Transcript Chapter 13 - Solutions

Solutions
Chapter 13
Solutions
• Solution - homogeneous mixture of solute
and solvent.
• Solvent - the component present in the
largest amount.
• Solutes - The other components
Factors that determine whether a
solution will form:
•
Entropy – usually solution formation
increases entropy (more disordered)
•
Energy – based on the strength of IMFs
– neither the solute or solvent can be
more attracted to itself than to the other
substance.
The Solution Process
– IMFs between the solvent particles and IMFs
between the solute particles must be equal to
or weaker than IMFs between the solute and
solvent.
Oil – nonpolar – dispersion forces
Water –Polar - hydrogen bonds
The hydrogen bonds in water are
stronger than any attractive force
between the two substances
The Solution Process
• Consider NaCl (solute) dissolving in water
(solvent)
Ion-dipole forces between the
ions and the water molecules
(in this case) are stronger than
the ionic force holding the ions
together and stronger than the
hydrogen bonds holding the
water molecules together.
(hydration)
“Rule of thumb”:
• Polar solvents dissolve polar solutes and
nonpolar solvents dissolve nonpolar
solutes ("like dissolves like")
More on Energy…
ΔHsolute = energy to separate solute particles
it’s always endothermic (+)
ΔHsolvent = energy to separate solvent particles
it’s always endothermic (+)
ΔHmix= depends on IMFs between solvent & solute
it’s always exothermic (-)
ΔHsoln = ΔHsolute + ΔHsolvent + ΔHmix
The solution process overall may be endothermic (cold
packs) or exothermic (hot packs) depending on the
magnitudes of each enthalpy value
• Miscible - substances that mix in any
proportions
– Example: Ethanol and water are miscible liquids.
• Immiscible - substances that do not mix
significantly.
– Example: Oil and water are immiscible.
• Solids (salt, sugar, etc) dissoved in a liquid
(water) are somewhere in between!
Saturated Solutions and Solubility
• As a solid dissolves, a solution forms (the
opposite process is crystallization)
• If crystallization and dissolution are in
equilibrium with undissolved solute, the
solution is saturated. (no further solute
can be dissolved)
Solubility
• The amount of solute required to form a
saturated solution.
• A solution with a concentration of
dissolved solute that is less than the
solubility is said to be unsaturated.
• A solution is said to be supersaturated if
more solute is dissolved than in a
saturated solution (heat, dissolve more,
cool)
Factors Affecting Solubility
• Solubility depends on:
– The nature of the solute & solvent
• More detail on the “like dissoves like” rule
– The temperature.
– The pressure (for gases).
• Consider the solubility of alcohols in water.
– Water and ethanol are miscible because the
broken hydrogen bonds in both pure liquids
are re-established in the mixture.
• However, not all alcohols are miscible with
water.
• The number of carbon atoms in a chain
affects solubility.
– The greater the number of carbons in the
chain (the larger the nonpolar part), the more
the molecule behaves like it’s nonpolar
• This decreases its solubility in water
– The greater the number of –OH groups (the
polar part), the more the molecule behaves
like it’s polar
• This increases its solubility in water
Substances with similar intermolecular
attractive forces tend to be soluble in
one another. (like dissolves like)
Pressure Effects
• The solubility of a gas in a liquid is a function of
the pressure of the gas over the solution.
• Solubilities of solids and liquids are not greatly
affected by pressure.
• With higher gas pressure, more molecules of
gas are close to the surface of the solution and
the probability of a gas molecule striking the
surface and entering the solution is increased.
Properties of Solutions
Lower pressure
above means
less solute dissolves
Higher pressure
above means
more solute dissolves
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• The higher the pressure, the greater the
solubility.
• The lower the pressure, the less the
solubility.
• Henry’s Law - The solubility of a gas is
directly proportional to the partial pressure
of the gas above the solution.
