Part One: Introduction. Solute = is a substance being dissolved, and usually present in small amounts.

Transcription

1 CHAPTER 12: SOLUTIONS Part One: Introduction A. Terminology. 1. Solution - 2. Consists of a solvent and one or more solutes. 3. Solute = is a substance being dissolved, and usually present in small amounts. 4. Solvent = Solvent is usually the most abundant species in the mixture. 5. Solubility = the amount of a given solute that dissolves in a given quantity of solvent. NaCl in water, e.g., solubility of NaCl in water = 3.6 g/ml. 6. Two fluids that mix in all proportions to form a solution are said to be (alcohol and water). Oil and water would be said to be immiscible.. Figure 12.1 B. Importance. 1. Extremely important in all life 2. A solid, liquid, or gas can act as either solvent or solute. Usually the solvent is a liquid. 3. Sea water is an aqueous solution of many salts & some gases. 4. Tap water is an aqueous with trace amounts of various ions, molecular compounds, and gases. (Distilled water is pure). Chapter 12 Page 1

2 5. Air is a solution of gases. 6. Solids solutions are called. One example is steel, which is mostly Fe with small amounts of additives like carbon, chromium, titanium, etc. Another example is dental fillings are solid solutions called by the special name amalgams =. A. Solubility and Saturated Solutions Part Two: The Solution Process 1. Solution formation is a process of. Example, dissolving table salt in water: NaCl(s) Na + (aq) + Cl - (aq) Figure 12.2 Chapter 12 Page 2

3 Ions are continually leaving the solid crystal into the solution and returning to the crystal, until eventually: a) all the solid is dissolved, OR b) the solution reaches its saturation point, at which some solid remains. Which case pertains depends on the of NaCl in water. Figure 12.3 B. Factors that influence solubility. 1. General rule = e.g., oil dissolves in gasoline because both are hydrocarbons, which are non-polar molecules water, however, is a polar molecule, such that oil does not dissolve in it, but alcohols do (which are also polar). Therefore, intermolecular forces are involved. 2. There is some natural tendency of all substances to mix, based on tendency toward a state of increased disorder (ENTROPY CHANGE) 3. This is balanced by the tendency of a system to have the lowest energy possible (ENERGY CHANGE). In other words, is heat absorbed or released in forming a solution? Release of energy (exothermic) favors solution formation. Chapter 12 Page 3

4 The Entropy effect and the Energy effect combine to determine the solubility. A process is favored by: a. b. 4. Energy change is called the heat of solution, ΔH solution. Depends on how strongly solute and solvent particles interact. 5. Relative strengths of these interactions determine the extent of solubility (refer to figure below) a. Strong solvent-solute attractions favor solubility. b. Weak solvent-solvent attractions favor solubility. c. Weak solute-solute attractions favor solubility. 6. Solution formation is always accompanied by an increase in the disorder of both solvent and solute, always favorable to solubility. C. Dissolving Solids in Liquids, viewed as a stepwise process (Section 12.2) Chapter 12 Page 4

5 1. Take an ionic solid. Must overcome crystal lattice energy (solute-solute interactions): 2. Must overcome solvent-solvent interactions. If solvent is water, must break Hydrogen bonds: 3. Must solvate the solute particles: Figure 12.8 Chapter 12 Page 5

6 4. Combined Steps 2 and 3 is called solvation, or in case of water, hydration. 5. Overall energy involved: ΔH solution = Figure Energy is not the only factor. Entropy increase upon solution is a favorable factor which can sometimes overcome unfavorable energy factor if latter is not too large. 7. Example: NaCl(s) + H 2 O(aq) Na + (aq) + Cl - (aq) ΔH solution is (+), endothermic (unfavorable). Still occurs due to slightly larger favorable entropy consideration. 8. Ionic compounds which have very large crystal lattice energies will be insoluble in H 2 O because solvation and entropy can t overcome energy holding lattice together. 9. Some ionic solids dissolve with release of heat (exothermic) - both energy and entropy favor solution process. CaCl 2, Na 2 SO 4,... Chapter 12 Page 6

7 10. Nonpolar solids (e.g., paraffin) do not dissolve in H 2 O because interactions with water (hydration) is very weak compared to water-water interactions. Weak interaction Strong Hydrogen Weak interaction btwn. in nonpolar solid bonding solvent- solvent & nonpolar (should be easy to solvent interactions. solute. Nothing gained break up). (difficult to break up) here except entropy of HOWEVER... mixing. D. Dissolving Liquids in Liquids Molecular solutions (Miscibility). (Section 12.2) 1. Same factors involved. Here, though, a solid does not need to be broken up, so no lattice energy to overcome. 2. Case 1: Polar or Hydrogen-bonding solute with polar or Hydrogen bonding solvent. Methanol dissolves in water. H-bonding. Moderately H-bonding. Moderately H-bonding. Nothing strong solute-solute strong solvent-solvent lost, entropy of interactions. interactions. mixing favors. 3. Case 2: Nonpolar solute with polar or Hydrogen-bonding solvent. CCl 4 does not dissolve in H 2 O. Weak solute-solute H-bonding. Moderately Weak solute-solvent interactions. (only strong solvent-solvent. interactions. (loss of London forces) solvent-solvent int.) (hydrophobic effect) Chapter 12 Page 7

