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Physical Properties of Solutions
Chapter 12
SolutionStoichiometry
end of Chapter 4
Problems: 12.12, 12.15, 12.16, 12.17, 12.18, 12.21, 12.22, 12.28, 12.36, 12.38, 12.51, 12.54, 12.55, 12.57, 12.60, 12.65, 12.76, 12.77, 12.1124.12, 4.60, 4.70, 4.72, 4.74, 4.77, 4.88
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12.1
A solution is a homogenous mixture of 2 or more substances
The solute is(are) the substance(s) present in the smaller amount(s)
The solvent is the substance present in the larger amount
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An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity.
A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity.
nonelectrolyte weak electrolyte strong electrolyte4.1
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A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature.
An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature.
A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature.
Sodium acetate crystals rapidly form when a seed crystal isadded to a supersaturated solution of sodium acetate.
12.1
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12.2
Three types of interactions in the solution process:• solvent-solvent interaction• solute-solute interaction• solvent-solute interaction
Hsoln = H1 + H2 + H3
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“like dissolves like”
Two substances with similar intermolecular forces are likely to be soluble in each other.
• non-polar molecules are soluble in non-polar solvents
CCl4 in C6H6
• polar molecules are soluble in polar solvents
C2H5OH in H2O
• ionic compounds are more soluble in polar solvents
NaCl in H2O or NH3 (l)
12.2
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Concentration UnitsThe concentration of a solution is the amount of solute present in a given quantity of solvent or solution.
Percent by Mass
% by mass = x 100%mass of solutemass of solute + mass of solvent
= x 100%mass of solutemass of solution
12.3
Mole Fraction (X)
XA = moles of A
sum of moles of all components
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Concentration Units Continued
M =moles of solute
liters of solution
Molarity (M)
Molality (m)
m =moles of solute
mass of solvent (kg)
12.3
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What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL?
m =moles of solute
mass of solvent (kg)M =
moles of solute
liters of solution
Assume 1 L of solution:5.86 moles ethanol = 270 g ethanol927 g of solution (1000 mL x 0.927 g/mL)
mass of solvent = mass of solution – mass of solute
= 927 g – 270 g = 657 g = 0.657 kg
m =moles of solute
mass of solvent (kg)=
5.86 moles C2H5OH
0.657 kg solvent= 8.92 m
12.3
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Convert % mass to Molarity
• What is the Molarity of a 95% acetic acid solution? (density = 1.049 g/mL)
If you assume 1 L, that amount of solution = 1049 g
95% of the solution is acetic acid
1049 g solution x 0.95 = 997 g solute
997 g X 1 mol/60.05 g = 16.6 mol solute
Since we assumed 1 L, that’s 16.6 mol / 1 L or
16.6 M
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Temperature and Solubility
Solid solubility and temperature
solubility increases with increasing temperature
solubility decreases with increasing temperature
12.4
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Fractional crystallization is the separation of a mixture of substances into pure components on the basis of their differing solubilities.
Suppose you have 90 g KNO3 contaminated with 10 g NaCl.
Fractional crystallization:
1. Dissolve sample in 100 mL of water at 600C
2. Cool solution to 00C
3. All NaCl will stay in solution (s = 34.2g/100g)
4. 78 g of PURE KNO3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g
12.4
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Temperature and Solubility
Gas solubility and temperature
solubility usually decreases with
increasing temperature
12.4
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Pressure and Solubility of Gases
12.5
The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law).
c = kP
c is the concentration (M) of the dissolved gas
P is the pressure of the gas over the solution
k is a constant (mol/L•atm) that depends onlyon temperature
low P
low c
high P
high c
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Chemistry In Action: The Killer Lake
Lake Nyos, West Africa
8/21/86CO2 Cloud Released
1700 Casualties
Trigger?
• earthquake
• landslide
• strong Winds
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Solution Stoichiometry (Chapter 4)
The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.
M = molarity =moles of solute
liters of solution
What mass of KI is required to make 500. mL ofa 2.80 M KI solution?
volume KI moles KI grams KIM KI M KI
500. mL = 232 g KI166 g KI
1 mol KIx
2.80 mol KI
1 L solnx
1 L
1000 mLx
4.5
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4.5
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Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.
Dilution
Add Solvent
Moles of solutebefore dilution (i)
Moles of soluteafter dilution (f)=
MiVi MfVf=4.5
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How would you prepare 60.0 mL of 0.2 MHNO3 from a stock solution of 4.00 M HNO3?
MiVi = MfVf
Mi = 4.00 Mf = 0.200 Vf = 0.06 L Vi = ? L
4.5
Vi =MfVf
Mi
=0.200 x 0.06
4.00= 0.003 L = 3 mL
3 mL of acid + 57 mL of water = 60 mL of solution
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Gravimetric Analysis
4.6
1. Dissolve unknown substance in water
2. React unknown with known substance to form a precipitate
3. Filter and dry precipitate
4. Weigh precipitate
5. Use chemical formula and mass of precipitate to determine amount of unknown ion
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TitrationsIn a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.
Equivalence point – the point at which the reaction is complete
Indicator – substance that changes color at (or near) the equivalence point
Slowly add baseto unknown acid
UNTIL
the indicatorchanges color
4.7
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What volume of a 1.420 M NaOH solution isRequired to titrate 25.00 mL of a 4.50 M H2SO4 solution?
