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SOLUTIONS CHAPTER 12. INTRODUCTION How do substances dissolve? Why do substances dissolve? What factors affect solubility? How do dissolved substances.

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Presentation on theme: "SOLUTIONS CHAPTER 12. INTRODUCTION How do substances dissolve? Why do substances dissolve? What factors affect solubility? How do dissolved substances."— Presentation transcript:


2 INTRODUCTION How do substances dissolve? Why do substances dissolve? What factors affect solubility? How do dissolved substances affect properties of the solution? (Colligative Properties)

3 I. Solution Terminology 1. Solution: homogeneous mixture of two or more substances Solvent: substance present in greatest amount substance doing the dissolving Solute: substance present in lesser amount substance which is dissolved


5 2. Solubility: amount of solute that will dissolve in a given volume of solvent at equilibrium. Soluble: a substance which dissolves in a solvent. Insoluble: a substance which does not dissolve in a solvent Miscible: when two liquids dissolve in each other in any proportion Immiscible: liquids will not dissolve in each other

6 3. Dissolution: the process of dissolving solute molecules in a solvent Solvation: the process of solvent molecules surrounding the solute molecules Hydration: solvation in water. Water molecules surround the solvent molecules. 4. Hydrophilic: “water loving”, polar part of a molecule attracted to water. 5. Hydrophobic: “water fearing”, nonpolar part of a molecule not attracted to water.

7 6. Saturated Solution: when the concentration of the solute equals its solubility. A saturated solution contains as much of a solute as the solubility allows. “A Dynamic Equilibrium” See next Figure 7. Unsaturated Solution: when concentration of the solute is less than its solubility. 8. Supersaturated Solution: when concentration of solute exceeds its solubility. Temporary state, unstable state

8 Formation of a Saturated Solution Solid begins to dissolve. As solid dissolves, some dissolved solute begins to crystallize. Eventually, the rates of dissolving and of crystallization are equal; no more solute appears to dissolve. Longer standing does not change the amount of dissolved solute.

9 II. Explanation of Solubility A. General Rule of Solubility “Likes Dissolve Likes” 1. For something to dissolve, the solute and solvent must have similar types of intermolecular forces. Polar or ionic compounds tend to be soluble in polar solvents. Nonpolar compounds tend to be soluble in nonpolar solvents.

10 B. Energies Affecting Solubility 1. Attractive forces between solute molecules must be broken. (energy requiring or unfavorable). 2. Attractive forces between solvent molecules must be broken. (energy requiring or unfavorable). 3. Attractive forces between solute and solvent molecules must be formed. (energy producing or favorable). Must be equal to or greater than unfavorable steps (above) for solvation to occur.


12 4. Entropy (degree of energy randomness or disorder) generally increases in solvation process (favorable). Especially important for gases.

13 C. Examples of Dissolving Liquids in Liquids 1. Water and Ethyl Alcohol are “Misible” Explain Diagram of interactions Energetics of Process and ΔH solution See following diagram


15 2. Hydrocarbons are “Miscible” Oil and Gasoline Benzene and Octane 3. Water and Hydrocarbons are “Immiscible” Water and Benzene Water and Gasoline

16 D. Examples of Dissolving Solids in Liquids 1. Ionic Compounds in Water In order to become solvated, the ionic interactions between the ions in the crystal structure must be broken. (unfavorable). The energy which holds the crystal together is known as the lattice energy. How is this energy overcome? 1. Water molecules surround ions and “hydrate” them. (favorable) 2. Usually an increase in entropy. (favorable)

17 Diagram of NaCl Dissolving in Water See next Diagram See Energy Diagram Why are some ionic compounds( like AgCl) not soluble in water?




21 2. Sugar (Glucose) Soluble in Water (Glucose, C 6 H 12 O 6, is a polar covalent compound).

22 E. Factors Affecting Solubility 1. Temperature Effect of temp. on solubility can best be explained using Le Chatelier’s Principle. When any change in concentration, temperature, pressure, or volume is imposed on a system at equilibrium, the system responds by attaning a new equilibrium condition that minimizes the impact of the imposed change. Equilibrium will shift to the left or right!!!

