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Gases dissolving in liquidsGases dissolving in liquids Pressure and temperature influence gas Pressure and temperature influence gas
solubilitysolubility Solubility directly proportional to gas pressureSolubility directly proportional to gas pressure
Henry’s LawHenry’s Law:: SSgg = k = kHHPPgg SSg g = gas solubility (M = mol/L) = gas solubility (M = mol/L) kkH H = Henry’s law constant (unique to each gas; = Henry’s law constant (unique to each gas;
M/mm Hg)M/mm Hg) PPgg = partial pressure of gaseous solute (mm = partial pressure of gaseous solute (mm
Hg)Hg)
Example Example 27.0 g of acetylene gas 27.0 g of acetylene gas
dissolves in 1.00 L of dissolves in 1.00 L of acetone at 1.00 atm acetone at 1.00 atm partial pressure of partial pressure of acetylene. acetylene.
If the partial pressure If the partial pressure of acetylene is of acetylene is increased to 6.00 atm, increased to 6.00 atm, what is the solubility of what is the solubility of acetylene in acetone in acetylene in acetone in mol/L? MW of acetylene mol/L? MW of acetylene = 26.037 g/mol= 26.037 g/mol
1.1. 27.0 g x (mol/26.037 g) 27.0 g x (mol/26.037 g) x (1/1.00 L) = 1.04 Mx (1/1.00 L) = 1.04 M
2.2. SSgg = k = kHHPPgg
3.3. 1.04 M = k1.04 M = kHH x 1.00 atm x 1.00 atm 1.1. kkHH = 1.04 M/atm = 1.04 M/atm
4.4. SSgg = (1.04 M/atm) x 6.00 atm = (1.04 M/atm) x 6.00 atm
1.1. = = 6.24 M6.24 M Could also solve this by:Could also solve this by:
– (S(Sg1g1/P/Pg1g1) = (S) = (Sg2g2/P/Pg2g2))
– How did I come up with How did I come up with this?this?
ProblemProblem
The partial pressure of oxygen gas, O2, in air at sea level is 0.21 atm. – Using Henry’s Law, calculate the molar
concentration of oxygen gas in the surface water (at 20°C) of a lake saturated with air given that the solubility of O2 at 20°C and 1.0 atm pressure is 1.38•10-3 M.
They call it “pop” in the They call it “pop” in the MidwestMidwest
Drinks carbonated under high pressureDrinks carbonated under high pressure– Above 90 atmAbove 90 atm
– Under COUnder CO22 atmosphere atmosphere
Once bottle opened, partial pressure of gas Once bottle opened, partial pressure of gas above soda plummetsabove soda plummets– COCO22 solubility decreases drastically solubility decreases drastically
– Gas bubbles out of solnGas bubbles out of soln Once the fizz is gone, it can never be regainedOnce the fizz is gone, it can never be regained
– Truly, one of the existential tragedies of this Truly, one of the existential tragedies of this universeuniverse
The bendsThe bends
Deeper diving has higher pressuresDeeper diving has higher pressures– Must use breathing tankMust use breathing tank– If it contains NIf it contains N22 then higher pressure forces N then higher pressure forces N22
to dissolve in higher amounts in bloodto dissolve in higher amounts in blood If ascension too fast, lower pressure causes If ascension too fast, lower pressure causes
NN22 to start bubbling out of blood too quickly to start bubbling out of blood too quickly – Rupturing of arteriesRupturing of arteries
Excruciatingly painful deathExcruciatingly painful death– Must be rushed to hyperbaric chamberMust be rushed to hyperbaric chamber
Tanks now don’t use NTanks now don’t use N22,, but Hebut He– Why?Why?