• Mathematically: Sg=kPg
– Note: The Henry's law constant (k) differs for
each solute-solvent pair and differs with
temperature.
– S is usually expressed in mol/L
• make sure units cancel when using this equation.
Henry’s Law Application
• Carbonated beverages are bottled where
PCO2> 1 atm.
• As the bottle is opened, PCO2 decreases
and the solubility of CO2 decreases
(solution fizzes)
Temperature Effects
• As temperature increases, solubility of solids
generally increases (there are some exceptions)
– Sugar dissolves better in warm water than in cold
water.
• As temperature increases, solubility of gases
generally decreases.
– Carbonated beverages go flat as they get warm.
– Thermal pollution: If lakes get too warm, O2 become
less soluble and fish suffocate
Concentration of Solutions
Concentration
Abbreviation
Mass Percent
%
Parts per million
ppm
Mole Fraction
X
Molarity
M
Molality
m
Definition
𝑴𝒂𝒔𝒔 𝑺𝒐𝒍𝒖𝒕𝒆
∗ 𝟏𝟎𝟎%
𝑴𝒂𝒔𝒔 𝑺𝒐𝒍′ 𝒏
𝑴𝒂𝒔𝒔 𝑺𝒐𝒍𝒖𝒕𝒆
∗ 𝟏𝟎𝟔
′
𝑴𝒂𝒔𝒔 𝑺𝒐𝒍 𝒏
𝑴𝒐𝒍𝒆𝒔 𝒐𝒏𝒆 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕
𝑻𝒐𝒕𝒂𝒍 𝑴𝒐𝒍𝒆𝒔
𝑴𝒐𝒍𝒆𝒔 𝑺𝒐𝒍𝒖𝒕𝒆
𝑳𝒊𝒕𝒆𝒓𝒔 𝒐𝒇 𝑺𝒐𝒍′ 𝒏
𝑴𝒐𝒍𝒆𝒔 𝑺𝒐𝒍𝒖𝒕𝒆
𝑲𝒊𝒍𝒐𝒈𝒓𝒂𝒎𝒔 𝑺𝒐𝑙𝑣𝑒𝑛𝑡
Colligative Properties
• Colligative properties depend on number of
solute particles.
• There are four colligative properties to
consider:
– Vapor pressure lowering (Raoult's Law).
– Boiling point elevation.
– Freezing point depression.
– Osmotic pressure.
Van’t Hoff Factor (i)
• The number of particles formed when a
solute dissolves
– NaCl = 2
– Al(NO3)3 = 4
– C12H22O11 = 1
This is a maximum!
Lowering the Vapor Pressure
• Nonvolatile solutes (with no measurable vapor
pressure themselves) reduce the ability of the
surface solvent molecules to escape the liquid.
• Therefore, vapor pressure is lowered.
• The amount of vapor pressure lowering depends
on the amount of solute.
• Raoult’s law quantifies the extent to which a
nonvolatile solute lowers the vapor pressure of
the solvent
PA=XAPAo
PA = Vapor Pressure of sol’n
XA = mole fraction of solvent
PAo= Vapor pressure of pure solvent
• Ideal solution: one that obeys Raoult’s law.
• Real solutions show ideal behavior when:
– The solute concentration is low.
– The solute and solvent have similarly sized
molecules.
– The solute and solvent have similar types of
intermolecular attractions.
Boiling-Point Elevation
• A nonvolatile solute lowers the vapor
pressure of a solution.
• Higher temperature is required to reach a
vapor pressure of 1 atm for the solution
• The molal boiling-point-elevation
constant, Kb, expresses how much Tb
changes with molality
Tb = Kb m i
• The boiling-point elevation is proportional
to the concentration of solute particles
• i = van’t hoff factor
– A 1 m solution of NaCl is 2 m in total solute
particles.
Freezing-Point Depression
• When a solution freezes, crystals of almost
pure solvent are formed first.
