1 / 26

Properties of solutions

Properties of solutions. Chapter 13. The Solution Process . A solution forms when one substance disperses uniformly throughout another. The reason substances dissolve is due to intermolecular forces. Ion-Dipole forces between water and ionic solids.

tessa
Download Presentation

Properties of solutions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Properties of solutions Chapter 13

  2. The Solution Process • A solution forms when one substance disperses uniformly throughout another. • The reason substances dissolve is due to intermolecular forces. • Ion-Dipole forces between water and ionic solids. • Dispersion forces between C6H14 and CCl4 • The interactions between solute and solvent molecules are know as solvation. • When the solvent is water they are know as hydration.

  3. Energy Changes in Solvation • NaCl dissolves in water because the water molecules have strong enough attractive forces for the Na+ and Cl- ions. • The water molecules must also move away from each other to make space for the Na+ and Cl- ions.

  4. Heat of Solvation

  5. Saturated Solutions and Solubility • As a solid dissolves the concentration of solute particles in the solution increases. • This leads to a certain amount of crystallization (the opposite of solvation) • When the solute concentration reaches a certain point the rate of solvation and the rate of crystallization will be equal • At this point the solution is said to be saturated.

  6. The amount of solute needed to form a saturated solution in a given quantity of solvent is known as the solubility of that solute. • The solubility of a compound is the maximum amount of that compound that will dissolve in a given amount of solvent at a specific temperature. • The solubility of NaCl at 0 oC is 35.7 g per 100 mL of water. • If we dissolve less than this amount the solution is said to be unsaturated. • Under certain conditions it is possible for a solution to contain more than the amount of solute needed to form a saturated solution.

  7. Factors Affecting Solubility • Along with the chemical nature of the solvent and solute the solubility of a substance depends on… • Temperature • Pressure (For gases only)

  8. Solute-Solvent Interactions • One factor that affects solubility is the chemical nature of both the solute and the solvent. • The stronger the attractions between solute and solvent molecules the higher the solubility. • If we try to dissolve a polar solute into a polar solvent it would work very well. • This is due to favorable dipole-dipole interactions between solute and solvent particles. • If we tried to dissolve a polar solute into a nonpolar solvent it would not work. • This is because there would be little to know attractive forces between solute and solvent particles.

  9. Pressure Effects • The solubility of solids and liquids is affected so little by ambient pressure that we do not need to account for it. • However the solubility of gases is highly affected by pressure. • The solubility of a gas increases as pressure increases. • Soda • Henry’s Law: • Sg = kPg

  10. Using Henry’s Law • Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4 atm over the liquid at 25 oC. The Henry’s Law constant for CO2 in water at this temperature is 3.1 x 10-2 mol/L-atm.

  11. Temperature Effects • The solubility of most solid solutes in water increases as the temperature of the solution increase. • The solubility of gases in water decreases with increasing temperature. • Poor fishies…

  12. Ways to express solution concentration • Mass Percentage : % Mass • Parts Per Million: ppm • Mole Fraction

  13. Molarity: M • Molality: m

  14. Practice Problems • A solution is made by dissolving 13.5 g of glucose (C6H6O6) in 0.100 kg of water. What is the mass percentage of solute in this solution? • 11.9 % • A 2.5 g sample of ground water was found to contain 5.4 μg of Zn2+. What is the concentration of Zn2+ in parts per million? • 2.2 ppm

  15. A solution is made by dissolving 4.35 g of glucose in 25.0 mL of water at 25 oC. Calculate the molality of the glucose in solution. • 0.964 m • An aqueous solution of hydrochloric acid contains 36% HCl by mass. Calculate the mole fraction of HCl in solution and the molality of HCl in solution. • Mole fraction = 0.22 • Molality = 15 m

  16. A solution contains 5.0 g of toluene (C7H8) dissolved in 225 g of benzene (C6H6) and has a density of 0.876 g/mL. Calculate the molarity of toluene in solution. • 0.2 M

  17. Colligative Properties • The physical properties of many solutions differ from those of the pure solvent. • Example: • Pure water freezes at 0 oC but aqueous solutions freeze at lower temperatures. • Pure water boils at 100 oC but aqueous solutions boil at higher temperatures.

  18. Vapor Pressure Lowering • We saw in a previous chapter that all liquids have a vapor pressure associated with them. • We find that the vapor pressure of a pure solvent is higher than the vapor pressure of a solution made using that solvent. • Raoult’s Law • PA = XAPoA

  19. Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25 oC. Calculate the vapor pressure at 25oC of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of water at 25 oC is 23.8 torr.

  20. Boiling Point Elevation and Freezing Point Depresson • The boiling point of a solution is higher than that of the pure solvent. • The change in boiling point is denoted as ΔTb • We can solve for the change in boiling point using the equation… • ΔTb = Kbm • Where Kb is the molal boiling point elevation constant and m is the molal concentration of the solution.

  21. Example • Automotive antifreeze consists of ethylene glycol (C2H6O2), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25.0% by mass solution of ethylene glycol in water. • Kb = 0.51oC/m • Kf = 1.86oC/m

  22. List the following aqueous solutions in order of their expected freezing point: 0.050 m CaCl2, 0.15 mNaCl, 0.10 mHCl, 0.050 m CH3COOH, 0.10 m C12H22O11.

  23. Osmosis • Osmosis is the movement of materials across a permeable membrane. • The pressure required to prevent osmosis of a pure solvent is called the osmotic pressure. • Isotonic – Two solutions with identical osmotic pressure. • Hypotonic vs Hypertonic • π = MRT

  24. The average osmotic pressure of blood is 7.7 atm at 25oC. What molarity of glucose would be isotonic with blood?

  25. Determining Molar Mass by Freezing Point Depresson • A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.25 g of the substance in 40.0 g of CCl4. The boiling point of the resultant solution was 0.357 oC higher than that of the pure solvent. Calculate the molar mass of the solvent.

More Related