Effects of temp on solubilityEffects of temp on solubility
Obviously, as temp Obviously, as temp increases, solubility increases, solubility decreasesdecreases
Since increasing Since increasing heat causes gases heat causes gases to dissolve out to dissolve out (endothermic)(endothermic) dissolving gases dissolving gases
is an exothermic is an exothermic processprocess
Another look at gas solubility: Another look at gas solubility: Le Châtelier’s PrincipleLe Châtelier’s Principle
Explains temperature relevance of Explains temperature relevance of solubilitysolubility
For systems in equilibrium, change in one For systems in equilibrium, change in one side causes system to counteract on other side causes system to counteract on other side:side:
Gas + liquid solvent Gas + liquid solvent sat. soln + heat sat. soln + heat– So add heat, rxn goes to left by kicking out gasSo add heat, rxn goes to left by kicking out gas– Add gas, rxn goes to right by saturating soln & Add gas, rxn goes to right by saturating soln &
giving off heatgiving off heat
Solubility of solids based on Solubility of solids based on temperaturetemperature
In general, In general, solubility increases solubility increases w/ increasing tempw/ increasing temp– But exceptionsBut exceptions– No general behavior No general behavior
pattern notedpattern noted
CrystalsCrystals
One can separate One can separate impure dissolved impure dissolved salts by reducing salts by reducing temperaturetemperature– Impurity or desired Impurity or desired
product crystallizes product crystallizes out at specific temp out at specific temp as solubility as solubility collapsescollapses
Colligative propertiesColligative properties
Vapor and osmotic pressures, bp, Vapor and osmotic pressures, bp, and mp are colligative propertiesand mp are colligative properties– Depend on relative # of solute and Depend on relative # of solute and
solvent particlessolvent particles
Vapor PressureVapor Pressure Remember: Remember:
– Equilibrium vapor pressureEquilibrium vapor pressure Pressure of vapor when liq and vapor in equilibrium Pressure of vapor when liq and vapor in equilibrium
at specific tempat specific temp Vapor pressure of soln lower than pure Vapor pressure of soln lower than pure
solvent vapor pressuresolvent vapor pressure Vapor pressure of solvent Vapor pressure of solvent relative # of relative # of
solvent molecules in solnsolvent molecules in soln– i.e., solvent vapor pressure i.e., solvent vapor pressure solvent mole solvent mole
fractionfraction
Raoult’s LawRaoult’s Law
PPsolutionsolution = X = Xsolventsolvent P P°°solventsolvent
So if 75% of molecules in soln are So if 75% of molecules in soln are solvent molecules (0.75 = solvent molecules (0.75 = XXsolventsolvent))– Vapor pressure of solvent (Vapor pressure of solvent (PPsolventsolvent) = 75% ) = 75%
of of PP°°solventsolvent
Problem Problem
The vapor pressure of pure acetone (CH3COCH3) at 30°C is 0.3270 atm. Suppose 15.0 g of benzophenone, C13H10O (MW = 182.217 g/mol), is dissolved in 50.0 g of acetone (MW = 58.09 g/mol). – Calculate the vapor pressure of acetone
above the resulting solution.
Solution Solution
0.2986atmatm3270.0913.0PXP
913.00.0823mol0.861mol
mol861.0X
mol861.058.09g
mol50.0g :solvent
mol0823.0182.217g
mol15.0g :solute
solventsolventsolution
solvent
ProblemProblem
The vapor pressure of pure liquid CS2 is 0.3914 atm at 20°C. When 40.0 g of rhombic sulfur (a naturally occurring form of sulfur) is dissolved in 1.00 kg of CS2, the vapor pressure falls to 0.3868 atm. – Determine the molecular formula of
rhombic sulfur.
Solution Solution
8
sulfur rhombic
sulfur rhombic
3
solvent
solvent
solventsolventsolution
S
898.7sulfur 32.066g
mol
mol
256gmol
256g
0.156mol
g0.40
156.0mol
mol13.1mol
13.1mol0.9882
mol1.1376.143g
molg101.001.00kg :solvent
9882.0X
atm3914.0Xatm3868.0
PXP
Limitations of Raoult’s LawLimitations of Raoult’s Law Doesn’t take into Doesn’t take into
consideration attractive consideration attractive forces in solnsforces in solns
For For ideal soln ideal soln (to right), (to right), forces between forces between solute/solvent molecules = solute/solvent molecules = forces w/in pure solventforces w/in pure solvent
– Thus, PThus, Ptottot = P = PAA + P + PBB
– Like graph to rightLike graph to right Fine for similarly Fine for similarly
constructed molecules constructed molecules (hydrocarbons)(hydrocarbons)– London dispersion forces London dispersion forces
are weakestare weakest
Solute-solvent Solute-solvent > solv-solv> solv-solv Decreases vapor Decreases vapor
pressurepressure– decreased volatilitydecreased volatility
Get lower vapor Get lower vapor pressure than pressure than calculated calculated
Ex:Ex:– CHClCHCl33 & C & C22HH55OCOC22HH55
H on former H-bonds H on former H-bonds to latterto latter
– Does it increase or Does it increase or decrease the latter’s decrease the latter’s IMF?IMF?
Solute-solvent Solute-solvent < solv-solv< solv-solv Increases vapor Increases vapor
pressurepressure– increased volatilityincreased volatility
Get higher vapor Get higher vapor pressure than pressure than calculated calculated
Ex:Ex:– CC22HH55OH and HOH and H22OO– Former disrupts H-Former disrupts H-
bonding of latterbonding of latter Does it increase or Does it increase or
decrease the latter’s decrease the latter’s IMF?IMF?
