1. Solutions Textbook Chapter 14

2. Definitions n A solution is a homogeneous mixture n A solute is dissolved in a solvent. –solute is the substance being dissolved –solvent is the liquid in which the solute is dissolved –an aqueous solution has water as solvent n A saturated solution is one where the concentration is at a maximum - no more solute is able to dissolve. –A saturated solution represents an equilibrium: the rate of dissolving is equal to the rate of crystallization. The salt continues to dissolve, but crystallizes at the same rate so that there “appears” to be nothing happening.

3. Definitions-continued n An unsaturated solution is one that has not reached the limit of solute that will dissolve n A concentrated solution is one with a relatively large amount of solute dissolved in a solvent n A dilute solution is one with a relatively small amount of solute is dissolved in the solvent

4. Solution and Concentration n 3 Ways of expressing concentration –Molarity(M): moles solute / Liter solution –Mass percent: (mass solute / mass of solution) * 100 –Mole Fraction(  A ) - moles solute / total moles solution

5. Concentration: Molarity Example If 0.435 g of KMnO 4 is dissolved in enough water to give 250. mL of solution, what is the molarity of KMnO 4 ? Now that the number of moles of substance is known, this can be combined with the volume of solution — which must be in liters — to give the molarity. Because 250. mL is equivalent to 0.250 L. As is almost always the case, the first step is to convert the mass of material to moles. 0.435 g KMnO 4 1 mol KMnO 4 = 0.00275 mol KMnO 4 158.0 g KMnO 4 M= mol = 0.00275 mol KMnO 4 = 0.0110 M L 0.250 L solution

6. % Mass Example: Example 1: A solution is prepared by mixing 1.00 g of ethanol, C 2 H 5 OH with 100.0 g of water. Calculate the mass percent of ethanol in this solution.

7. When a solution is diluted, solvent is added to lower its concentration. The amount of solute remains constant before and after the dilution: moles BEFORE = moles AFTER M 1 V 1 = M 2 V 2 Suppose you have 0.500 M sucrose stock solution. How do you prepare 250 mL of 0.348 M sucrose solution ? Concentration 0.500 M Sucrose 250 mL of 0.348 M sucrose Dilution A bottle of 0.500 M standard sucrose stock solution is in the lab. Give precise instructions to your assistant on how to use the stock solution to prepare 250.0 mL of a 0.348 M sucrose solution. A bottle of 0.500 M standard sucrose stock solution is in the lab. Give precise instructions to your assistant on how to use the stock solution to prepare 250.0 mL of a 0.348 M sucrose solution.

8. Example: What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H 2 SO 4 solution? Givens: M 1 = 16 mol/L M 2 = 0.10 mol/L V 1 = ? V 2 = 1.5 L M 1 V 1 = M 2 V 2 V 1 = M 2 V 2 = (0.10 mol/L)(1.5 L) M 1 16 mol/L V 1 = 9.4 x 10 -3 L or 9.4 mL

9. Factors Affecting Solubility 1. Nature of Solute / Solvent 1. Nature of Solute / Solvent. - Like dissolves like (IMF) 2. Temperature - i) Solids/Liquids- Solubility increases with Temperature Increase K.E. increases motion and collision between solute / solvent. ii) gas - Solubility decreases with Temperature Increase K.E. result in gas escaping to atmosphere. 3. Pressure Factor - i) Solids/Liquids - Very little effect Solids and Liquids are already close together, extra pressure will not increase solubility. ii) gas - Solubility increases with Pressure. Increase pressure squeezes gas solute into solvent.

10. Solubilities of Solids vs Temperature Solubilities of several ionic solid as a function of temperature. MOST salts have greater solubility in hot water. A few salts have negative heat of solution, (exothermic process) and they become less soluble with increasing temperature.

11. Temperature & the Solubility of Gases The solubility of gases DECREASES at higher temperatures

13. Colligative Properties Dissolving solute in pure liquid will change all physical properties of liquid, Density, Vapor Pressure, Boiling Point, Freezing Point, Osmotic Pressure Colligative Properties are properties of a liquid that change when a solute is added. number identity The magnitude of the change depends on the number of solute particles in the solution, NOT on the identity of the solute particles.

14. Vapor Pressure Lowering for a Solution n The diagram below shows how a phase diagram is affected by dissolving a solute in a solvent. n The black curve represents the pure liquid and the blue curve represents the solution. n Notice the changes in the freezing & boiling points.

16. Normal Boiling Process Extension of vapor pressure concept: Normal Boiling Point: BP of Substance @ 1atm When solute is added, BP > Normal BP Boiling point is elevated when solute inhibits solvent from escaping. Elevation of B. pt. Express by Boiling point Elevation equation

17. Boiling Point Elevation  T b = (T b -T b °) = i ·m ·k b Where,  T b = BP. Elevation T b = BP of solvent in solution T b ° = BP of pure solvent m = molality, k b = BP Constant Some Boiling Point Elevation and Freezing Point Depression Constants Normal bp (°C) K b Normal fp (°C) K f Solvent pure solvent (°C/m) pure solvent (°C/m) Water 100.00 +0.5121 0.01.86 Benzene80.10 +2.535.50 4.90 Camphor 207 +5.611 179.75 39.7 Chloroform 61.70 +3.63 - 63.5 4.70 (CH 3 Cl) (CH 3 Cl) Some Boiling Point Elevation and Freezing Point Depression Constants Normal bp (°C) K b Normal fp (°C) K f Solvent pure solvent (°C/m) pure solvent (°C/m) Water 100.00 +0.5121 0.01.86 Benzene80.10 +2.535.50 4.90 Camphor 207 +5.611 179.75 39.7 Chloroform 61.70 +3.63 - 63.5 4.70 (CH 3 Cl) (CH 3 Cl)

18. When solution freezes the solid form is almost always pure. Solute particles does not fit into the crystal lattice of the solvent because of the differences in size. The solute essentially remains in solution and blocks other solvent from fitting into the crystal lattice during the freezing process. Freezing Point Depression n Normal Freezing Point: FP of Substance @ 1atm n When solute is added, FP < Normal FP n FP is depressed when solute inhibits solvent from crystallizing.

19. Freezing Point Depression Phase Diagram and the lowering of the freezing point.  T f = i ·m ·k f Where,  T f = FP depression i = van’t Hoff Factor m = molality, k f = FP Constant Generally freezing point depression is used to determine the molar mass of an unknown substance. Derive an equation to find molar mass from the equation above.