4.7
WRITE THE CHEMICAL EQUATION!
volume acid moles acid moles base volume base
H2SO4 + 2NaOH 2H2O + Na2SO4
4.50 mol H2SO4
1000 mL solnx
2 mol NaOH
1 mol H2SO4
x1000 ml soln
1.420 mol NaOHx25.00 mL = 158 mL
M
acid
rx
coef.
M
base
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Chemistry in Action: Metals from the Sea
CaCO3 (s) CaO (s) + CO2 (g)
Mg(OH)2 (s) + 2HCl (aq) MgCl2 (aq) + 2H2O (l)
CaO (s) + H2O (l) Ca2+ (aq) + 2OH (aq)-
Mg2+ (aq) + 2OH (aq) Mg(OH)2 (s)-
Mg2+ + 2e- Mg
2Cl- Cl2 + 2e-
MgCl2 (l) Mg (l) + Cl2 (g)
Now back to Chapter 12…Now back to Chapter 12…
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Colligative Properties of Nonelectrolyte Solutions
Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.
Vapor-Pressure Lowering
Raoult’s law
If the solution contains only one solute:
X1 = 1 – X2
P 10 - P1 = P = X2 P 1
0
P 10 = vapor pressure of pure solvent
X1 = mole fraction of the solvent
X2 = mole fraction of the solute12.6
P1 = X1 P 10
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Fractional Distillation Apparatus
12.6
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Boiling-Point Elevation
Tb = Tb – T b0
Tb > T b0 Tb > 0
T b is the boiling point of the pure solvent
0
T b is the boiling point of the solution
Tb = Kb m
m is the molality of the solution
Kb is the molal boiling-point elevation constant (0C/m)
12.6
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Freezing-Point Depression
Tf = T f – Tf0
T f > Tf0 Tf > 0
T f is the freezing point of the pure solvent
0
T f is the freezing point of the solution
Tf = Kf m
m is the molality of the solution
Kf is the molal freezing-point depression constant (0C/m)
12.6
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12.6
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What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g.
Tf = Kf m
m =moles of solute
mass of solvent (kg)= 2.41 m=
3.202 kg solvent
478 g x 1 mol62.01 g
Kf water = 1.86 0C/m
Tf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C
Tf = T f – Tf0
Tf = T f – Tf0 = 0.00 0C – 4.48 0C = -4.48 0C
12.6
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Colligative Properties of Nonelectrolyte Solutions
Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.
12.6
Vapor-Pressure Lowering P1 = X1 P 10
Boiling-Point Elevation Tb = Kb m
Freezing-Point Depression Tf = Kf m
Osmotic Pressure () = MRT
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Colligative Properties of Electrolyte Solutions
12.7
0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions
Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.
0.1 m NaCl solution 0.2 m ions in solution
van’t Hoff factor (i) = actual number of particles in soln after dissociation
number of formula units initially dissolved in soln
nonelectrolytesNaCl
CaCl2
i should be
12
3
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Which would you use for the streets of Bloomington to lower the freezing point of ice and why? Would the temperature make any difference in your decision?
a) sand, SiO2
b) Rock salt, NaCl
c) Ice Melt, CaCl2
Change in Freezing Change in Freezing Point Point
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Boiling-Point Elevation Tb = i Kb m
Freezing-Point Depression Tf = i Kf m
Osmotic Pressure () = iMRT
Colligative Properties of Electrolyte Solutions
12.7
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Change in Freezing Change in Freezing Point Point
Common Applications of Common Applications of Freezing Point Freezing Point DepressionDepression
Propylene glycol
Ethylene glycol – deadly to small animals
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Change in Boiling Change in Boiling Point Point
Common Applications of Common Applications of Boiling Point ElevationBoiling Point Elevation
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At what temperature will a 5.4 molal At what temperature will a 5.4 molal solution of NaCl freeze?solution of NaCl freeze?
SolutionSolution
∆∆TTFPFP = K = Kff • m • i • m • i
∆ ∆TTFPFP = (1.86 = (1.86 ooC/molal) • 5.4 m • 2C/molal) • 5.4 m • 2
∆ ∆TTFP FP = 20.1= 20.1 ooCC
FP = 0 – 20.1 = -20.1FP = 0 – 20.1 = -20.1 ooCC
Freezing Point Freezing Point DepressionDepression
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The Cleansing Action of Soap
12.8
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Osmotic Pressure ()
12.6
Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one.
A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules.
Osmotic pressure () is the pressure required to stop osmosis.
dilutemore
concentrated
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HighP
LowP
Osmotic Pressure ()
= MRT
M is the molarity of the solution
R is the gas constant
T is the temperature (in K) 12.6
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A cell in an:
isotonicsolution
hypotonicsolution
hypertonicsolution
12.6
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Chemistry In Action: Desalination
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A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance.
Colloid versus solution
• collodial particles are much larger than solute molecules
• collodial suspension is not as homogeneous as a solution
12.8
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Colloids
• Brownian motion • Tyndall Effect
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Suspensions
• These are mixed, but not dissolved in each other
• Will settle over time
• Particles are bigger than 1 micrometer (larger than colloid)
• Examples: dust in air, muddy water