Week 3 Sections 13.3- 13.5
Factors that FAVOR solubility:
13.3
1. Strong solute-solvent interactions
Factors Affecting Solubility
Solute-Solvent Interactions
Pressure Effects
Temperature Effects
13.4
Ways of Expressing Concentration
Mass Percentage, ppm, and ppb
Mole Fraction, Molarity, and Molality
Conversion of Concentration Units
13.5
Colligative Properties
Lowering the Vapor Pressure
Boiling-Point Elevation
Freezing-Point Depression
Osmosis
Determination of Molar Mass
2. Weak solute-solute interactions
3. Weak solvent-solvent interactions
More often we’ll settle for the solute-solvent interactions
being similar to the solute-solute and solvent-solvent
interactions.
A general rule: Like dissolves like.
i.e. polar and polar
non-polar and non-polar
[13.6 Colloids]
Realize there is an inherent tendency for the two isolated materials to form solution,
regardless of the energetics!!! This represents an entropy factor.
Consider the Solubilities (mol/100g water) of Alcohols
CH3OH
CH3CH2OH
CH3(CH2)2OH
CH3(CH2)3OH
CH3(CH2)4OH
CH3(CH2)5OH
CH3(CH2)6OH
∞
∞
∞
0.11
0.030
0.0058
0.0008
Fig 13.12 Structure of glucose—note red O atoms in OH groups
C6H12O6 or C5H5OCH2(OH)5
1
Fig. 25.26 Sucrose and Lactose Molecules.
Olestra has fatty acids at each of the OH in sucrose.
Fig 25.27 The Starch Molecule, note the blue atoms
Fig 25.28 Cellulose Molecule, note the blue atoms
Table 13.2 Gas solubilities in water
at 20 oC with 1 atm gas pressure (Table 13.2)
Solubility/Molarity
Fig 25.29 Starch (a) and Cellulose (b) in 3-D representation
He
0.40 x 10-3
N2
0.69 x 10-3
CO
1.04 x 10-3
O2
1.38 x 10-3
Ar
1.50 x 10-3
Kr
2.79 x 10-3
CO2
3.1 x 10-2
NH3
~ 53
2
Table 13.2 Gas solubilities in water
at 20 oC with 1 atm gas pressure (Table 13.2)
Solubility/Molarity
Fig 13.14 Henry’s Law,
Cg = k Pg
See Table 13.2 for k in M/atm
He
0.40 x 10-3
N2
0.69 x 10-3
CO
1.04 x 10-3
O2
1.38 x 10-3
Ar
1.50 x 10-3
Kr
2.79 x 10-3
CO2
3.1 x 10-2
NH3
~ 53
Consider N2 dissolved in water at 4.0 atm.
Note k = 0.69 x 10-3 mol/L-atm
Also in Table 13.2
Cg = k Pg
= (0.69 x 10-3 mol/L-atm)(4.0 atm)
= 2.76 x 10-3 mol/L
at normal atmospheric conditions, however, Pg = 0.78 atm
Cg = (0.69 x 10-3 mol/L-atm)(0.78 atm)
= 0.538 x 10-3 mol/L
Note that (2.76 - 0.54) x 10-3 mol/L = 2.22 x 10-3 mol/l
Thus for 1.0 L of water, 0.0022 mol of nitrogen would
be released = 0.0022 x 22.4L = 0.049 L = 49 mL !
To read about nitrogen narcosis, see http://www.gulftel.com/~scubadoc/narked.html
and http://www.diversalertnetwork.org/medical/articles/index.asp
Fig 13.18 Temperature Effect on Solubility of Gases.
Demonstrations
The ammonia fountain
Add acetone to a sat’d CuSO4 sol’n
Heat water and calcium acetate
Heat water and potassium nitrate
Fig 13.17 Effect of T on Solubilities of Ionic Cmpds.
Are these exothermic or endothermic processes?
