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Solution - a homogeneous mixture of two or more chemical substances. Solvent - the major component of a solution ( or sometimes the only liquid component). Solute - a minor component of a solution (the stuff which is dissolved in the solution). - PowerPoint PPT Presentation
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Solution - a homogeneous mixture of two or more chemical substances
Solvent - the major component of a solution ( or sometimes the only liquid component)
Solute - a minor component of a solution (the stuff which is dissolved in the solution)
Solutions: Concentration Expressions
Molarity: Moles of solute per liter of solution
Molality: Moles of solute per Kg of solvent
Mole Fraction: Moles of solute divided by totalmoles of solute and solvent.
Thermodynamics of Solubility
What is H° of solution?
Thermodynamics of Solubility
NaCl(s) Na+(g) + Cl-(g) Ho1 = 786 kJ/mol
H2O(l) + Na+(g) + Cl-(g) Na+(aq) + Cl-(aq)
Hohyd = Ho
2 + Ho3 = -783 kJ/mol
H° = +3 kJ/mol
So why do salts with H positive dissolve?
Entropy G = H - TS
NaCl S° = 46 J/Kmol
But some S° values are negative!
M
O
O
O O
O
O
H
H
H
H
HH
H
H
HH
H H
H
O H
H
OH
XH
OH
H O
H
The higher the charge on the ion and the smallerthe size of the ion, the greater the interaction. Thisleads to a greater ordering of the water.
Hydration of a salt - MX
The solvation of the ions leads to anordering of the water solution. Thus andecrease in the entropy of the water.
Chemical species that interact favorablywith water are call hydrophilic.
Chemical species that do not interact favorablywith water are called hydrophobic.
Non polar molecules such as hexane donot dissolve in water.
Oil floats on the top of water.
The enthalpy for dissolving hexane in wateris actually negative. So why doesn’t it dissolve?
HO
H
H
OH H
OH
HO
H
H
OHH
OH
H
CH H
H
When a non polar molecule is dissolved in water the water molecules form a highly ordered clatharate structure
The structure hereis a two dimensionalrepresentation of thetrue structure whichthree dimensional.
This ordering of the watermolecules leads to a negative S.
This is the basis of the hydrophobic effect
Solubilities of alcohols in water
Methanol infinite
Ethanol infinite
1-propanol infinite
1-butanol 80 g/L
1-pentanol 22 g/L
1-hexanol 5 g/L
SO O
ONa
Surfactants
The amount of gas dissolved in a liquid solutionis directly proportional to the pressure.
Partial Pressure = P = kHHenry’s Law
kH (atm) for water
CH4 4.13 x 102
O2 4.34 x 104
N2 8.57 x 104
In the atmosphere how much O2 will dissolve in H2O?
P = kHkH O2 4.34 x 104 atm
P(O2) in atm .21atm
.21atm = 4.34 x 104 atm x
= 4.84x 10-6 mol fraction of O2
1000 g H2O/18 g/mol = 55.6 mol H2O
4.84x 10-6 x 55.6 = 2.69 x 10-4 mol O2
2.69 x 10-4 mol x 32 g/mol = .0086 g O2
.0086 g O2 / liter H2O
at normal atmospheric pressure
In the actual atmosphere
.21 x 32.0 g / 22.4 = .30 g O2 / liter air
Aren’t you glad you are not a fish!
What will happen if you have two beakers in a closed system? One contains 1 liter of 1M NaCl,the other 1 liter of pure H2O.
Raoult’s Law
Vapor pressure ofsolution with a nonvolatile solute.
Pobs = solvPosolv
Raoult’s Law for a mixture of two volatile liquids.
Ptotal = PA + PB = APoA + BPo
B
Raoult’s says that the presence of a solutewill lower the vapor pressure of a solution ascompared to the pure solvent.
What will this do to the freezing point andboiling point of the solution?
Freezing Point Depression
T = Kf msolute
Boiling Point Elevation
T = Kb msolute
Calculate the boiling point of a solution containing6.16 g of sucrose, C12H22O11, dissolved in 30.g of H2O.
T = Kb msolute
Kb = 0.51 C°kg/mol m.wt. sucrose 342 g/mol
mol = 6.16 g/(342g/mol) = .018 mol
T = 0.51 C°kg/mol x .60 mol/kg =.31 C°
Boiling point of solution = 100.31 C°
m = .018 mol / .030 kg = .60 mol/kg
A solution of .90g of glucose in 20.0g of water hasa freezing point of -0.465 C°. What is mwt. of glucose?
Kf = 1.86 C°kg/mol
m = T/Kf = .465 C° / (1.86 C°kg/mol) =.25 m
.25 mol/kg = X mol/.020kg
X mol of glucose = .005 mol
.90 g/ .005 mol = mwt. / 1mol = 180 g / mol
Osmosis
Osmotic Pressure isthe pressure that just stops the osmosis.
= MRT
= MRT
= MRT M = / RT
A 2.2 gram sample of polyethylene was disolved intoluene to give 100ml solution. The osmoticpressure at 25°C was measured to be 1.10 x 10-2atm.What is the mwt. of the polyethylene?
M = 1.10 x 10-2 / (.082 x 298) = 4.5 x 10-4 M
.10 L x 4.5 x 10-4 mol/L = 4.5 x 10-5 mol
2.2 g / 4.5 x 10-5 mol = mwt = 49,000 g/mol
49,000/ 28 = 1750 units CH2CH2n
ReverseOsmosis
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