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The Solution Process

The Solution Process. In order for a solute to dissolve in a solvent, there has to be an attraction between the two that is comparable in magnitude to the strengths of the intermolecular forces already existing in the solute and in the solvent. The Solution Process. Factors Affecting Solubility.

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The Solution Process

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  1. The Solution Process In order for a solute to dissolve in a solvent, there has to be an attraction between the two that is comparable in magnitude to the strengths of the intermolecular forces already existing in the solute and in the solvent.
  2. The Solution Process
  3. Factors Affecting Solubility The extent to which one substance dissolves in another depends on: the nature of the solute the nature of the solvent the temperature of the solution the partial pressure of any gases in solution
  4. Solute-Solvent Interactions In general, when other factors are comparable, the stronger the attractions between solute and solvent molecules, the greater the solubility.
  5. Solute-Solvent Interactions Polar liquids tend to dissolve readily in polar solvents (due to dipole-dipole interactions between polar molecules). Acetone and water are both polar and are miscible, i.e., they mix in all proportions. Nonpolar liquids tend to be insoluble in polar liquids. Gasoline and water are immiscible.
  6. Solute-Solvent Interactions Nonpolar liquids tend to dissolve readily in nonpolar solvents (due to dispersion forces between the molecules). Gasoline and hexane are both nonpolar and are miscible. The overall rule is “like dissolves like.”
  7. Solute-Solvent Interactions When solute and solvent contain both polar and nonpolar components, you can estimate the solubility by looking at how much of each molecule is polar and how much is nonpolar. As the length of the carbon chain in an alcohol (R-OH) increases, its solubility in water decreases and its solubility in hexane (C6H14) increases.
  8. Factors Affecting Solubility The extent to which one substance dissolves in another depends on: the nature of the solute the nature of the solvent the temperature of the solution the partial pressure of any gases in solution
  9. Temperature and Solubility The solubility of most solids in water increases as the temperature of the solution increases.
  10. Temperature and Solubility The solubility of all gases in water decreases with increasing temperature.
  11. Factors Affecting Solubility The extent to which one substance dissolves in another depends on: the nature the solute the nature of the solvent the temperature of the solution the partial pressure of any gases in solution
  12. Pressure and Solubility: Henry’s Law The solubility of a gas in any solvent increases as the pressure over the solvent increases. Henry’s Law: Sg = kPg where Sg is the solubility of the gas in the solvent k is the Henry’s Law constant. It depends on the identity of the gas, the identity of the solvent, and the temperature. Pg is the partial pressure of the gas over the solution.
  13. Pressure and Solubility: Henry’s Law The solubility of a gas in any solvent increases as the pressure over the solvent increases. Henry’s Law: Sg = kPg
  14. Pressure and Solubility: Henry’s Law Henry’s Law: Sg = kPg Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over the liquid at 25°C. The Henry’s Law constant for CO2 and water at 25°C is 0.031 mol/L-atm. Sg = 0.031 mol/L-atm (4.0 atm) =0.12 mol/L
  15. Solutions Unsaturated - The solvent is capable of dissolving more solute. Dilute, Concentrated – qualitative descriptions of unsaturated solutions. Saturated – The solvent has so much solute in it that it is not capable of dissolving any more. The concentration of a saturated solution has a definite value, but that value changes with temperature. Supersaturated - A metastable solution that, through careful cooling of a saturated solution, holds MORE solute than a saturated solution. Introduction of a crystal of the solute or any mechanical disturbance of the solution will result in spontaneous crystallization of the excess solute.
  16. Saturated vs. Unsaturated Solubility = concentration of saturated solution Line on chart represents limit of solubility for that compound. 1. At 50°C, 40 g of KCl are added to 100g of water. Does all of the KCl dissolve? Is solution saturated? What is the concentration of the solution in g salt / 100 g of water? 2. At 50°C, 50 g of KCl are added to 100g of water. Does all of the KCl dissolve? Is solution saturated? What is the concentration of the solution in g salt / 100 g of water?
  17. Units of Concentration Concentration Symbol Definition unit mass percent % mass of solute A x 100 mass of solution parts per million ppmmass of solute A x 106 mass of solution parts per billion ppbmass of solute A x 109 mass of solution ppm and ppb are handy for expressing concentrations of very dilute solutions.
  18. Units of Concentration Concentration Symbol Definition unit mole fraction XAmol of A total mol Molarity* M mol solute L of solution Molality mmol solute kg of solvent *Varies with temperature, because the density of the solvent (and thus its volume) changes with temperature.
  19. Units of Concentration That Are Temperature Invariant mole fraction XA Molality m mass percent % parts per million ppm parts per billion ppb These units of concentration do NOT change with temperature.
  20. Calculating Concentrations - Example A solution contains 5.0 g of toluene (C7H8) and 225 g of benzene (C6H6) and has a density of 0.876 g/mL. For the solution, calculate a) its concentration in mass % b) the mole fraction of toluene c) the molarity of toluene d) the molality of toluene
  21. Calculating Concentrations - Example A solution contains 5.0 g of toluene (C7H8) and 225 g of benzene (C6H6) and has a density of 0.876 g/mL. a) mass %:5.0 g toluene x 100 (225 + 5.0) g solution b) mole fraction of toluene: mol toluene = 5.0 / 92.13 = 0.05427 mol mol benzene = 225 / 78.11 = 2.8806 mol Xtoluene = 0.05427 (0.05427 + 2.8806) = 2.2 % = 0.018 ** Notice the lack of units for these concentrations. **
  22. Calculating Concentrations - Example A solution contains 5.0 g of toluene (C7H8) and 225 g of benzene (C6H6) and has a density of 0.876 g/mL. Lets do molality first. d) the molality of toluene:0.05427 mol toluene 0.225 kg solvent c) the molarity of toluene: The volume of the solution is (225 g + 5.0 g) = 262.56 mL 0.876 g/mL molarity = 0.05427 mol toluene 0.26256 L of solution = 0.24 m = 0.21 M
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