Transcript Measuring Solutions

Solutions
Occur in all phases
 The
solvent does the dissolving.
 The solute is dissolved.
 There are examples of all types of
solvents dissolving all types of
solvent.
 We will focus on aqueous solutions.
Energy of Making Solutions
of solution ( DHsoln ) is the energy
change for making a solution.
 Most easily understood if broken into
steps.
 1.Break apart solvent
 2.Break apart solute
 3. Mixing solvent and solute
 Heat
1. Break apart Solvent
 Have
to overcome attractive forces.
DH1 >0
2. Break apart Solute.
 Have
to overcome attractive forces.
DH2 >0
3. Mixing solvent and solute
 DH3 depends
on what you are mixing.
 If molecules can attract each other
DH3 is large and negative.
 Molecules can’t attract- DH3 is small
and negative.
 This explains the rule “Like dissolves
Like”
of DH3 helps determine whether
Solute and
a solution will form
 Size
Solvent
E
n
e
r
g
y
Solvent
Reactants
DH1
DH2
DH3
Solution
DH3
Solution
Ways of Measuring
 Molarity
= moles of solute
Liters of solution
 % mass = Mass of solute
x 100
Mass of solution
 Mole fraction of component A
cA =
nA
nA + nB
Ways of Measuring
 Molality
=
moles of solute
Kilograms of solvent
 Molality
is abbreviated m
- read but don’t focus on it.
 It is molarity x number of active
pieces
 Normality
Vapor Pressure of Solutions
A
nonvolatile solute lowers the
vapor pressure of the solution.
 The
molecules of the solvent
must overcome the force of
both the other solvent
molecules and the
solute molecules.
Raoult’s Law:
Psoln
 Vapor
= csolvent x Psolvent
pressure of the solution =
mole fraction of solvent x
vapor pressure of the pure solvent
 Applies only to an ideal solution
where the solute doesn’t contribute
to the vapor pressure.
 Water
has a higher vapor
pressure than a solution
Aqueous
Solution
Pure water
 Water
evaporates faster from for
water than solution
Aqueous
Solution
Pure water
 The
water condenses faster in the
solution so it should all end up
there.
Aqueous
Solution
Pure water
Practice Problem
A
solution of cyclopentane with a
nonvolatile compound has vapor
pressure of 211 torr. If vapor
pressure of the pure liquid is 313
torr, what is the mole fraction of the
cyclopentane?
 Psoln = XcpPcp
 211 torr = Xcp (313 torr)
 .674
Try one on your own
 Determine
the vapor pressure of a
solution at 25 C that has 45 grams of
C6H12O6, glucose, dissolved in 72
grams of H2O. The vapor pressure of
pure water at 25 C is 23.8 torr.
 Psolution= Xsolvent Psolvent
 Psolution = .941(23.8 torr)
 Psolution = 22.4 torr
Ideal solutions
 Liquid-liquid
solutions where both are
volatile.
 Modify Raoult’s Law to
PA + PB = cAPA0 + cBPB0
 Ptotal = vapor pressure of mixture
 PA0= vapor pressure of pure A
 If this equation works then the solution
is ideal.
 Ptotal =
Vapor Pressure
P of pure A
P of pure B
χA
χb
Deviations
 If
solvent has a strong affinity for
solute (H bonding).
 Lowers solvent’s ability to escape.
 Lower vapor pressure than expected.
 Negative deviation from Raoult’s law.
 DHsoln is large and negative
exothermic.
 Endothermic DHsoln indicates positive
deviation.
Vapor Pressure
Positive deviationsWeak attraction between
solute and solvent
Positive ΔHsoln
χA
χb
Vapor Pressure
Negative deviationsStrong attraction between
solute and solvent
Negative ΔHsoln
χA
χb
Problem #1
 The
vapor pressure of a solution
containing 53.6g of glycerin C3H8O3
in133.7g ethanol C2H5OH is 113 torr at
40C. Calculate the vapor pressure of
pure ethanol at 40C assuming that the
glycerin is a non volatile, nonelectrolyte
solute in ethanol.
Answer to #1
Psoln = Xeth Peth
113torr = 2.90mol/3.48mol (Peth)
135.6 torr = Peth
Problem #2
 At
a certain temperature, the vapor
pressure of pure benzene C6H6 is
0.930atm. A solution was prepared by
dissolving 10.0g of a nondissociating,
nonvolatile solute in 78.11g of benzene
at that temperature. The vapor pressure
was found to be 0.900atm. Assuming
the solution behave ideally, determine
the molar mass of the solute.
