration Units Mass percentage Mole fraction ppm and ppb Molarity Molality Concentration - amount of solute present in a given quantity of solvent or solution:

8. Concentration Units Mass percentage (w/w) % by mass = x 100% mass of solute mass of solute + mass of solvent = x 100% (w/w) mass of solute mass of solution Mole Fraction (X) X A = moles of A sum of moles of all components

9. (a)Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2.50 kg of bleaching solution? Answer: (a) 2.91% (b) 90.5 g of NaOCl Practice Exercise p 543

10. An aqueous solution of hydrochloric acid contains 36% HCl by mass. Calculate the mole fraction of HCl in the solution. Sample Exercise 13.6 Calculation of Mole Fraction Assume we have 100. g of solution:

11. Parts per million (ppm) Parts per billion (ppb) Concentration Units Continued

12. M = moles of solute liters of solution Molarity (M) Molality (m) m = moles of solute mass of solvent (kg)

13. What is the molality of a solution made by dissolving 36.5 g of naphthalene (C 10 H 8 ) in 425 g of toluene (C 7 H 8 )? Answer: 0.670 m Practice Exercise p 544

14. Conversion of Concentration Units If we know the density of the solution, we can calculate the molality from the molarity and vice versa. Fig 13.19 Calculating molality and molarity

15. What is the molality of a 5.86 M ethanol (C 2 H 5 OH) solution whose density is 0.927 g/mL? m =m = moles of solute mass of solvent (kg) M = moles of solute liters of solution Assume 1 L of solution: 5.86 moles ethanol = 270 g ethanol 927 g of solution (1000 mL) (0.927 g/mL) mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = 0.657 kg m =m = moles of solute mass of solvent (kg) = 5.86 moles C 2 H 5 OH 0.657 kg solvent = 8.92 m