8 4. Case 3: Nonpolar solute and nonpolar solvent. CCl 4 does dissolve in benzene. Weak solute-solute Weak solvent-solvent Weak solute-solvent interactions. (only interactions. (only (London). Entropy London forces) London forces) of mixing favors this. 5. Generalization - E. Gases in Liquids. (Section 14.4) 1. Same factors as before. 2. Polar gases dissolve in polar liquids, nonpolar gases dissolve in nonpolar liquids. 3. CO 2 and O 2 are nonpolar, so only slightly dissolve in H 2 O. 4. CO 2 more so because of ionization: 5. HCl and NH 3 are polar covalent gases, dissolve strongly in H 2 O. Part Three: Effects of Environmental Variables on Solubility (Section 12.3) A. Temperature Effect. (Section 12.3) 1. Explained by LeChatelier s Principle = 2. When heat of solution is exothermic: Chapter 12 Page 8

9 3. When heat of solution is endothermic: 4. NaCl solubility as T ; Na 2 SO 4 solubility as T. Figure O 2 dissolves in H 2 O exothermically: T O 2 solubility. Basis of thermal pollution. B. Pressure Effect. 1. Henry s Law = 2. Example: Carbonated beverage stored with pressurized CO Henry s Law: Chapter 12 Page 9

10 Part Four: Expressions of Concentration (Section 12.4) A. Molarity. 1. Molarity = B. Mass Percentage of solute. 1. Mass % solute = 2. Example problem: A certain solution of an alcohol and water is 2.0 M alcohol and has a density of 0.95 g/ml. The alcohol has a molar mass of 126 g/mol. What is the mass percent alcohol in this solution? C. Molality. 1. m = 2. Example: What is the molality of a solution in which 1.0 gram glucose (C 6 H 12 O 6 ) has been added to 100 ml of H 2 O? 3. Note that: Why? Because under these conditions: D. Mole Fraction. 1. X A = Chapter 12 Page 10

11 2. Unitless. 3. Example: What is X glucose and X water in previous problem? A. Definition. Part Five: Colligative Properties of Solution (Sections 12.5 through 12.8) 1. Colligative property = 2. Examples: a. b. c. d. 3. Depends on: B. Vapor Pressure Lowering - Raoult s Law. (Section 12.5) 1. A solution containing a nonvolatile solute has a vapor pressure than the pure solvent. ( nonvolatile solute means solute has no vapor pressure of its own) 2. Physical Picture: 3. V.P. lowering depends on fraction of surface blocked by solute particles, and thus, mole fraction of solute and solvent, hence: Chapter 12 Page 11

12 4. Raoult s Law: 5. Also could write: 6. Solutions obeying this perfectly are called ideal solutions. 7. Slight deviations occur when: -- attractions differ from -- attractions and -- attractions. C. Boiling Point Elevation. (Section 12.6) 1. Closely related to vapor pressure lowering, because boiling occurs when Each solvent has its own characteristic K b, independent of what the solute is. See Table K b = C/molal for H 2 O 5. Problem: What is the normal boiling point of a glucose solution in which 270 g glucose are added to 1.0 L of H 2 O? D. Freezing Point Depression. (Section 12.6) Chapter 12 Page 12

13 1. 2. K f = 1.86 C/molal for H 2 O. 3. What is freezing point of the preceding glucose solution? 4. This is how antifreeze works, as well as salt on icy roads. E. Determination of Molar Mass by Colligative Properties. (Section 12.6) 1. Add known mass of solute to known mass of solvent. Measure f.p. lowering ΔT f. Calculate Molar Mass (M). 2. Problem: When g of sulfur is finely ground and melted with 4.38 g of camphor, the freezing point of the camphor is lowered by 5.47 C. What is the molecular weight of sulfur? What is its molecular formula? Given K f = 40 C/m Chapter 12 Page 13

14 F. Colligative Properties of Ionic Solutions. (Section 12.8) 1. When solute is an electrolyte (breaks into ions in solution), colligative properties are because of more particles present. Collig. Prop. depend on total concentration of ions produced. 2. Deviations from ideal behavior can be large since ions interact with each other in solution and thus are not independent. e.g. You don t get quite a doubling of b.p. elevation when you dissolve 1 mole of NaCl vs. 1 mole of glucose because the Na + and Cl - ions interact. 3. Effect of ionization is expressed by the van t Hoff factor i: i = ΔT actual ΔT non elec i = 2 if NaCl had no ion-ion interactions. i = 1.83 in 1.0 m NaCl ΔT b = i K b m modified F.P. equation for ionic solutes G. Osmotic pressure. (Section 12.7) 1. Osmosis = 2. Semipermeable membrane = allows solvent to flow but not solutes. (e.g., cell membranes are semipermeable) 3. osmotic pressure π Chapter 12 Page 14

15 4. Working equation: Part Six: Colloids (Sections 12.9) A. Definition. 1. Remember our classification of mixtures as: a. homogeneous b. heterogeneous 2. Colloids are somewhere between these two extremes. 3. Collodial dispersion = dispersed particles are small enough to remain suspended in the solvent, so no settling takes place, but yet are large enough to make the sample cloudy. 4. Cloudiness is due to = scattering of light by suspended particles when they become large enough that they are comparable to the wavelength of the light. B. Types of colloids - study Table Soap molecules solubilize oils in water by forming Colloids with the oil. oil-soluble end (Hydrophobic) water-soluble end (Hydrophilic) Chapter 12 Page 15

16 Figure a micelle particle Chapter 12 Page 16

17 Figure how soap cleans fabrics Figure cell membranes Chapter 12 Page 17

18 NOTES: Chapter 12 Page 18