3
Prepare 0.500 L of a 0.100 M aqueous solution of KHCO3 (MW = 100.12 g)
Ways of expressing concentration:
a) percent, ppm, ppb
usually m/m
b) mole fraction = XA , XB sum of Xi = 1
c) molarity = M or mol/Lsolution
depends on T and density of soln
preparation requires dilution
d) molality = m or mol/kgsolvent
independent of T
easily prepared
Prepare 0.500 L of a 0.100 M aqueous solution of KHCO3 (MW = 100.12 g)
(0.100 mol KHCO3/L of soln)(0.500 L) = 0.0500 mol KHCO3
Note:
molarity = M = n/V
therefore
n = MV
(a) Prepare 0.500 L of a 0.100 M aqueous solution of KHCO3 (MW = 100.12 g)
requires 5.01 g of KHCO3 dissolved and diluted to 0.500 L
(b) Use this solution as a ‘stock’ solution to prepare a final solution of
0.0400 M concentration. What is the final volume of this solution?
Prepare 0.500 L of a 0.100 M aqueous solution of KHCO3 (MW = 100.12 g)
(0.100 mol KHCO3/L of soln)(0.500 L) = 0.0500 mol KHCO3
(0.0500 mol KHCO3)(100.12g/mol KCHCO3) = 5.01 g KHCO3
(a) Prepare 0.500 L of a 0.100 M aqueous solution of KHCO3 (MW = 100.12 g)
requires 5.01 g of KHCO3 dissolved and diluted to 0.500 L
(b) Use this solution as a ‘stock’ solution to prepare a final solution of
0.0400 M concentration. What is the final volume of this solution?
Since n = MV,
M1V1 = M2V2
so, (0.100 M)(0.500 L) = n = (0.0400 M)( V2 )
or
V2 =
M 1V1 (0.100 M )(0.500 L)
=
= 1.25L
M2
(0.0400M )
4
Consider a solution prepared by dissolving 22.4 g MgCl2
in 0.200 L of water. Assume the density of water is
1.000 g/cm3 and the density of the solution is
1.089 g/cm3.
Calculate
mole fraction
molarity
molality
A 9.386 M solution of H2SO4 has a density of 1.509 g/cm3.
Calculate
molality
% by mass
mole fraction of H2SO4
Week 3 Sections 13.3- 13.5
13.3
13.4
13.5
Factors Affecting Solubility
Solute-Solvent Interactions
Pressure Effects
Temperature Effects
Ways of Expressing Concentration
Mass Percentage, ppm, and ppb
Mole Fraction, Molarity, and Molality
Conversion of Concentration Units
Colligative Properties
Lowering the Vapor Pressure
Boiling-Point Elevation
Freezing-Point Depression
Osmosis
Determination of Molar Mass
Colligative Properties
Solution properties that depend only on the
total # of ‘particles’ present.
Vapor Pressure
Boiling Point
Freezing Point
Osmotic Pressure
[13.6 Colloids]
Note that VP of solution is lower than that of pure solvent.
5
Raoult’s Law
Vapor Pressure lowering
PA = XA PAo
PA = vapor pressure over solution
XA = mole fraction of component A (solvent)
PAo = vapor pressure of pure
component A (solvent)
also PA = (1 – XB) PAo
where XB = mol fraction of B (solute)
At 25 oC, the vapor pressure of benzene is 0.1252 atm,
i.e. PAo = 0.1252 atm. If 6.40 g of naphthalene (C10H8,
128.17 g/mol) is dissolved in 78.0 g of benzene (78.0 g/mol),
calculate the vapor pressure of benzene over the solution
At 25 oC, the vapor pressure of benzene is 0.1252 atm,
i.e. PAo = 0.1252 atm. If 6.40 g of naphthalene (C10H8,
128.17 g/mol) is dissolved in 78.0 g of benzene (78.0 g/mol),
calculate the vapor pressure of benzene over the solution
Raoult’s Law says
PA = XA PAo
At 25 oC, the vapor pressure of benzene is 0.1252 atm,
i.e. PAo = 0.1252 atm. If 6.40 g of naphthalene (C10H8,
128.17 g/mol) is dissolved in 78.0 g of benzene (78.0 g/mol),
calculate the vapor pressure of benzene over the solution
Raoult’s Law says
PA = XA PAo
therefore we need XA
therefore we need XA
nA = ?
nA = 78.0 g benzene (1 mol/78.0 g) = 1.00 mol benzene
nB = 6.40 g naph. (1 mol/128.17 g) = 0.0499 mol naph.
nB = ?
let A = benzene
and PA = Pbenzene = (1.00/1.0499)(0.125) = 0.1193 atm
note: ∆P = PA – PAo = XA PAo – PAo = PAo (Xa-1)
or ∆P = -XB PAo
6
Boiling Point Elevation and Freezing Point Depression
Boiling Point Elevation
∆Tb = Tb’ – Tb = Kb m
where m is the molal concentration
Freezing Point Depression
∆Tf = Tf’ – Tf = - Kf m
where m is the molal concentration.
