8.2 Colligative Properties Objective 2 Chemistry
Essential Questions What are the colligative properties? How does adding a solute affect them? How do you find the new boiling and freezing points after the addition of the solute?
Concentration Measurements There are two main ways to measure concentration are molarity (M) and molality (m)
Colligative Properties Colligative properties are physical properties that are affected by the concentration of the solute in the solution Vapor Pressure Reduction Boiling Point Elevation Freezing point depression
Colligative Properties (cont) Ionic compounds split into multiple particles which increase the effect compared to covalent compounds The more ions it breaks into the greater the effect Example: NaCl breaks into Na+ and Cl-, but CaCl2 would break into Ca2+ and 2 Cl-.
Vapor Pressure Reduction In a liquid, molecules are constantly moving from the liquid phase to the gas phase and back again Vapor pressure is the pressure of the gas “pushing” on the liquid
Vapor Pressure Reduction (cont) By adding a solute, it prevents some of the molecules escaping into the gas phase Because there are fewer molecules in the gas phase, the pressure is reduced
Vapor Pressure Reduction (cont) The decrease in a solution’s vapor pressure is proportional to the number of particles the solute makes in solution. Which would have the lowest pressure?
Boiling Point Elevation A liquid boils when its vapor pressure is equal to the external pressure because at this point the molecules can escape into the gas phase Adding a solute reduces vapor pressure and increases the energy needed for the vapor pressure to equal the external pressure The increase in energy results in the boiling point increasing which is called Boiling Point Elevation
Boiling Point Elevation (cont) Boiling point elevation (ΔTb) is proportional to the molality of the solution ΔTb=Kbm Where: Kb is the molal boiling point constant (every solvent has a different Kb value) m is the molality of the solution ΔTb is the elevation of the boiling point For soluble ionic compounds m is multiplied by the number of ions formed
Boiling Point Elevation (cont) By how much will the boiling point of water be elevated if 100g of sucrose (C12H22O11) is added to 500 g of water? The solution has a concentration of .584 m C12H22O11 and the Kb for water is 0.512 °C/m. Use the formula to solve for the change in temp. ΔTb=Kbm ΔTb= (.512 °C/m)(.584m) = 0.30 C What would be the new boiling point? Add the ΔTb to the regular boiling point 100C + 0.3 C = 100.30 C
Boiling Point Elevation (cont) Find the boiling point for .237 m NaCl in 6325 g CHCl3. Kb = 3.63 °C/m and boiling point = 61.2 °C Find the boiling point for .431 m NH3 in 4353 g H2O Find the boiling point 5.65 m SrBr2 in 542 g C6H6. Kb = 2.53 °C/m and boiling point = 80.1 °C
What happens? What happens to ice when you add salt? The ice melts Why do you think it melts?
Freezing Point Depression As a substance freezes the particles take on an orderly pattern, but when a solute is added the pattern is disrupted This disruption means more kinetic energy must be withdrawn before the substance can freeze so it freezes at a lower temperature This reduction in temperature causes a lower freezing point called Freezing Point Depression
Freezing Point Depression (cont) Freezing Point depression (ΔTf) is proportional to the molality of the solution ΔTf=Kfm Where: Kf is the molal freezing point constant (every solvent has a different Kf value) m is the molality of the solution ΔTf is the depression of the freezing point For soluble ionic compounds m is multiplied by the number of ions formed
Freezing Point Depression (cont) Calculate the freezing point depression of a solution of 100g of antifreeze, ethylene glycol (C2H6O2), in 0.5 kg of water. The solution has a concentration of 3.22 m C2H6O2 and the Kf for water is 1.86 °C/m. Substitute into formula ΔTf=Kfm ΔTf=(1.86 °C/m)(3.22m) = 6.01°C What would be the new freezing point? Subtract ΔTf from the regular freezing point 0.0 °C - 6.01°C = - 6.01°C
Freezing Point Depression (cont) Find the freezing point for 1.19 m Li2O in 6325 g C6H6. Kf = 5.12 °C/m and freezing point = 5.5 °C Find the freezing point for 3.2 m NO2 in 4353 g H2O Find the freezing point .554 m CaBr2 in 542 g C2H5OH. Kf = 1.99 °C/m and freezing point = -114.6 °C
Essential Questions What are the colligative properties? How does adding a solute affect them? How do you find the new boiling and freezing points after the addition of the solute?
8.2 Tracked Assignment Worksheet