23 a. Gas Dissolving in Liquid: Exothermic Gas + solvent  saturated solution ΔH = - _ What is effect on solubility of increasing temp? b. Solids Dissolving in Liquid: May be Exothermic or Endothermic? KOH(s) → K + (aq) + OH - (aq) ΔH soln = -57.6 kJ/mol H 2 O NH 4 NO 3 (s) → NH 4 + (aq) + NO 3 - (aq) ΔH soln = +25.7 kJ/mol H 2 O

24 F. Pressure Effects on Dissolving Gases ( Remember to use “Likes Dissolve Likes” to predict what gases will dissolve in a liquid). Solubility of a gas in a liquid increases as the pressure of the gas increases. Explain why. See Figure


26 Quantitative Relationship For Pressure and Solubility of a Gas Henry’s Law S g = k H P g S = solubility of gas P g = partial pressure of gas k H = constant (different for each gas) (see Table 12.4 for values)

27 Problem Solving: Nitrogen comprises 78% of the atmosphere and has a “k H ” value of 8.42 x 10 -7 M / mm Hg in water. a. How many grams of N 2 will be present in 1.00 L of water that is in equilibrium with air at atmospheric pressure, 760mm Hg? b. How many grams of N 2 would be present if the atmospheric pressure is increased to 10 atmospheres?

28 III. Concentration Units Other concentration terms useful for describing properties of dilute solutions. A. Volume Percent = B. Mass Percent =

29 Example Problem: What is the mass percent of methanol in a solution made up by adding 27.5 mL of methanol to 500 g of water? (Density of methanol = 0.791 g / mL)

30 C. Mole Fraction = A solution is prepared by adding 0.76 g NaCl, 0.21 g KOH, and 54 g of water. a. What is the mass percent of NaCl? b. What is the mole fraction of NaCl? c. What is the mole fraction of water? d. What is the mole percent of water?

31 D. Parts Per Million =ppm 1 g solute / 1,000,000 g solution 1 mg solute / 1000 g solution Parts per Billion = ppb 1 g solute / 1,000,000,000 g solution 1 mg solute / 1,000,000 g solution The mass percent of NaCl = 0.043% in an aqueous solution. What is the ppm NaCl in the solution?

32 E. Molality Recall Molarity = M = mol solute / L solution Volume is affected by temperature changes, and therefore another concentration term, molality, will be used in problem solving with temperature changes. Molality = m =

33 Example Problem: What is the molality of a solution prepared by dissolving 7.53 g of methanol (CH 3 OH, 32.0 g/mol) in 200.0 g of water? If the density of the solution is 0.987 g/mL, what is the molarity?

34 IV. Colligative Properties A colligative property of a solution depends only on the concentration of the solute particles (molecules or ions), but not on the identity of the solute. * All colligative properties are entropy driven!! Explain (A 2.0 molal NaCl solution has the same colligative properties as a 4.0 molal glucose nonelectrolyte) solution, or 1.0 molal Na 3 PO 4 solution).

35 # Moles Particles # Moles Particles Molality Per Formula Unit Per Kg Solvent 2m NaCl 2 4 4m Glucose 1 4 1m Na 3 PO 4 4 4

36 Examples of Colligative Properties 1.Vapor Pressure Lowering 2.Boiling Point Elevation 3.Melting Point Depression (Lowering) 4.Osmotic Pressure (Osmosis)

37 A.Vapor Pressure Lowering 1. The vapor pressure of any pure solvent will be lowered by the addition of a nonvolatile solute to the solvent. 2. Explanation a. Pure solvent  vapor has greater increase in entropy than solution  vapor. Solvent in solution is in a more disordered state and thus less increase in entropy when going to gas. *** b. Consider ability of solvent molecules in a solution to escape to gas.


39 3. Raoult’s Law Simple relationship between the vapor pressure of one component (solvent in this case) and its mole fraction in solution. P solution = X solvent P 0 solvent P solution = vapor pressure of solvent gas over solution X solvent = mole fraction of solvent in the solution P 0 solvent = vapor pressure of pure solvent

40 Example Vapor Pressure Problem Calculate the vapor pressure of water at 90 o C in a solution prepared by dissolving 5.00 g glucose (C 6 H 12 O 6 ) in 100 g water. The vapor pressure of water at 90 0 C is 525.8 mm Hg. (Glucose is a nonelectrolyte, nonvolatile covalent compound.

