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Henrys Law, Freezing Point Depression, Boiling Point Elevation and Raoults Law Wow, That is a Mouthful.

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Presentation on theme: "Henrys Law, Freezing Point Depression, Boiling Point Elevation and Raoults Law Wow, That is a Mouthful."— Presentation transcript:

1 Henrys Law, Freezing Point Depression, Boiling Point Elevation and Raoults Law Wow, That is a Mouthful

2 Henrys Law The solubility of a gas is directly proportional to the gas pressure S g = k h P g When the partial pressure of the solute above a solution drops, the solubility of the gas in the solution drops as well to maintain the equilibrium. This can be used to calculate the molar solubility of a gas. What is the concentration of O 2 in a fresh water stream in equilibrium with air at 25 o C and 1.0 atm. (Hint there is 21% O 2 in the air)

3 Does this always fit? As concentrations and partial pressures increase, deviations from Henry's law become noticeable. This behavior is very similar to the behavior of gases, which are found to deviate from the ideal gas law as pressures increase and temperatures decrease. For this reason, solutions which are found to obey Henry's law are sometimes called ideal dilute solutions

4 Colligative Properties Colligative properties are properties that depend only on the relative number of particles and not on what the actual substance is Remember that

5 Changing Vapor Pressure (Raoults Law) Vapor pressure at a given temperature is the pressure that the vapor exerts when the rate of molecules leaving the surface is equal to the rate of them re-condensing. But what happens when something is now dissolved in the solvent. 2 things- 1. less solvent molecules at the surface. 2. different sets of attractive forces

6 Lets look at each one individually 1. less solvent molecules at the surface. Therefore less chance the water leaves, the vapor pressure is lowered. This makes sense based on Henrys law. The vapor pressure of the solvent will be proportional to the mole fraction in the liquid. P solvent = X solvent K If we also look at a pure solvent P o, then P o = X solvent K, Therefore P solvent = X solvent P o This is Raoults Law

7 Raoults Law Raoults law assumes that the solution is ideal. Therefore, the forces between solute and solvent molecules must be the same as the solvent to solvent. If solvent-solute interactions are stronger than solvent-solvent, the actual vapor pressure will be lower than calculated If solvent-solute interactions are weaker than solvent-solvent, the actual vapor pressure will be higher than calculated

8 Try a problem Assume you dissolve 10.0g of sugar (C 12 H 22 O 11 ) in 225mL (225g) of water and warm the water to 60 o C. What is the vapor pressure of the water over this solution? The normal vapor pressure of water at 60 o C is 149.4 torr.

9 Raoults Law Cont. Adding a nonvolatile solute to a solvent will lower the vapor pressure. P solvent = P solvent - P o solvent P solvent = (X solvent P o solvent ) – P o solvent = -(1- X solvent )P o solvent X solvent +X solute = 1 P solvent = -X solute P o solvent

10 Why does this matter? Well remember that vapor pressure determines the boiling point of a liquid. If you add solute it will change the solvents vapor pressure, therefore the boiling point changes. This is called boiling point elevation!

11 Boiling Point Elevation The Boiling point elevation, Δt bp, is directly proportional to the molality of the solute Δt bp = K bp m solute Δt bp = K bp m solute K bp is called the molal boiling point elevation constant by solvent and is ( o C/m) How many grams of ethylene glycol, HOCH 2 CH 2 OH, do you have to add to 125 g of water to increase the bp by 1 o C? (The K bp Water = +0.5121 o C/m

12 What is another use? Molar mass by boiling point elevation! A solution prepared from 1.25 g of oil of wintergreen (methyl salicylate) in 99.0 g of benzene has a boiling point of 80.31 o C. Determine the molar mass of the compound. (Benzenes normal bp is 80.10 and the Kbp is +2.53 o C/m) Answer is 150 g/mol

13 Freezing Point Depression Very similar to boiling point Δt fp = K fp m solute Δt fp = K fp m solute The reason for this is very similar in changes in vapor pressure equilibrium There are more atoms of pure solvent going from solid to liquid than from liquid to solid.

14 What about for electrolytes? We would assume that adding NaCl or such to water would have twice the effect That is pretty much true. To see the real effect, we need a vant Hoff factor Δt fp, measured/ Δt fp calculated i = Δt fp, measured/ Δt fp calculated As the Δt fp calculated is if no ionization occurred. The i is not a perfect for the number of ions, but is closest to it for dilute solutions due to the intermolecular attractive forces. Δt fp, measured= K fp m solute i Δt fp, measured= K fp m solute i

15 Calculate the freezing point of 525 g of water that contains 25.0 g of NaCl. Assume i is 1.85 for NaCl.


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