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CHAPTER 16

CHAPTER 16. SOLUTIONS. Theme of the Chapter Solutions vs. Pure Liquids. Henry’s law. as the pressure of the gas above the liquid increases, so does the solubility of a gas in the liquid ex: carbonated soda. Henry’s Law. As Pressure increases Solubility increases

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CHAPTER 16

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  1. CHAPTER 16 SOLUTIONS

  2. Theme of the Chapter Solutions vs. Pure Liquids

  3. Henry’s law as the pressure of the gas above the liquid increases, so does the solubility of a gas in the liquid • ex: carbonated soda

  4. Henry’s Law • As Pressure increases • Solubility increases • This is a direct proportion

  5. Henry’s Law ex 1: If the solubility of a gas in water is .77g/L at 350 kPa of pressure, what is the solubility, in grams per liter at 100 kPa of pressure?

  6. Molarity 1. Tells how concentrated a solution is. 2. the # of moles of solute dissolved in 1L of solution

  7. Molarity ex1: How many moles of solute are present in 1.5L of .20M Na2SO4? -we are looking for moles of solute

  8. How to Make Dilutions If you have a solution and the molarity is too high, you need to dilute the solution to lower the molarity (concentration). Use this equation to make a more dilute solutiion:

  9. How to Make Dilutions Ex 1: How would you prepare 200. mL of 0.60M MgSO4 from a stock solution of 3.0 M MgSO4? • The dilute solution is 0.60M MgSO4, and we want to make 200.mL of it. • The concentrated solution is 3.0M MgSO4. We need to remove some of this stock solution and then add some distilled water to it to dilute it. So, lets figure out how much distilled water to add. M1V1 = M2V2 = (200.mL)(.60M) = (3.0M)(V2) V2 = 40mL therefore, 1. you would take 40mL of the stock solution (3.0M), and since we need 200mL, 2. you would add 160mL (200mL -40mL = 160mL) of distilled water in order to make a .60M solution So, now we have 200.mL of our dilution 0.60M solution

  10. How to Make Dilutions 1. Add 40 mL of stock to empty container 2. Add 160 mL of distilled water to empty container (need 200mL, added 40 mL, need 160 mL to top off) stock solution 3.0M MgSO4 Dilute solution 0.60M MgSO4

  11. Colligative Properties How are solutions different from pure liquids? When you add a solute to a liquid the Freezing point , Boiling point, and Vapor Pressure of the solution change

  12. Colligative Properties Water: The freezing point of pure water is 0°C. The normal boiling point of water is 100°C. But if you make a solution using water as the solvent, the freezing point of that solution will not be 0°C nor will the boiling point be 100°C. A SALT WATER SOLUTION: The freezing point will be lower than 0°C, boiling point will be higher than 100°C, & the vapor pressure will decrease.

  13. Colligative Properties So… 1. Freezing point depression 2. Boiling point elevation 3. Vapor pressure lowering

  14. Vapor Pressure Lowering The pressure over a liquid or solution that results when molecules escape the liquid and enter the gaseous phase. • In order for a molecule in the liquid phase to enter the gaseous phase, the molecule must be at the surface of the liquid and have enough kinetic energy to break the surface.  The addition of a nonvolatile hinders this situation in two ways: • The solute particles occupy space at the surface. • The solute particles introduce a new set of attractive forces with the solvent molecules.

  15. The more dissolved solute the greater the changes. • Ex: the more salt you add to salt water the higher the Boiling Point, the lower the vapor pressure and freezing point • Which will have a higher boiling point? 1M solution of CuCl2 vs. 1M solution C6H12O6 1M solution of CuCl2 Because there are more ions in solution that hold onto the water molecules

  16. Freezing Point Depression

  17. Vapor Pressure Lowering

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