41 B. Boiling Point Elevation 1. Why does addition of a nonvolatile solute to a solvent elevate the boiling point? 2. Equation: ΔT b = K b mi ΔT b = amount the boiling point is raised (elevated) K b = constant dependent on solvent only m = molality of solute (mol solute / kg solvent) i = van’t Hoff factor, (# particles per formula unit of solute)

42 Formula Unit of Compound i CH 3 OH1 NaCl2 Na 2 SO 4 3 K 3 PO 4 4 What is the boiling point for the following aqueous solutions? The K b for water is 0.52 o C kg mol -1. a. Pure water b. 1 molal CH 3 OH solution c. 1 molal NaCl solution

43 3. Molar Mass From Boiling Point Elevation A solution is prepared by dissolving 11.0 g of a nonvolatile organic solute in 100 g of chloroform solvent. The boiling point of the solution was 84.30 o C. Given: B.P. of chloroform = 61.20 o C. K b for chloroform = 3.63 o C / m What is the molar mass of the unknown solute? (Det. ΔT, m, #mol solute, molar mass) 4. Significant of BP Elevation

44 C. Freezing Point Depression 1. Explanation For Lower Freezing Point For pure solvent, the speed of the particles decreases until they can occupy space in the growing crystal clusters. The presence of solute molecules gets in the way of the solvent molecules joining the growing crystal clusters. 2. Equation: ΔT f = K f mi

45 3. Problems a. What is the freezing point of a 0.0222 m CaCl 2 aqueous solution? K f for water = 1.86 o C/m. b. A solution made by dissolving 3.46 g of a nonelectrolyte solute in 85.0 g benzene froze at 4.13 o C. F.P. of benzene = 5.45 o C. K f for benzene = 5.07 O C/m. 1) Calculate the molality of the solution. 2) Calculate the molar mass of the solute. 4. Significance

46 D. Osmotic Pressure 1. Definitions Osmosis is defined as the passage of solvent molecules through a semipermeable membrane from a solution of lower solute concentration to a solution of higher solute concentration. Semipermeable membranes allow only certain small molecules to pass through. Osmotic pressure is the pressure that must be applied to the solution to stop osmosis. Fig. 12.16 or the following:


48 3. Equation: Π = M RTi Π = osmotic pressure M = concentration of solution in molarity (M) R = gas constant (0.0821 L atm / K mol T = temperature in Kelvin scale i = van’t Hoff factor ** Note similarity to Ideal Gas Law

49 4. Problems a. Calculate the osmotic pressure for a 3.75 x 10 -4 M aqueous solution of the polymer dextran at 25 o C. Dextran is a nonelectrolyte. b. Calculate the molar mass of dextran if a solution of 5.0 g of dextran in 1.0 L of water gives an osmotic pressure of 2.9 x 10 -3 atm. at 25 o C. (Det. M, #mol solute, molar mass) 5. Significance / Practical Applications

50 Ordinarily a patient must be given intravenous fluids that are isotonic—have the same osmotic pressure as blood. Practical Applications of Osmosis External solution is hypertonic; produces osmotic pressure > π int. Net flow of water out of the cell. External solution is hypotonic; produces osmotic pressure < π int. Net flow of water into the cell. Red blood cell in isotonic solution remains the same size.

51 Reverse osmosis (RO): reversing the normal net flow of solvent molecules through a semipermeable membrane. Pressure that exceeds the osmotic pressure is applied to the solution. RO is used for water purification. Pressure greater than π is applied here … … water flows from the more concentrated solution, through the membrane. Practical Applications of Osmosis (cont’d)

52 V. Additional Reading Responsible for Terms and Significance 1. Reverse osmosis 2.Isotonic Hypotonic Hypertonic 3.Colloids and Tyndall Effect (These are found in Sections 12.7 and 12.8)

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