Nonvolatile solute added to Nonvolatile solute added to solventsolvent
SaltsSalts– Lower vapor pressure of solventLower vapor pressure of solvent– Make solvent less volatileMake solvent less volatile
Nonvolatile solute added to Nonvolatile solute added to solventsolvent
Raises bpRaises bp Lowers mpLowers mp
– Why?Why? Adding more Adding more
nonvolatile solute or nonvolatile solute or increasing solute increasing solute molalitymolality– decreases vapor decreases vapor
pressure even morepressure even more Phase diagram to rightPhase diagram to right
– Pure water (black)Pure water (black)– Adulterated water Adulterated water
(pink)(pink)
Bp and molality relationshipBp and molality relationship
TTbpbp = K = Kbpbp mmsolutesolute
KKbp bp = molal boiling pt elevation = molal boiling pt elevation constant for solventconstant for solvent ((°C/°C/mm))
Bp elevation, Bp elevation, TTbpbp, directly , directly proportional to solute molalityproportional to solute molality
Antifreeze Antifreeze Propylene glycolPropylene glycol
– 1,2-propanediol1,2-propanediol– Formerly used ethylene Formerly used ethylene
glycol glycol Phased outPhased out Poisonous Poisonous
– Lowers melting pt Lowers melting pt – Increases boiling ptIncreases boiling pt– Reduces risk of radiator Reduces risk of radiator
“boiling over”“boiling over”– Appreciated during the Appreciated during the
summer months in the summer months in the desertdesert
Example Example
Pure toluene (C7H8) has a normal boiling point of 110.60°C.
A solution of 7.80 g of anthracene (C14H10) in 100.0 g of toluene has a boiling point of 112.06°C.
– Calculate Kb for toluene.
1.1. TTbpbp = K = Kbpbp mmsolutesolute
2.2. TTbpbp = 112.06 = 112.06°C - 110.60°C = 1.46°C
3. 7.80g x (mol/178.23g) = 4.38 x 10-2 mol
4. (4.38 x 10-2 mol/0.1000 kg) = 0.438 m
5. 1.46°C/0.438 m = 3.33°C/m
Freezing point depressionFreezing point depression
Similarly, Similarly, TTfpfp = K = Kfpfp mmsolutesolute
KKfp fp = molal fp depression constant = molal fp depression constant ((°C/°C/mm))
Antifreeze & CaClAntifreeze & CaCl22
ProblemProblem
Barium chloride has a freezing point of 962°C and a Kf of 108 °C/m.
A solution of 12.0 g of an unknown substance dissolved in 562 g of barium chloride gives a freezing point of 937°C.– Determine the molecular weight of the
unknown substance.
Solution Solution
mol
g92
0.13mol
12.0g
0.13solute moles
solvent 0.562g
solute moles23.0
23.0
C108C25T
C25C937C962T
m
mm
Solutions containing ions: Solutions containing ions: their colligative propertiestheir colligative properties
Colligative properties based on amount of Colligative properties based on amount of solute/solventsolute/solvent
Molality of ions depend on Molality of ions depend on number of number of constituents in cmpdconstituents in cmpd
Different for ionic vs. covalent cmpdsDifferent for ionic vs. covalent cmpds Ex:Ex:1.1. NaCl ionizes into two ionsNaCl ionizes into two ions
– So 0.5So 0.5 m m NaCl has 0.5 x 2 NaCl has 0.5 x 2 mm = 1 = 1 mmtottot
2.2. Benzene doesn’t ionizeBenzene doesn’t ionize– So 0.5 So 0.5 mm benzene = 0.5 benzene = 0.5 mmtottot
Using equation w/out above factor will lead to Using equation w/out above factor will lead to values that are offvalues that are off
How to correct for it: the How to correct for it: the van’t Hoff factorvan’t Hoff factor
ii = = the number of solute particles after dissolving Colligative properties are larger for Colligative properties are larger for
electrolytes than for nonelectrolytes of the electrolytes than for nonelectrolytes of the same molalitysame molality– Why? (Hint: solve the below)Why? (Hint: solve the below)
Give the Give the ii-values for: methanol, CaSO-values for: methanol, CaSO44, BaCl, BaCl22
TTfpfp (measured) = K (measured) = Kfpfp mm ii
Problem Problem
How many grams of Al(NOHow many grams of Al(NO33))33 must be must be added to 1.00 kg of water to raise added to 1.00 kg of water to raise the boiling point to 105.0the boiling point to 105.0°C°C
KKbb = 0.51 °C/ = 0.51 °C/mm MW = 212.9962 g/molMW = 212.9962 g/mol
solutionsolution
needed )Al(NO g530mol
212.9962g2.5mol
2.5solute moles
solvent 1.00kg
solute moles5.2
5.2
4C0.51
C5.0
C5.0C100.0-C105.0T
33
m
mm
OsmosisOsmosis
Net movement of water (solvent) Net movement of water (solvent) from area of lower solute from area of lower solute concentration to area of higher concentration to area of higher solute concentration across a semi-solute concentration across a semi-permeable membranepermeable membrane– Bio101Bio101
More… More… Pressure of column of soln = pressure of water Pressure of column of soln = pressure of water
moving through membranemoving through membrane Osmotic pressureOsmotic pressure = pressure made by column = pressure made by column
of soln = diff of heightsof soln = diff of heights = cRT= cRT c = mol/L = Mc = mol/L = M R = 0.08206 L R = 0.08206 L atm/(mol atm/(mol K) K) ideal gas law ideal gas law T = in KelvinT = in Kelvin = atm= atm Useful for measuring MM of biochemical Useful for measuring MM of biochemical
macromoleculesmacromolecules– Proteins and carbsProteins and carbs
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