[note the definition and the negative sign!!!]
for H2O, Kb = 0.052 oC/m and Kf = 1.86 oC/m
Consider a water solution which has 0.500 mol
of sucrose in 1.000 kg of water. Therefore it
has a concentration of 0.500 molal or 0.500 mol/kg.
(a) When 5.50 g of biphenyl (C12H10, 154.2 g/mol) is
dissolved in 100 g benzene (C6H6, 78.0 g/mol), the
BP increases by 0.903 oC. Calculate Kb for benzene.
recall Kb = 0.52 oC/m and Kf = 1.86 oC/m
What is the boiling point and freezing point
of this solution?
7
(a) When 5.50 g of biphenyl (C12H10, 154.2 g/mol) is
dissolved in 100 g benzene (C6H6, 78.0 g/mol), the
BP increases by 0.903 oC. Calculate Kb for benzene.
Î Kb = 2.53 K kg/mol
(b) When 6.30 g of an unknown hydrocarbon is dissolved
in 150.0 g of benzene, the BP of the solution increases
by 0.597 oC.
A sample of sea water contains the following in 1.000 L
of solution. Estimate the freezing point of this solution.
ClNa+ = 4.58 mol
SO42Mg2+ = 0.052 mol
HCO3Ca2+ = 0.010 mol
K+
= 0.010
Brneutral species = 0.001 mol
= 0.533 mol
= 0.028 mol
= 0.002 mol
= 0.001 mol
Sum of species = 1.095 mol
What is the MW of the unknown substance?
Consider Exercise 13.9
Consider Exercise 13.9
List the following aqueous solutions in increasing order
of their expected freezing points.
List the following aqueous solutions in increasing order
of their expected freezing points.
0.050 m CaCl2
0.15 m NaCl
0.10 m HCl
0.050 m HOAc
0.10 m C12H22O11
0.050 m CaCl2
0.15 m NaCl
0.10 m HCl
0.050 m HOAc
0.10 m C12H22O11
x3
x2
x2
x1
x1
=
=
=
=
=
0.150
0.30
0.20
0.050
0.10
These calculations assume total dissociation of the
salts and zero dissociation of the last two.
This effect of the dissociation of electrolytes is usually
taken into account through the van Hoff i factor.
∆Tb = i Kb m
Osmosis (get started)
where it is defined as
i=
∆Tf(actual)
Kf meffective
meffective
-------------= ------------= -----------Kf mideal
mideal
∆Tf(ideal)
In real systems, these i factors are NOT integers,
but rather fractions whose values depend on
concentration.
8
Osmotic
Pressure:
a fascinating
behavior.
Yet it is the
result of a
very simple
tendency to
equalize the
concentrations
of solutions.
The critical part is the membrane!!!
Osmotic Pressure
or π = (n/V) R T
πV=nRT
or π = M R T
π = ρ g h, where ρ = density of solution
g = 9.807 m s-2
h = height of column
π = ρ g h, where ρ = density of solution
g = 9.807 m s-2
h = height of column
If h = 0.17 m of a dilute aqueous soln with ρ = 1.00 g/cm3
π = [(1.00g/cm3)(10-3 kg/g)(106 cm3/m3)](9.807 m/s2)(0.17 m)
= 1.7 x 103 kg m-1 s-1 = 1.7 x 103 Pa
A chemist dissolves 2.00 g of protein in 0.100 L of
water. The observed osmotic pressure is 0.021 atm at
25 oC. What is the MW of the protein?
or = (1.7 x 103 Pa) / (1.013 x 105 Pa/atm) = 0.016 atm
9
Fractional Distillation – How it’s done
Real solutions do not behave ideally!
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