Chapter 11 - Solutions
PS 11.1 Notes
--TERMS-Solute - substance being dissolved
Solvent - substance doing the dissolving
Solution - Homogeneous mixture of solute and solvent
Saturated - contains all that will dissolve at conditions
Unsaturated - contains less than what will dissolve
Super Saturated - contains more than what should be dissolved
Electrolytes - conduct electricity when in solution
Non electrolytes - do not conduct electricity when in solution
***To conduct solution you need _________________***
Electrolyte:
Fe(NO3)3 (s) --->
Non-electrolyte: C12H22O11 (s) --->
Solution Preparation:
1) Dissolving a solid
a) add ___ grams of solid to a _________________
b) Add water _______________________________
2) Dilution of a more concentrated solution.
(moles stays the same, volume changes)
M1V1 = M2V2 (1 = conc., 2 = dilute)
a) Measure __ mL of concentrated solution.
b) Add water _______________________________
Ex. A How would you make 250. mL of a 0.400 M solution of NaCl?
Ex. B Make the above solution by diluting 2.00 M NaCl
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CONCENTRATION UNITS:
Mass percent solute =
mass solute
total mass solution
Parts Per Million (ppm) =
Part Per Billion (ppb) =
x 102 =
mass solute
x 106 =
total mass solution
mass solute
x 109 =
total mass solution
MOLE FRACTION (XA):
XA =
number of moles A
total number of moles all components
NOTE: XA + XB + .... = 1.00
MOLALITY (m):
Molality (m) =
MOLARITY (M):
Molarity (M) =
number of moles solute
number of kg solvent
number moles solute
number liters solution
Normality - (N) the number of equivalents per liter of solution
(depends of the reaction taking place)
Acid/Base Rxn:
Redox Rxn:
Mass Percent
mass A
total .massso ln
Mole Fraction (XA)
moles A
moles tot
Molarity (M)
moles solute
Liters so ln
Molality (m)
moles solute
kg solvent
USEFUL EQUATIONS/HINTS
Break the problem up into steps.
a. What do you know?
b. What, specifically, are you trying to find?
c. How can you get there? What do you need?
ex. A solution that is 20.0% by mass C6H12O6 (d=
1.13 g/mL). If you’re trying to find molality,
molarity, and mole fraction. You must make an
assumption.
Assume 100. grams solution; this means 20.0 g
C6H12O6 AND 80.0 g H2O. ALL these facts will
become important at some point so write them all
down and LABEL!!!
100. g soln
20.0 g C6H12O6
80.0 g H2O
2
Amount of Solute is
Increased
Decreased
Amount of Solvent is
Increased
Decreased
Amount of solution is
Increased
Decreased
Molarity of
Solute
Moles of
solute
Mass of
solute
Volume of
Solution
Sample Problems
1A) Find: The molarity of concentrated ammonium hydroxide. The mass density of the
solution is 0.900 g/mL and mass % of NH4OH is 57.6%.
Hint: assume a convenient amount
e.g.
Assume 1 mL solution and use 0.900 g/mL to start
OR
Assume 100 grams solution and use 57.6% NH4OH to start.
1B) Find Molality of the above.
2A) Find the mass % of concentrated H2SO4. The density of the solution is 1.84 g/mL and it
is 18.0 M.
(Assume useful volume and use given to figure out its composition.)
2B) Find the mole fractions of 2A, i.e., XH2SO4 and XH2O
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Principles of Solubility & Energies of Solution Formation
“LIKE DISSOLVES LIKE”
• Two substances with intermolecular forces of about the same type and magnitude are likely to
be very soluble in one another.
• Items soluble in water tend to be either ionic or possible hydrogen bonding.
• Non-electrolytes without hydrogen bonding are generally soluble in non-polar or slightly polar
solvents such as CCl4 or benzene. (gasoline)
Heat of solution = the enthalpy change associated with dissolving a solute in a solvent; the sum of
the energies needed to expand both solvent and solute in a solution and
the energy released from the solvent-solute interaction.
ΔHsoln = ΔH1 solute-solute + ΔH2 solvent-solvent + ΔH3 solute-solvent
Solvation – also sometimes called dissolution, is the process of attraction and association of
molecules of a solvent with molecules or ions of a solute.
Hydration – the interaction between solute particles and water molecules
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Heat of hydration = the enthalpy change associated placing gaseous molecules or ions in water;
the sum of the energies needed to expand the solvent and the energy
released from the solvent-solute interaction.
ΔHhyrdation = ΔH2 solvent-solvent + ΔH3 solute-solvent
Ex. (Given Lattice energy of sodium chloride is -786 kJ/mol)
NaCl(s)
H2O(l) + Na+(g) + Cl-(g)
 Na+(g) + Cl-(g)
 Na+(aq) + Cl-(aq)
ΔH1 =
ΔHhyd = ΔH2 + ΔH3 = -783 kJ
ΔHsoln =
kJ
What determines how strongly an ion is hydrated?
PS 11.2 Notes
TEMPERATURE EFFECTS UPON SOLUBILITY
• An increase in temperature always favors the endothermic process.