Answer #2
Psoln = XbenzenePbenzene
.900atm = Xbenzene (.930atm)
Xbenzene = .9677
Xsolute = .0323
MM = 10.0g/.0323mol = 310g/mol
Problem #3
A solution of NaCl in water has a vapor
pressure of 19.6 torr at 25C. What is the
mol fraction of solute particle in this
solution if the vapor pressure of water is
23.8 torr at 25C?
Answer #3
Psoln = XwaterPwater
19.6torr = Xwater(23.8torr)
.824 = Xwater therefore Xsolute = .176
Problem #3
For the same problem as #3:
What is the vapor pressure of the solution
at 45C if the vapor pressure of water is
71.9 torr at 45C?
Answer #3
Psoln = .824(71.9torr)
Psoln = 59.2 torr
Problem #4
A solution is made from 0.0300mol
CH2Cl2 and 0.0500mol CH2Br2 at 25C.
Assuming that the solution is ideal,
calculate the % composition of the vapor
at 25C.
PCH2Cl2 = 133 torr
PCH2Br2 = 11.4 torr
Answer #4
Psoln = XCH2Cl2 P + XCH2Br2 P
Psoln = (.03/.08)(133torr) + (.05/.08)(11.4 torr)
Psoln = 57.0 torr
XCH2Cl2 = 49.9 / 57 = .875 = 87.5%
XCH2Br2 = 7.13 / 57 = .125 = 12.5%
Colligative Properties
 Because
dissolved particles affect
vapor pressure - they affect phase
changes.
 Colligative properties depend only
on the number - not the kind of
solute particles present
 Useful for determining molar mass
Water
1 atm
Vapor Pressure
of pure water
Vapor Pressure
of solution
1 atm
Freezing and
boiling points
of solvent
1 atm
Freezing and boiling
points of solution
1 atm
DTf
DTb
Boiling point Elevation
 Because
a non-volatile solute lowers
the vapor pressure it raises the boiling
point.
 The equation is: DT = Kbmsolute
 DT
is the change in the boiling point
 Kb is a constant determined by the
solvent.
 msolute is the molality of the solute
Freezing point Depression
 Because
a non-volatile solute lowers the
vapor pressure of the solution it lowers the
freezing point. Freezing occurs when Pv (l)
= Pv(s). Because the Pv(soln) is reduced,
the temp must be lowered to form a solid.
 The equation is: DT = -Kfmsolute
 DT is the change in the freezing point
 Kf is a constant determined by the solvent
 msolute is the molality of the solute
Electrolytes in solution
 Since
colligative properties only
depend on the number of molecules.
 Ionic compounds should have a
bigger effect.
 When they dissolve they dissociate.
 Individual Na and Cl ions fall apart.
 1 mole of NaCl makes 2 moles of ions.
 1mole Al(NO3)3 makes 4 moles ions.
 Electrolytes
have a bigger impact on
on melting and freezing points per
mole because they make more pieces.
 Relationship is expressed using the
van’t Hoff factor i
i = Moles of particles in solution
Moles of solute dissolved
 The expected value can be determined
from the formula of the compound.
 The
actual value is usually less
because
 At any given instant some of the ions
in solution will be paired up.
 Ion pairing increases with
concentration.
 i decreases with increasing
concentration.
 We can change our formulas to
DT = iKm
Problem #1
A 2.00 gram sample of a large
biomolecule was dissolved in 15.0 grams
of CCl4 . The boiling point of this solution
was found to be 77.85C. Calculate the
molar mass of the biomolecule.
Kb = 5.03C kg / mol
BPCCl4 = 76.5C
Answer #1
DT = Kb m
(77.85 – 76.5C) = 5.03C Kg/mol x m
.268mol/Kg = m
.268 mol/Kg = Xmol / .015kg
= .00402 mol
2.00g / .00402mol = 498g/mol
Problem #2
The freezing point of t-butanol is 25.5C
and Kf = 9.1C Kg/mol. Usually t-butanol
absorbs water on exposure to air. If the
freezing point of a 10.0 gram sample of tbutanol is 24.59C, how many grams of
water are present in the sample?
Answer #2
DT = -Kf m
(24.59 – 25.5C) = -9.1 C Kg/mol x m
.1 mol/Kg = m
.1 m = X mol H2O /.01kg = .001mol H2O
.018g H2O
Problem #3
Calculate the freezing point and the
boiling point for each:
.050m MgCl2
.050m FeCl3
Kf = 1.86C Kg/mol
Kb = 0.51C Kg/mol
Predict which will have higher BP and
which will have the lower FP
Answer #3
Dissociate each:
MgCl2
Mg2+ + 2ClFeCl3
Fe3+ + 3Cl-
i= 2.7
i= 3.4
MgCl2 Tf = -.25C Tb = 100.069C
FeCl3 Tf = -.32C Tb = 100.087C