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Solutions

Solutions. Chapter 13. Solutions. Solution - homogeneous mixture of solute and solvent . Solvent - the component present in the largest amount . Solutes - The other components. Factors that determine whether a solution will form:.

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Solutions

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  1. Solutions Chapter 13

  2. Solutions • Solution - homogeneous mixture of solute and solvent. • Solvent - the component present in the largest amount. • Solutes - The other components

  3. Factors that determine whether a solution will form: • Entropy – usually solution formation increases entropy (more disordered) • Energy – based on the strength of IMFs – neither the solute or solvent can be more attracted to itself than to the other substance.

  4. The Solution Process • IMFs between the solvent particles and IMFs between the solute particles must be equal to or weaker than IMFs between the solute and solvent. Oil – nonpolar – dispersion forces Water –Polar - hydrogen bonds The hydrogen bonds in water are stronger than any attractive force between the two substances

  5. The Solution Process • Consider NaCl (solute) dissolving in water (solvent) • Ion-dipole forces between the ions and the water molecules (in this case) are stronger than the ionic force holding the ions together and stronger than the hydrogen bonds holding the water molecules together. (hydration)

  6. “Rule of thumb”: • Polar solvents dissolve polar solutes and nonpolar solvents dissolve nonpolar solutes ("like dissolves like")

  7. More on Energy… ΔHsolute = energy to separate solute particles it’s always endothermic (+) ΔHsolvent= energy to separate solvent particles it’s always endothermic (+) ΔHmix= depends on IMFs between solvent & solute it’s always exothermic (-) ΔHsoln = ΔHsolute + ΔHsolvent + ΔHmix The solution process overall may be endothermic (cold packs) or exothermic (hot packs) depending on the magnitudes of each enthalpy value

  8. Miscible - substances that mix in any proportions • Example: Ethanol and water are miscible liquids. • Immiscible - substances that do not mix significantly. • Example: Oil and water are immiscible. • Solids (salt, sugar, etc) dissoved in a liquid (water) are somewhere in between!

  9. Saturated Solutions and Solubility • As a solid dissolves, a solution forms (the opposite process is crystallization) • If crystallization and dissolution are in equilibrium with undissolved solute, the solution is saturated. (no further solute can be dissolved)

  10. Solubility • The amount of solute required to form a saturated solution. • A solution with a concentration of dissolved solute that is less than the solubility is said to be unsaturated. • A solution is said to be supersaturatedif more solute is dissolved than in a saturated solution (heat, dissolve more, cool)

  11. Factors Affecting Solubility • Solubility depends on: • The nature of the solute & solvent • More detail on the “like dissoves like” rule • The temperature. • The pressure (for gases).

  12. Consider the solubility of alcohols in water. • Water and ethanol are miscible because the broken hydrogen bonds in both pure liquids are re-established in the mixture. • However, not all alcohols are miscible with water.

  13. The number of carbon atoms in a chain affects solubility. • The greater the number of carbons in the chain (the larger the nonpolar part), the more the molecule behaves like it’s nonpolar • This decreases its solubility in water • The greater the number of –OH groups (the polar part), the more the molecule behaves like it’s polar • This increases its solubility in water Substances with similar intermolecular attractive forces tend to be soluble in one another. (like dissolves like)

  14. Pressure Effects • The solubility of a gas in a liquid is a function of the pressure of the gas over the solution. • Solubilities of solids and liquids are not greatly affected by pressure. • With higher gas pressure, more molecules of gas are close to the surface of the solution and the probability of a gas molecule striking the surface and entering the solution is increased.

  15. Properties of Solutions Higher pressure above means more solute dissolves Lower pressure above means less solute dissolves

  16. The higher the pressure, the greater the solubility. • The lower the pressure, the less the solubility.

  17. Henry’s Law - The solubility of a gas is directly proportional to the partial pressure of the gas above the solution. • Mathematically: Sg=kPg • Note: The Henry's law constant (k) differs for each solute-solvent pair and differs with temperature. • S is usually expressed in mol/L • make sure units cancel when using this equation.

  18. Henry’s Law Application • Carbonated beverages are bottled where PCO2> 1 atm. • As the bottle is opened, PCO2 decreases and the solubility of CO2 decreases (solution fizzes)

  19. Temperature Effects • As temperature increases, solubility of solids generallyincreases (there are some exceptions) • Sugar dissolves better in warm water than in cold water. • As temperature increases, solubility of gases generally decreases. • Carbonated beverages go flat as they get warm. • Thermal pollution: If lakes get too warm, O2 become less soluble and fish suffocate

  20. Concentration of Solutions

  21. Colligative Properties • Colligative properties depend on number of solute particles. • There are four colligative properties to consider: • Vapor pressure lowering (Raoult's Law). • Boiling point elevation. • Freezing point depression. • Osmotic pressure.

