Chpt 11 - Solutions
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Concentrations
Energy of solutions
Solubility
Colligative Properties
Colloids
• HW: Chpt 11 - pg. 531-538, #s 12, 14, 20, 24,
29, 34, 41, 42, 44, 46, 52, 66, 72, 77, 81 Due
Mon Dec. 3
Various Types of Solutions
Example
Air, natural gas
Vodka, antifreeze
Brass
Carbonated water (soda)
Seawater, sugar solution
State of
Solution
Gas
Liquid
Solid
Liquid
Liquid
State of
Solute
Gas
Liquid
Solid
Gas
Solid
State of
Solvent
Gas
Liquid
Solid
Liquid
Liquid
Hydrogen in platinum
Solid
Gas
Solid
Solvent is majority component. Solute is minority
component, usually the substance dissolved in the
solvent (liquid).
Solution composition
moles of solute
Molarity (M ) =
liters of solution
Mass (weight) percent =
Mole fraction (c A ) =
Molality (m ) =
mass of solute
´ 100%
mass of solution
moles A
total moles of solution
moles of solute
kilogram of solvent
Molarity
You have 1.00 mol of sugar in 125.0 mL
of solution. Calculate the concentration
in units of molarity.
8.00 M
You have a 10.0 M sugar solution. What
volume of this solution do you need to
have 2.00 mol of sugar?
0.200 L
Molarity (M) example
Consider separate solutions of NaOH and
KCl made by dissolving 100.0 g of each
solute in 250.0 mL of solution.
Calculate the concentration of each
solution in units of molarity.
10.0 M NaOH
5.37 M KCl
Mass percent (%)
What is the percent-by-mass
concentration of glucose in a
solution made my dissolving 5.5 g
of glucose in 78.2 g of water?
6.6%
Mole fraction (A)
A solution of phosphoric acid was made
by dissolving 8.00 g of H3PO4 in 100.0
mL of water. Calculate the mole fraction
of H3PO4. (Assume water has a density
of 1.00 g/mL.)
0.0145
Molality (m)
A solution of phosphoric acid was made
by dissolving 8.00 g of H3PO4 in 100.0
mL of water. Calculate the molality of
the solution. (Assume water has a
density of 1.00 g/mL.)
0.816 m
Solution Formation Schematic
Solution Formation Process
1. Separating the solute into its individual
components (expanding the solute).
2. Overcoming intermolecular forces in
the solvent to make room for the
solute (expanding the solvent).
3. Allowing the solute and solvent to
interact to form the solution.
Solution Formation Energies
• Steps 1 and 2 require energy, since forces must be
overcome to expand the solute and solvent.
• Step 3 usually releases energy.
• Steps 1 and 2 are endothermic, and step 3 is often
exothermic.
• Enthalpy change associated with the formation of the
solution is the sum of the ΔH values for the steps:
ΔHsoln = ΔH1 + ΔH2 + ΔH3
• ΔHsoln may have a positive sign (energy absorbed) or
a negative sign (energy released).
Exo vs. Endo Hsoln
Demo NH4NO3 and NaOH examples
Explain why water and oil (a long chain
hydrocarbon) do not mix. In your explanation,
be sure to address how ΔH plays a role.
H1
H2
H3
Hsoln
Outcome
Polar solute, polar
solvent
Large
Large
Large,
negative
Small
Solution
forms
Nonpolar solute, polar
solvent
Small
Large
Small
Large,
positive
No solution
forms
Nonpolar solute,
nonpolar solvent
Small
Small
Small
Small
Solution
forms
Polar solute, nonpolar
solvent
Large
Small
Small
Large,
positive
No solution
forms
Solubility Factors
• Structural Effects:
 Polarity
(like dissolves like)
• Pressure Effects:
 Henry’s law (for dissolved gases)
• Temperature Effects:
 Affecting aqueous solutions
Pressure effects
• Henry’s law:
C
k
P
=
=
=
C = kP
concentration of dissolved gas
constant
partial pressure of gas solute
above the solution
• Amount of gas dissolved in a solution is
directly proportional to the pressure of
the gas above the solution.
