Henry's Law, Freezing Point Depression, Boiling

# Henry's Law, Freezing Point Depression, Boiling - Henry’s...

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Unformatted text preview: Henry’s Law, Freezing Point Depression, Boiling Point Elevation and Raoult’s Law Wow, That is a Mouthful Henry’s Law Henry’s The solubility of a gas is directly proportional to The the gas pressure the Sg= khPg When the partial pressure of the solute above a When solution drops, the solubility of the gas in the solution drops as well to maintain the equilibrium. solution This can be used to calculate the molar solubility This of a gas. of What is the concentration of O2 in a fresh water stream in equilibrium with air at 25oC and 1.0 atm. stream and (Hint there is 21% O2 in the air) (Hint Does this always fit? Does As concentrations and partial pressures As 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 ideal Colligative Properties Colligative Colligative properties are “properties that Colligative depend only on the relative number of particles and not on what the actual substance is” substance Remember that Changing Vapor Pressure (Raoult’s Law) (Raoult’s Vapor pressure at a given temperature is Vapor the pressure that the vapor exerts when the rate of molecules leaving the surface is equal to the rate of them re-condensing. equal But what happens when something is now But dissolved in the solvent. dissolved 2 things- 1. less solvent molecules at the thingssurface. 2. different sets of attractive forces forces Lets look at each one individually Lets 1. less solvent molecules at the surface. 1. Therefore less chance the water leaves, the vapor pressure is lowered. This makes sense based on Henry’s law. The vapor pressure of the solvent will be proportional to the mole fraction in the liquid. to Psolvent= XsolventK If we also look at a pure solvent Po, then Po= XsolventK, Therefore Psolvent= Xsolvent Po then solvent This is Raoult’s Law Raoult’s Law Raoult’s Raoult’s law assumes that the solution is ideal. Raoult’s 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 lower If solvent-solute interactions are weaker than If solvent-solvent, the actual vapor pressure will be higher than calculated higher Try a problem Try Assume you dissolve 10.0g of sugar Assume (C12H22O11) iin 225mL (225g) of water and n (C warm the water to 60oC. What is the warm C. vapor pressure of the water over this solution? The normal vapor pressure of water at 60oC iis 149.4 torr. s water Raoult’s Law Cont. Raoult’s Adding a nonvolatile solute to a solvent Adding will lower the vapor pressure. will ∆Psolvent= Psolvent- Posolvent ∆Psolvent= (XsolventPosolvent) – Posolvent = -(1Xsolvent)Posolvent Xsolvent+Xsolute = 1 ∆Psolvent= -XsolutePosolvent -X Why does this matter? Why Well remember that vapor pressure Well determines the boiling point of a liquid. If you add solute it will change the solvent’s vapor pressure, therefore the boiling point changes. boiling This is called boiling point elevation! Boiling Point Elevation Boiling The Boiling point elevation, Δtbp, iis directly s The proportional to the molality of the solute proportional Δtbp= Kbpmsolute Kbp is called the molal boiling point elevation constant by solvent and is (oC/m) constant How many grams of ethylene glycol, How HOCH2CH2OH, do you have to add to 125 g of HOCH OH, water to increase the bp by 1oC? (The KbpWater water Water = +0.5121 oC/m What is another use? What Molar mass by boiling point Molar elevation! elevation! A solution prepared from 1.25 g of oil of solution wintergreen (methyl salicylate) in 99.0 g of benzene has a boiling point of 80.31oC. benzene C. Determine the molar mass of the compound. (Benzene’s normal bp is 80.10 and the Kbp is +2.53 oC/m) Answer is 150 g/mol Freezing Point Depression Freezing Very similar to boiling point Δtfp= Kfpmsolute The reason for this is very similar in The changes in vapor pressure equilibrium changes There are more atoms of pure solvent There going from solid to liquid than from liquid to solid. to What about for electrolytes? What We would assume that adding NaCl or such to water We would have twice the effect would That is pretty much true. That pretty To see the real effect, we need a van’t Hoff factor i = Δtfp, measured/ Δtfp calculated measured/ As the Δtfp calculated is if no ionization occurred. The i is As not a perfect for the number of ions, but is closest to it for dilute solutions due to the intermolecular attractive forces. forces. Δtfp, measured= Kfpmsolute i solute Calculate the freezing point of 525 g of water that contains 25.0 g of NaCl. Assume i is 1.85 for NaCl. is ...
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