6/22/11
Last time:
Chem 103
Lecture 1b*
Solutes and Solutions
*Week 1 Wednesday(b)
Henry’s Law
Solubility of a gas in a liquid is proportional to the partial pressure of the gas
on the liquid surface:
sg = kHPg where sg= solubility of gas; Pg = partial P of gas
Example: O2 has kH = 1.26 x 10-3 M atm-1 (at 25°C)
If the O2 mole fraction (XO2) in air is 0.21,
a) what is PO2 if the atmospheric pressure is 1.0 atm?
PO2 = XO2Patm=0.21(1.0 atm) = 0.21 atm.
b) what is the molar solubility of O2 in an aquarium at Patm =1.0 atm?
sO2 = kO2PO2 = 1.26 x 10-3 M atm-1 (0.21 atm)=2.65 x 10-4M
Sample Problem
What is the molality (m) of a 1.20 M NaCl solution whose density (d) is
1.0200 g/mL? (FW of NaCl = 58.5 g/mol)
Solution:
Recall that M = mol solute/L soln, m = mol solute/kg solvent
Assume 1.0 L: therefore, #mol NaCl = 1.20,
Also: mass soln = 1.0200 g/mL(1000mL/L)(1kg/1000g) = 1.0200 kg
kg solvent = kg solution – kg solute = kg solution – kg NaCl
But kg NaCl = 1.20 mol (58.5 g/mol)(1 kg/1000 g) = 0.0702 kg
So, kg solvent = 1.0200 kg - 0.0702 kg = 0.9498 kg;
Hence, m = 1.20 mol/.9498 kg = 1.26 m
Orientation
Contact information
What is a solution.
Enthalpy change in dissolution process
Effect of temperature on solubility of solids and gases.
Today:
1. 
2. 
3. 
a) 
b) 
c) 
d) 
Henry’s Law
Units of concentration
Colligative properties:
Freezing point depression
Boiling point elevation
Vapor pressure lowering
Osmotic pressure
Concentration units
Review of concentration units:
1 molar (M) = 1 mole solute/ liter soln
1 molal (m) = 1 mole solute/ kg solvent
1 % (m/v) = (1 g solute/g soln )x 100%
(same as =g solute/100g soln)
1 ppm = (1 g solute/g soln )x 106
(same as =g solute/106g soln)
(for dilute aqueous solutions: 1 ppm = 1mg solute/L soln)
Mole fraction, X1 = #mol 1 / total moles = n1 /(Σni)
Example Problem #2
What is the ppm Ca2+ in a 2.50 L solution containing 1.20g CaCl2?
(MW’s: Ca=40.1 g/mol; CaCl2 = 111.1 g/mol)
Solution: recall that ppm ≈ 1 mg/L
g Ca2+ = 1.20g CaCl2 (1mol)(1 mol Ca2+)(40.1g Ca2+) = 0.433 g
111.1g mol CaCl2 mol Ca2+
ppm Ca2+ = 0.433g Ca2+ (1000 mg) = 173 mg Ca2+ = 173 ppm Ca2+
2.50L
g
L
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Colligative properties
Boiling point elevation
Colligative properties = physical properties of solutions
which depend upon the concentration of solute
particles.
4 main examples we are concerned with in this class:
a)  Boiling point elevation
b)  Freezing point depression
c)  Vapor Pressure lowering
d)  Osmotic pressure
∆Tb = ikbmsolute
Where ∆Tb = change in boiling point
kb = molal boiling pt elev’n const.
m = molality of solute particles
i = Van’t Hoff factor (accounts for #ions)
So, to get boiling pt, Tb: Tb = Tb° + ∆Tb
For water, kb = 0.512 K/m ; Tb° = 100.°C
Checking your understanding:
a) Is kb a constant of the solvent or the solute?
kb is a constant of the solvent.
b) What’s i =? for NaCl solution?
i = 2 for NaCl solution because there are 2 moles of ions per
mole of NaCl, namely 1 mole each of Na+ and Cl-.
