Transcript Properties of Solutions - Dallas School District

Ch 13: Solutions
Role of Disorder in
Solutions
• Disorder (Entropy) is a factor
• Solutions mix to form maximum
disorder
Two Ways to Form Solutions
1. Physical Dissolving (Solvation)
• NaCl(s)  Na+(aq) + Cl-(aq)
• C12H22O11(s)  C12H22O11(aq)
• Particles are surrounded by solvent molecules
• Can evaporate water/solvent to get original
compound back
Types of Reactions
2. Chemical reaction
• Ni(s) + 2HCl(aq)  NiCl2(aq) + H2(g)
• Evaporating solvent gives the products
Solubility – Maximum amount of a solute that can
dissolve in 100 mL of a solution
Ex: NaCl
35.7 g/100mL
– Saturated solution – Contains the max. amount of
solute with some undissolved solid,
– Unsaturated – more solute will dissolve.
– Supersaturated – More than the max is dissolved by
heating and slowly cooling.
Like Dissolves Like: Miscibility
• Polar dissolves polar (dipole-dipole Forces) and
ionic (ion:dipole)
– Water and Ammonia
• Non-Polar dissolves non-polar (London Forces)
– Soap and grease
Would acetone (shown below) dissolve in water?
:O:
||
CH3CCH3
Acetone
Using you knowledge of “like dissolves like”, explain
the following trends in solubility.
Alcohol
CH3OH
Solubility in H2O
(mol/100 g H2O at 20oC)
∞
CH3CH2OH
∞
CH3CH2CH2OH
∞
CH3CH2CH2CH2OH
0.11
CH3CH2CH2CH2CH2OH
0.030
CH3CH2CH2CH2CH2CH2OH
0.0058
CH3CH2CH2CH2CH2CH2CH2OH
0.0008
Pressure Effects
• Solubilty of a gas increases with pressure of gas
over the liquid (soda bottle)
• Henry’s Law
Sgas = kPgas
Henry’s Law
The Henry’s law constant for CO2 is 0.031 mol/Latm.
a. Calculate the concentration of CO2 in a soda
bottle pressurized to 4.00 atm of CO2.
b. After the bottle has been opened, the
concentration drops to 9.3 X 10-6 M. Calculate
the partial pressure of CO2 over the soda.
Temperature Effects
• Solubility of most solids
increases with temperature
• Solubility of most gases
decreases with temperature
(warm soda)
– Warm water is deoxygenated
– Problem with thermal pollution of
lakes
Ways of Expressing Concentration
Mass % = mass of compound in soln X
total mass of soln
Parts Per Million
ppm = mass of component X 106
total mass of soln
100
Concentration: Ex 1
13.5 g of C6H12O6 is dissolved in 0.100 kg of water.
Calculate the mass percentage.
mass % = 13.5 g
X 100 = 11.9%
(100 g + 13.5 g)
Concentration: Ex 2
A 2.5 g sample of groundwater is found to contain
5.4 mg of Zn2+. What is the concentration of the
Zn2+ ion in ppm.
Concentration: Ex 2
A 2.5 g sample of groundwater is found to contain
5.4 mg of Zn2+. What is the concentration of the
Zn2+ ion in ppm.
5.4 mg | 1X10-6g =
5.4 X 10-6 g
| 1 mg
ppm = mass of component X 106
total mass of soln
ppm = 5.4 X 10-6 g X 106 = 2.2 ppm
2.5g
Concentration: Ex 3
Calculate the mass percentage of NaCl in a solution
containing 1.50 g of NaCl in 50.0 g of water.
ANS: 2.91 %
Concentration: Ex 4
Bleach is 3.62 % NaOCl. What mass of NaOCl is
contained in 2500 g of bleach?
ANS: 90.5 g NaOCl
Mole Fraction
Mole Fraction
X = moles of component
total moles of all components
What is the mole fraction of HCl if 36.5 grams is
dissolved in 144 grams of water?
ANS: 0.111
Molality
Molality = moles of solute
kilograms of solvent
Why not use Molarity?
• Molarity varies with temperature
• Total volume of a solution changes with
temperature (liquid expands)
• Mass does not change with temperature
Molality: Ex 1
A solution is made by dissolving 4.35 grams of
C6H12O6 in 25.0 mL of water. Calculate the
molality of the glucose.