Gas solubility in liquid
Soda pop’s carbonated water has the carbon dioxide forced
into the solution under pressure. When the can is opened
Patm is much lower than Pcan so CO2 leaves -> pop goes flat.
Temperature effects
• Although the solubility of most solids in
water increases with temperature, the
solubilities of some substances
decrease with increasing temperature.
• Predicting temperature dependence of
solubility is very difficult.
• Solubility of a gas in solvent typically
decreases with increasing temperature.
Temp solubility
charts
Colligative Properties
• Depend only on the number, not on the
identity, of the solute particles in an
ideal solution:
 Vapor pressure lowering
http://www.chem.purdue.edu/gchelp/solutions/colligv.html
 Boiling-point elevation
 Freezing-point depression
 Osmotic pressure
Vapor Pressure of solutions
If the Pvap of the solvent (water) > Pvap of the solution, equilibrium is reached
when the solvent evaporates and the solvent is absorbed by solution. It does
this to lower the Pvap towards its equilibrium value.
Raoult’s Law
• Nonvolatile solute lowers the vapor
pressure of a solvent.
Psoln = csolvPsolv
• Raoult’s Law:
Psoln
=
observed vapor
pressure of solution
solv =
mole fraction of solvent
Posolv =
vapor pressure of pure solvent
Raoult’s Law - ideal solution
Ideal solution occurs with a
nonvolatile solute in solution
Also the vapor pressure is
then proportional to the mole
fraction of the solvent using
(total moles of ions of solute)
in the solvent
Boiling Point elevation
• Nonvolatile solute elevates the boiling
point of the solvent.
• ΔT = Kbmsolute
ΔT = boiling-point elevation
Kb = molal boiling-point elevation
constant
msolute= molality of solute particles
Freezing Point depression
• When a solute is dissolved in a solvent,
the freezing point of the solution is lower
than that of the pure solvent.
• ΔT = Kfmsolute
ΔT = freezing-point depression
Kf = molal freezing-point depression
constant
msolute= molality of solute particles
Phase Diagram of solutions
Boiling Point - Freezing Point
explanation
http://en.wikipedia.org/wiki/Boilingpoint_elevation
http://en.wikipedia.org/wiki/Freezingpoint_depression
Boiling Pt Elev Problem
A solution was prepared by dissolving 25.00
g glucose in 200.0 g water. The molar mass
of glucose is 180.16 g/mol. What is the boiling
point of the resulting solution (in °C)?
Glucose is a molecular solid that is present as
individual molecules in solution. Kb =
0.51oC.kg/mol
100.35 °C
Osmotic Pressure
• Osmosis – flow of solvent into the
solution through a semipermeable
membrane. (Kidney dialysis uses this Principle).
P = MRT
P
M
R
T
=
=
=
=
osmotic pressure (atm)
molarity of the solution
gas law constant
temperature (Kelvin)
Osmotic Pressure graphic
Osmotic Pressure Problem
When 33.4 mg of a compound is
dissolved in 10.0 mL of water at 25°C,
the solution has an osmotic pressure of
558 torr. Calculate the molar mass of
this compound.
Strategy: need Temp in K, Pressure in atm, use R with atm unit to
get molarity. Then use vol get moles, then mass get molar
mass.
111 g/mol
Colloids
• Intermediate mixture - a heterogeneous
mixture with particle size between a
suspension and a solution
• A suspension of tiny particles in some
medium.
• Tyndall effect – scattering of light by particles.
• Suspended particles are single large
molecules or aggregates of molecules or ions
ranging in size from 1 to 1000 nm.
Types of colloids
Tyndall Effect graphic
Freezing Pt problem
You take 20.0 g of a sucrose (C12H22O11) and
NaCl mixture and dissolve it in 1.0 L of water.
The freezing point of this solution is found to
be -0.426°C. Kf = 1.86oC.kg/mol
Assuming ideal behavior, calculate the mass
percent composition of the original mixture,
and the mole fraction of sucrose in the
original mixture.
72.8% sucrose and 27.2% sodium chloride;
mole fraction of the sucrose is 0.313
Derivation of
Colligative Properties
• Specific derivation of the partial
derivatives and derivation for colligative
properties are found on the website.
http://www.chem.arizona.edu/~salzmanr/4
80a/480ants/colprop/colprop.html