  22. Van’t Hoff Factor (i) • The number of particles formed when a solute dissolves • NaCl = 2 • Al(NO3)3 = 4 • C12H22O11 = 1 This is a maximum!

  23. Lowering the Vapor Pressure • Nonvolatile solutes (with no measurable vapor pressure themselves) reduce the ability of the surface solvent molecules to escape the liquid. • Therefore, vapor pressure is lowered.

  24. The amount of vapor pressure lowering depends on the amount of solute. • Raoult’s law quantifies the extent to which a nonvolatile solute lowers the vapor pressure of the solvent PA=XAPAo PA = Vapor Pressure of sol’n XA = mole fraction of solvent PAo= Vapor pressure of pure solvent

  25. Ideal solution: one that obeys Raoult’s law. • Real solutions show ideal behavior when: • The solute concentration is low. • The solute and solvent have similarly sized molecules. • The solute and solvent have similar types of intermolecular attractions.

  26. Boiling-Point Elevation • A nonvolatile solute lowers the vapor pressure of a solution. • Higher temperature is required to reach a vapor pressure of 1 atm for the solution

  27. The molal boiling-point-elevation constant, Kb, expresses how much Tb changes with molality Tb= Kbmi • The boiling-point elevation is proportional to the concentration of solute particles • i = van’thoff factor • A 1 msolution of NaCl is 2 m in total solute particles.

  28. Freezing-Point Depression • When a solution freezes, crystals of almost pure solvent are formed first. • Solute molecules are usually not soluble in the solid phase of the solvent - sol’n become more concentrated (lower X of solvent in sol’n). • This lowers the vapor pressure for the solution • the triple point occurs at a lower temperature

  29. The melting-point (freezing-point) curve is a vertical line from the triple point. • The solution freezes at a lower temperature than the pure solvent. • The decrease in freezing point (Tf) is directly proportional to molality. Tf= Kfmi • Kf is the molal freezing-point- depression constant.

  30. Osmosis • Semipermeable membranes permit passage of some components of a solution. • Often they permit passage of water but not larger molecules or ions. • Osmosis is the net movement of a solvent from an area of low solute concentration to an area of high solute concentration.

  31. Consider a U-shaped tube with a two liquids separated by a semipermeable membrane. • The rate of movement of solvent from the less concentrated solution is faster than the rate of movement in the opposite direction. • As solvent moves across the membrane, the fluid levels in the arms become uneven. • The vapor pressure of solvent is higher in the arm with more solvent. • Eventually the pressure difference due to the difference in height of liquid in the arms stops osmosis.

  32. Osmotic Pressure

  33. Properties of Solutions The process of reverse osmosis uses mechanical pressure to make the solvent (water) flow in the opposite direction. This process can be used to desalinate ocean water to make drinking water.

  34. Osmotic pressure, π • The pressure required to prevent osmosis. • Osmotic pressure obeys a law similar in form to the ideal-gas law n=moles V= volume M= molarity R= the ideal gas constant T= absolute temperature pV = nRT OR

  35. Solutions are said to be isotonic if they have the same osmotic pressure. • Hypotonic solutions have a lower π and are less concentrated • Hypertonic solutions have a higher π and are more concentrated

  36. If a cucumber is placed in NaCl solution, it will lose water to shrivel up and become a pickle. • A dry carrot placed in water becomes firm as water enters via osmosis. • Eating large quantities of salty food causes retention of water and swelling of tissues (edema). • Water moves into plants, to a great extent, through osmosis. • Salt may be added to meat (or sugar added to fruit) as a preservative. • Salt prevents bacterial infection: A bacterium placed on the salt will lose water through osmosis and die.

  37. Determination of Molar Mass Camphor (C10H16O) melts at 179.8°C, and it has a particularly large freezing-point-depression constant, Kf=40.0°C/m. When 0.186 g of an organic substance of unknown molar mass is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7°C. What is the molar mass of the solute?

  38. ΔTf = kmi (179.8-176.7) = (40.0)(m)(1) 3.1 = 40.0m m = .0775 mol solute/kg camphor

  39. A sample of 2.05 g of polystyrene of uniform polymer chain length was dissolved in enough toluene to form 0.100 L of solution. The osmotic pressure of this solution was found to be 1.21 kPa at 25°C. Calculate the molar mass of the polystyrene.

  40. (1.21/101.325)(.100L) = n(.0821)(298) n = 0.0000488 mol

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