4.35 g
1 mol
180.2 g
Molality = 0.0241 mol
0.0250 kg
=
0.0241 mol
=
0.964 m
Molality: Ex 2
Calculate the molality of a solution made by
dissolving 36.5g C10H8 in 425 grams of toluene
(solvent).
ANS: 0.671 m
Molality: Ex 3
A solution of HCl contains 36 percent HCl by mass.
Calculate the mole fraction and molality of HCl.
Pretend 100 g
36 g HCl
64 g H2O
36 g HCl
1 mol HCl =
36.5 g HCl
64 g H2O 1 mol H2O =
18 g H2O
0.99 mol HCl
3.6 mol HCl
XHCl = 0.99 mol HCl
= 0.22
0.99 mol + 3.6mol
Molality = 0.99 mol HCl
0.064 kg H2O
= 15 m
Molality: Ex 4
A commercial bleach solution contains 3.62 percent
NaOCl by mass. Calculate the mole fraction and
molality of NaOCl.
ANS: XNaOCl = 0.00900, 0.505 m
Molality: Ex 5
The density of a solution of 5.0 g of toluene (C7H8)
and 225 g of benzene (C6H6) is 0.876 g/mL.
Calculate the molality and molarity of the
toluene.
Molality: Ex 5
The density of a solution of 5.0 g of toluene (C7H8)
and 225 g of benzene (C6H6) is 0.876 g/mL.
Calculate the molality and molarity of the
solution.
Molality
5.0 g 1 mol
=
0.054 mol
92.0 g
m = 0.054 mol/ 0.225 kg = 0.24 m
Molarity
D = mass/V
V = mass/D
V = 230 g
= 263 mL
0.876 g/ml
M = 0.054 mol = 0.21 M
0.263 L
Molality: Ex 6
A solution containing equal masses of glycerol
(C3H8O3) and water has a density of 1.10 g/mL.
Calculate
a) molality (10.9 m)
b) mole fraction (XC3H8O3 = 0.163)
c) molarity of glycerol in the solution (5.97 M)
Colligative Properties: Vapor Pressure
Lowering
Non-volatile solutes lower the
vapor pressure of the solvent
Raoult’s law
PA = XAPoA
PA = Vapor pressure
XA = Mole fraction of solvent
PoA = Pressure of pure solvent
Raoult’s Law: Ex 1
What is the vapor pressure of a solution made by
adding 50.0 mL of glycerin (C3H8O3) to 500.0 mL of
water? The density of glycerin is 1.26 g/mL and
the vapor pressure of pure water is 23.8 torr.
MassC3H8O3 = (50.0 mL)(1.26 g/mL ) = 63.0 g
MolesC3H8O3 = 63.0 g/92.1 g/mol = 0.684 mol
MolesH2O = 500.0 g/18 g/mol = 27.8 mol
XH2O =
27.8 mol
= 0.976
(27.8 mol + 0.684 mol)
PA = XAPoA
PA = (0.976)(23.8 torr) = 23.2 torr
Raoult’s Law: Ex 2
The vapor pressure of water at 110oC is 1070 torr. A
solution of ethylene glycol and water has a vapor
pressure of 1 atm at 110oC. What is the mole
fraction of ethylene glycol in the solution?
ANS: 0.290
Colligative Properties: Boiling Point
Elevation
• Non-volatile solute
raises the boiling
point of a solution
• Shifts the phase
diagram
• The pressure of the
solution reaches
atmospheric pressure
at a higher temp.
DTb = iKbm
Colligative Properties: Freezing Point
Depression
• Solutions freeze at a lower temperature than pure
solvent
• Salt water freezes lower (-2oC) than distilled water
(0oC)
DTf = iKfm
i = Van’t Hoff factor
m = molality of the nonvolatile solute
• The more ions produced, the greater the freezing
point depression or boiling point elevation
• C12H22O11
(i=1)
• NaCl = Produces two ions (i=2)
• CaCl2 = Produces three ions (i=3)
Pure Solvent
Solution
Boiling Point
Boiling Point
Freezing Point
Freezing Point
Colligative: Ex 1
Ethylene Glycol, C2H6O2, is used in antifreeze. What
will be the freezing and boiling point of a 25.0
mass percent solution of ethylene glycol and
water?