Freezing point depression
∆Tf = ikfmsolute
Where ∆Tf = change in freezing pt;
i = van’t Hoff factor (remember?).
kf = molal freezing pt depression constant
and, m = molality of solute particles
To get Tf : Tf = T°f - ∆Tf
For water, kf = 1.853 K/m
Sample probs continued…
What is MW of solute D if a sol’n of 10.0g of D in 20 g of
solvent A has Tf = 6.0°C? (kf= 2.0 °Cm-1)?
Answer: first determine m: from ΔTf = kfm; m= ΔTf/kf
m = (10.0-6.0)°C /2.0°Cm-1 = 2.0 m
Since m =2.0 m = 2.0 mol D/kg A = 10.0g D/MW / 0.020kg A
Solve for MW: MW = (10.0g D)/ (2.0 mol D/kg A )(0.020kg A)
MW = 250 g D/mol D
Sample problems
Solvent A normally freezes at 10.0°C. A 3.0 m
solution of B (nonionic) in A freezes at 4.0°C.
What is kf=?
Answer: kf = ΔTf/m = (10.0-4.0)°C/3.0m = 2.0 °Cm-1
What is the freezing point, Tf, of a 2.0 m solution of
a solute D (nonionic) in A?
Answer: ΔT = kfm = 2.0 °Cm-1 (2.0m) = 4.0°C
So Tf = 10°C - 4°C = 6°C
Vapor pressure lowering
P1 = X1P1° (Raoult’s Law)
Where P1 = vapor pressure of a solvent (1) in a
liquid solution.
X1 = mole fraction of that solvent in the solution
P1° = vapor P of solvent when it is pure solvent.
(N.b. for a solution, X1 < 1 , and so, P1 < P1° )
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Sample problem…
Osmotic pressure
P1 = X1P1° (Raoult’s Law)
π = iMRT (across semiperm.memb.)
Where π = osmotic pressure (π = pi = Greek p)
i = “ van’t Hoff factor”, a correction factor to
account for actual # particles in solution
M = molarity of solute
R = 0.0821 atmL/molK
T = temperature in K
Consider 2 liquids, A and B. The vapor pressures of
pure A and pure B are 25.0 torr and 45.0 torr
respectively. What is the total vapor pressure
inside an airtight chamber containing only 2
liquids, 1.5 moles of A and 2.5 moles of B?
Answer: Ptotal = PA + PB = XAPA + XBPB
Where: XA = 1.5/(1.5+2.5)=0.375 and XB=2.5/(1.5+2.5)=0.625
Ptot = (0.375)(25.0 torr) + (0.625)(45.0 torr)
=9.38 torr + 28.1 torr = 37.5 torr
Osmotic pressure problem
The equation is: π = i c R T where c = moles/L (molarity)
(can be derived from pV=nRT => p =(n/V)RT )
Example. A solution prepared by adding 50. g of solute to
make 1.0 L solution at 300 K has π = .821 atm. What is the
MW of the solute (assuming it is a nonelectrolyte)?
g solute/MWsolute
RTi
Vsolution
g solute
RTi
or, MW =
"Vsolution
atm # L
50.g
(0.0821
)(300K)(1)
MW =
mol # K
(.821atm)(1.0L)
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MW =1.50x10 g/mol
! = cRTi =
Red blood cells and π
Red blood cells (RBC’s) are “semipermeable bags”, which
must maintain the same concentration within and without, or
else! (i.e. solution surrounding it must be isotonic)
Water leaves cell: crenation
If solution is more concentrated (hypertonic) than the internal
concentration, what happens?
If solution is less concentrated (hypotonic) than the internal
concentration, what happens?
Water enters cell:hemolysis
Reverse osmosis:
desalination
If the applied pressure is
high enough (and the
membrane robust enough!),
it is possible to reverse the
8,4,%0,)3$+1
1$/8,4
Applied P=100
atm > π=25 atm
Example: desalination
plants - common in middle
east, Florida,…
Pure H2O
Salt water
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