1. Solutions Chapter 16 Copyright © The McGraw-Hill Companies, Inc.

2. 16.1 A solution is a homogenous mixture of 2 or more substances The solute what is being dissolved… usually the one with the smaller amount(s) The solvent is the substance present in the larger amount, doing the dissolving

3. A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature. A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature. Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate. 16.1

4. Three types of interactions in the solution process: solvent-solvent interaction solute-solute interaction solvent-solute interaction  H soln =  H 1 +  H 2 +  H 3

5. solid solubility usually increases with temperature, however, gas solubility generally decreases… 16.1 Temperature and Solubility

6. “like dissolves like” Two substances with similar intermolecular forces are likely to be soluble in each other. non-polar molecules are soluble in non-polar solvents CCl 4 in C 6 H 6 polar molecules are soluble in polar solvents C 2 H 5 OH in H 2 O ionic compounds are more soluble in polar solvents NaCl in H 2 O or NH 3 (l) 16.1

7. Which of the following in each pair is likely to be the more soluble in hexane, C 6 H 14 ? (A)cyclohexane ( C 6 H 12 ) or glucose ( C 6 H 12 O 6 ) (B)propionic acid ( CH 3 CH 2 COOH ) or sodium propionate ( CH 3 CH 2 COONa ) (C)Hydrochloric acid ( HCl ) or ethyl chloride ( CH 3 CH 2 Cl )

8. The Cleansing Action of Soap

9. A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance. Colloid versus solution collodial particles are much larger than solute molecules collodial suspension is not as homogeneous as a solution 12.8

10. Pressure and Solubility of Gases 12.5 The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law). S 1 = S 2 P 1 P 2 S 1 is the solubility of a gas at one pressure (P 1 ) low P low c high P high c S 2 is the solubility of a gas at one pressure (P 2 )

11. Concentration Units The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Percent by Mass (or volume) % by mass = x 100% mass of solute mass of solute + mass of solvent = x 100% mass of solute mass of solution 16.2 Mole Fraction (X) X A = moles of A sum of moles of all components

12. % by mass What is the mass percentage of Iodine, I 2, in a solution containing 0.0650 mol I 2 in 120.0g of CCl 4 ?

13. Percent by volume What is the percent by volume of vinegar if 25mL of pure vinegar (acetic acid) are diluted by 150mL of water?

14. Mole fraction What is the mole fraction of NaOH in water if 15.00g NaOH are dissolved in 1000.0g of water?

15. Concentration Units Continued M = moles of solute liters of solution Molarity (M) *note the capital letter Molality (m) *note the lower case m = moles of solute mass of solvent (kg) 16.2

16. Molarity What is the molarity of a solution formed by dissolving 1.80 mol of KCl in 1.600 L of water?

17. To dilute concentrations already in molarity, use this equation: M 1 V 1 = M 2 V 2 Ex: How can I make 500.0 mL of a 1.0 M HCl solution using 12 M stock solution? M 1 = 12 M V 1 = ? M 2 = 1.0 M V 2 = 500.0 mL V 1 = M 2 V 2 /M 1 = (1.0M)(500.0mL)/(12M) V 1 = 42mL * I will need 42 mL stock and 458 mL water (500mL- 42mL)

18. 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 x 0.927 g/mL) mass of solvent = mass of solution – mass of solute = 927 g sol’n – 270 g eth = 657 g solvent = 0.657 kg solvent m =m = moles of solute mass of solvent (kg) = 5.86 moles C 2 H 5 OH 0.657 kg solvent = 8.92 m 16.2

19. Boiling-Point Elevation  T b = T b – T b 0 T b > T b 0  T b > 0 T b is the boiling point of the pure solvent 0 T b is the boiling point of the solution  T b = K b m m is the molality of the solution K b is the molal boiling-point elevation constant ( 0 C/m) for a given solvent 12.6

20. Freezing-Point Depression  T f = T f – T f 0 T f > T f 0  T f > 0 T f is the freezing point of the pure solvent 0 T f is the freezing point of the solution  T f = K f m m is the molality of the solution K f is the molal freezing-point depression constant ( 0 C/m) for a given solvent

22. What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g.  T f = K f m m =m = moles of solute mass of solvent (kg) = 2.41 m = 3.202 kg solvent 478 g x 1 mol 62.01 g K f water = 1.86 0 C/m  T f = K f m = 1.86 0 C/m x 2.41 m = 4.48 0 C  T f = T f – T f 0 T f = T f –  T f 0 = 0.00 0 C – 4.48 0 C = -4.48 0 C