Pretend 100 grams of solution
25 grams of C2H6O2
75 grams of H2O (0.075 kg)
25 grams of C2H6O2 = 0.403 moles
m = 0.403 moles
0.075 kg H2O
=
5.37 m
o
DTb = iKbm = (1)(0.52oC/m)(5.37 m) = 2.8oC
DTf = iKfm = (1)( 1.86oC/m)(5.37 m) =10.0oC
Boiling Point = 102.8oC
Freezing Point = -10.0oC
Colligative: Ex 2
Calculate the freezing point of a solution containing
0.600 kg of CHCl3 and 42.0 g of C10H18O. Kf for
CHCl3 is 4.68oC/m and the normal freezing point is
-63.5 oC.
ANS: -65.6oC
Colligative: Ex 3
Rank the following aqueous solutions in order of
their expected freezing points:
0.050 m CaCl2
0.15 m NaCl
0.10 m HCl
0.050 m HC2H3O2 (acetic acid)
0.10 m C12H22O11 (sugar)
0.050 m CaCl2
0.15 m NaCl
0.10 m HCl
0.050 m HC2H3O2
0.10 m C12H22O11
(0.15 m in particles)
(0.30 m in particles)
(0.20 m in particles)
(just above 0.05 m)
(0.10 m in particles)
Lowest FP
Highest FP
NaCl < HCl < CaCl2 < C12H22O11 < HC2H3O2
Colligative: Ex 4
Rank the following in order of the increase in boiling
point that they will produce in 1 kg of water
1 mol Co(NO3)2
2 mol KCl
3 mol C2H6O2 (a very, very weak electrolyte(acidic))
1 mol Co(NO3)2
2 mol KCl
3 mol C2H6O2
(3 mol particles)
(4 mol of particles)
(3+ mol of particles)
Lowest BP
Highest BP
Co(NO3)2 < C2H6O2 < KCl
Freezing Pt Depression: Ex 5
What would be the molality of salt water if it freezes
at 0 oF? Kf = 1.86 oC/m.
ANS: 4.78 m
Colligative Properties: Osmotic Pressure
• Osmosis – movement of solvent from high
concentration to low concentration
• semipermeable membrane – allows to passage of
some particles but not others
Cucumber
Skin cell after
in salt water
soaking in a tub
• Note that solvent moves both
ways
• Solute too large to pass
through membrane
• Net movement is to try to
dilute the side with solutes
• Osmotic Pressure (p)
– pressure required
to prevent osmosis
PV = inRT
p V = inRT
p = inRT
V
p = iMRT
M = molarity
Osmotic Pressure: Ex 1
The average osmotic pressure of blood is 7.7 atm at
25oC. What concentration of glucose will be
isotonic with blood?
(0.31 M)
Osmotic Pressure: Ex 2
What is the osmotic pressure at 20oC of a 0.0020 M
sucrose, C12H22O11, solution? Express your answer
both in atmosphere and in torr.
ANS; 0.048 atm, 37 torr
Molar Mass: Ex 1
A solution of an unknown nonelectrolyte was
prepared by dissolving 0.250 g in 40.0 g of CCl4.
The boiling point of the resulting solution was
0.357oC higher than that of the pure solvent. Kb
for CCl4 is 5.02 oC/m. Calculate the molar mass of
the unknown.
DTb = iKbm
m = DTb /iKb
m = (0.357oC)/(1 X 5.02 oC/m) = 0.0711 m
m = mol of solute
kilograms of solvent
molsolute = (m)(kg of solvent)
molsolute = (0.0711 m)(0.0400 kg) = 0.00284 mol
Molar Mass =
0.250 g
0.00284 mol
= 88.0 g/mol
Molar Mass: Ex 2
Camphor, C10H16O, melts at 179.8oC and has a Kf of
40.0 oC/m. When 0.186 g of an unknown
substance is dissolved in 22.01 g of liquid
camphor, the freezing point is 176.7oC. What is
the molar mass of the solute?
ANS: 110 g/mol
Molar Mass: Ex 3
A solution contains 3.50 mg of protein dissolved in
water to form 5.00 mL of a solution. The osmotic
pressure at 25oC was found to be 1.54 torr.
Calculate the molar mass of the protein.
1.54 torr 1 atm
760 torr
=
0.00203 atm
p = iMRT
M = p/iRT
M = 0.00203 atm
= 8.28X10-5 M
(1)(0.0821 L-atm/mol-K)(298)
M = moles
liter
moles = (M)(liters) = (8.28X10-5 M)(0.00500L)
moles = 4.14 X 10-7 mol
Molar mass =
3.50 X 10-3 g
= 8454 g/mol
4.14 X 10-7 mol
Molar Mass: Ex 4
A 2.05 g sample of a plastic was dissolved in enough
toluene to form 100 mL of solution. The osmotic
pressure of this solution is 1.21 kPa at 25oC.