• Solute molecules are usually not soluble in
the solid phase of the solvent - sol’n
become more concentrated (lower X of
solvent in sol’n).
• This lowers the vapor pressure for the
solution
– the triple point occurs at a lower temperature
• The melting-point (freezing-point) curve is a
vertical line from the triple point.
• The solution freezes at a lower temperature than
the pure solvent.
• The decrease in freezing point (Tf) is directly
proportional to molality.
Tf = Kf m i
• Kf is the molal
freezing-pointdepression constant.
Osmosis
• Semipermeable membranes permit passage
of some components of a solution.
• Often they permit passage of water but not
larger molecules or ions.
• Osmosis is the net movement of a solvent
from an area of low solute concentration to
an area of high solute concentration.
• Consider a U-shaped tube with a two liquids
separated by a semipermeable membrane.
• The rate of movement of solvent from the less
concentrated solution is faster than the rate of
movement in the opposite direction.
• As solvent moves across the membrane, the
fluid levels in the arms become uneven.
• The vapor pressure of solvent is higher in the
arm with more solvent.
• Eventually the pressure difference due to the
difference in height of liquid in the arms stops
osmosis.
Osmotic Pressure
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Properties of Solutions
The process of reverse osmosis uses mechanical pressure to make
the solvent (water) flow in the opposite direction. This process can
be used to desalinate ocean water to make drinking water.
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Osmotic pressure, π
• The pressure required to prevent osmosis.
• Osmotic pressure obeys a law similar in
form to the ideal-gas law
n =moles
V= volume
M= molarity
R= the ideal gas constant
T= absolute temperature
pV = nRT
OR
n
p    RT  MRT
V 
• Solutions are said to be isotonic if they
have the same osmotic pressure.
• Hypotonic solutions have a lower π and
are less concentrated
• Hypertonic solutions have a higher π and
are more concentrated
• If a cucumber is placed in NaCl solution, it will
lose water to shrivel up and become a pickle.
• A dry carrot placed in water becomes firm as
water enters via osmosis.
• Eating large quantities of salty food causes
retention of water and swelling of tissues
(edema).
• Water moves into plants, to a great extent,
through osmosis.
• Salt may be added to meat (or sugar added to
fruit) as a preservative.
• Salt prevents bacterial infection: A bacterium
placed on the salt will lose water through
osmosis and die.
Determination of Molar Mass
Camphor (C10H16O) melts at 179.8°C, and it
has a particularly large freezing-pointdepression constant, Kf=40.0°C/m. When
0.186 g of an organic substance of
unknown molar mass is dissolved in 22.01
g of liquid camphor, the freezing point of
the mixture is found to be 176.7°C. What
is the molar mass of the solute?
• ΔTf = kmi
(179.8-176.7) = (40.0)(m)(1)
3.1 = 40.0m
m = .0775 mol solute/kg camphor
.0775 𝑚𝑜𝑙 𝑠𝑜𝑙𝑢𝑡𝑒
0.02201 𝑘𝑔 𝑐𝑎𝑚𝑝ℎ𝑜𝑟 ∗
= .001706 mol solute
𝑘𝑔 𝑐𝑎𝑚𝑝ℎ𝑜𝑟
.186 𝑔 𝑠𝑜𝑙𝑢𝑡𝑒
= 110 𝑔/mol solute
0.001706 𝑚𝑜𝑙 𝑠𝑜𝑙𝑢𝑡𝑒
A sample of 2.05 g of polystyrene of uniform
polymer chain length was dissolved in
enough toluene to form 0.100 L of solution.
The osmotic pressure of this solution was
found to be 1.21 kPa at 25°C. Calculate the
molar mass of the polystyrene.
𝜋𝑉 = 𝑛𝑅𝑇
(1.21/101.325)(.100L) = n(.0821)(298)
n = 0.0000488 mol
2.05 𝑔
= 42,000 𝑔/𝑚𝑜𝑙
0.0000488𝑚𝑜𝑙