NaNO3(s)
CO2 (g)
<====> Na+ (aq) + NO3- (aq)
<====> CO2 (aq)

Dissolving solids is ____________ ___________________.

Dissolving gases is ____________ ____________________.
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PRESSURE EFFECTS UPON SOLUBILITY
• Pressure has little or no effect on solubility of solids or liquids in water
• Solubility of gases (concen. of dissolved gas) are directly proportional to the
partial pressure of the gas.
Cg = k Pg
(Henry’s Law)
http://dwb4.unl.edu/Ch
em/CHEM869J/CHE
M869JImages/graphhenry.gif
The solubility of pure nitrogen in blood @ body temp, 37 °C, and 1.00 atm, is 6.2 x 10 -4 M. If a
diver breathes air (XN2 = 0.78) at a depth where total pressure is 2.5 atm, calculate the
concentration of nitrogen in his blood.
Colligative Properties of Solutions
• Colligative properties are properties that vary due to concentration rather than
type.
• These properties are limiting laws that are more accurate in dilute solution.
~98% true at 1.0 M
Vapor Pressure Lowering
proportional to mole fraction (concentration) solvent
Higher Boiling Point
Lower Freezing Point
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A.) Vapor Pressure Lowering
• Vapor pressure of liquid solvent decreases as more non-volatile solute is present.
Raoult’s Law:
B.) Boiling point elevation and freezing point depression result from vapor
pressure lowering.
Kf and Kb (see Table 11.5 pg. 505)
kf = molal freezing point constant (for a solvent!)
kb = molal boiling point constant (for a solvent!)
.
ΔTB = kB m i
ΔTF = kF
.
m i
van’t Hoff factor, i
• Colligative properties are dependent on number of particles in solution.
Compound
C6H12O6
NaCl
AlCl3
CaBr2
i
• Thus substances that form ions (electrolytes) have a greater effect than those
that remain as molecules (non electrolytes).
ΔTB and ΔTf can be used to calculate solute molecular masses.
(If given the change in Tf or Tb and the mass (g) of solute and mass (kg) solvent.)
ΔTF = kF
.
m
I
m = nsolute = gsolute / MMsolute
kgsolvent
kgsolvent
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EXAMPLE PROBLEMS – COLLIGATIVE PROPERTIES
Ex. 1 - Given: 155 g sugar (sucrose) dissolved in 425 g H2O at 50 °C
(V.P. H2O @ 50 °C = 92.5 mm Hg) Molar mass of sucrose is C12H22O11 = 342 g/mol.
Find: A) Mole fraction of sugar
B) V.P. of solution
Answers:
Ex 1 – a. xsugar= 0.0188 b. Psoln = 90.7 mmHg
Ex 2 – Tf = -4.46°C Tb = 101.2°C
Ex 3 – MM = 106 g/mol
Ex. 2 - Estimate F.P. and B.P. of a 0.600 m Aluminum Chloride water solution.
kF H2O = 1.86°C/m; kB H2O = 0.52°C/m
Ex. 3 - A student dissolves 1.50 g of a newly prepared compound in 75.0 g of cyclohexane. She
measure the f.p. to be 2.70 °C. Pure cyclohexane is 6.50 °C. The kf of cyclohexane is
20.2 °C/m. Find the molar mass of the compound.
A Deeper understanding of Principles – nonideal solutions & ion pairing
• A liquid-liquid solution that obeys Raoult’s law is called an ideal solution.
--> Nearly ideal behavior is often observed when the solute-solute, solvent-solvent, and solutesolvent interactions are very similar.
a.k.a. when the solute and solvent are very similar
• Negative deviation:
The vapor pressure of the solution is ________ than calculated (_________ solute-solvent
interactions) - usually occurs when ΔHsoln (
)
• Positive deviation:
The vapor pressure of the solution is ________ than calculated (_________ solute-solvent
interactions) - usually occurs when ΔHsoln (
)
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ion pairing:
PS 11.3 Notes
Osmosis - Results from vapor pressure lowering of solutions vs. pure solvent.
• Water moves from a region of high vapor pressure (pure water) to low vapor
pressure (solution).
• In osmosis, this is done through a “semipermeable” membrane.
• These membranes allow __________ to pass through but not ______ molecules.
• Osmotic pressure, π, is defined as being equal to the external pressure, P, just
sufficient to prevent osmosis from occurring.
π = nRT = MRTi
V
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FIND: Osmotic pressure, π, for a 1.50 M solution of sugar at 27 °C.
Determining Molar Mass from Osmotic Pressure:
The Uses and Aspects of Osmosis A) Use to find molecular weights - very useful for high molecular weight compounds
- proteins.
Example - A solution contains 1.0 g of hemoglobin dissolved to 0.100 L at 20
°C. It produces 2.75 mm Hg osmotic pressure. Find the
approximate molar mass of hemoglobin.
B) Reverse Osmosis - Used in water treatment
By applying an external pressure greater than the osmotic pressure the process
can be reversed.
C) Solutions added to living systems must be isotonic with the system.
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