Calculate the molar mass of the plastic. (1 atm =
101.325 kPa).
ANS: 42,000 g/mol
Colloids
Solutions/
Homogeneous
Ions/Molecular size solute
particles
Never separate
Colloids
Suspensions/
Heterogeneous
Mixture
Medium particles (10 to 2000 Å) Larger particles (like dirt in
water)
2 or more separate phases
Separates quickly
• Examples
– Fog
– Smoke
– Whipped Cream
– Milk
• Tyndall effect – scattering of light
Stabilization of Colloids
• Hydrophobic/hydrophilic imf
– Biomolecules
Emulsifying agents
– Soap
– Sodium stearate(used to digest fats)
6. Larger Noble gases have greater London Forces
(greater dipole: induced dipole forces)
16.a) Ion:Dipole
b) Dipole: induced dipole
c) Hydrogen bonding
(weakest) b < c < a (strongest)
30.a) glucose (OH’s allow h-bonding)
b) sodium propionate (has an ion)
c) HCl (small and polar)
34.S = kP
k = S/P = 1.38 X 10-3M/0.21 atm
k =6.57 X 10-3 M/atm
Partial Pressure of O2 at the higher elevation:
650 torr = 0.855 atm
PO2 = (0.855 atm)(0.21)
PO2 = 0.180 atm
S = kP
S=(6.57 X 10-3 )(0.180 atm)
S = 1.18 X 10-3 M
36a) 7.2% I2
b) 7.9 ppm Sr2+
38a) 0.0285
b) 5.66%
c) 0.638 m
40a) 0.125 M b) 0.140 M
c) 0.630 M
42a) 4.34 m
b) 3.1 g S8
44a) 27.7%
b) 0.0377
c) 2.18 m
d) 1.92 M
46a) 0.0439
b) 0.498 m
c) 0.417 M
48a) 0.278 mol b) 6.25X10-5 mol c) 0.00329 mol
50a) 21.8 g
b) 7.7 g/112.3g c) 209 g
d) 11 mL of 6.0 M HCl
52) 15 M NH3
62.a) 222 torr
b) 150 g
64. a) 0.75
b) 0.47
66. 10% sucrose < 10% glucose < 10% NaNO3
68. 0.030 m phenol < KBr = 0.040 m glycerin
70.a) -115.2, 78.8
b) -78, 72.4
c) -9.3, 102.6
72.-18 oC
74.2.8 atm
76.180 g/mol
78. 380 g/mol
Write Net Ionic Equations for:
MgCO3(s) + HNO3(aq) 
H2SO4(aq) + 2KOH(aq) 
NaHCO3(aq) + HCl(aq) 
A solution of nickel(II)sulfate is stored in a zinc
coated bucket.
a. Write the net ionic reaction that occurs.
b. Suppose the nickel(II)sulfate was stored in a
copper bucket. Would this be a better choice?
c. Write the net ionic reaction that occurs between
nickel(II)sulfate and barium nitrate.
1.
2.
3.
4.
5.
Lots of ice, little water
Stir and take temperature (~2 min)
Add medicine cup of rock salt
Stir and take temperature (~2 min)
Clean Up
1. Rinse Foam cup, medicine cup, and thermometer
2. Place in drying rack
A 80.5 g sample of ascorbic acid (C6H8O6) is dissolved in 210.0 grams of
water. The resulting solution has a density of 1.22 g/mL.
a) Calculate the molarity of the solution
b) Calculate the molality of the solution.
c) Calculate the freezing point of the solution. (Assume ascorbic acid
is a very weak electrolyte, Kf = 1.86 oC/m)
d) Would you expect the solution to have a higher or lower vapor
pressure than pure water?
e) Calculate the vapor pressure of the solution if the vapor pressure of
pure water is 17.5 torr.
Benzene (C6H6) freezes at 5.50 oC and has a Kf of 5.12 oC/m. 50.0
grams of an unknown solute is dissolved in 100.0 g of benzene. The
resulting solution freezes at -8.72 oC.
a. Calculate the molality of the solution. (2.78 m)
b. Calculate the molar mass of the solute. (180)
c. Why is benzene a good choice for this experiment?
d. Comment on the polarity of the unknown molecule.
e. What is the hybridization of the carbon atoms in benzene?