Post on 04-Jan-2016
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Chapter 13 – Properties of Solutions
Homework:33, 34, 35, 38, 39, 41, 44, 46, 47, 50, 55, 59, 63, 65, 66, 67,
69, 72, 74
13.4 – Ways of Expressing Concentration Concentration of a solution can be
expressed qualitatively or quantitatively Terms dilute and concentrated are
qualitative statements Dilute meaning a relatively small concentration of
solute Concentrated meaning a relatively large
concentration of solute Our focus will be on the quantitative
descriptions of concentration
Mass Percentage One of the simplest quantitative
methods of expressing concentration Given by
So if I had a solution of 36% HCl by mass It would contain 36 g of HCl for each 100 g
of solution Note: NOT per 100g of solvent
100solution of mass total
solutionin component of mass component of % Mass
ppm Oftentimes, very dilute solutions are given in
terms of parts per million (ppm) ppm is given by
What this means If I have a solution whose concentration is 1 ppm It has 1 g of solute for every million (106) g of
solution Or 1 mg of solute per kg of solution
610solution of mass total
solutionin component of masscomponent of ppm
Because density of water is 1 g/mL 1 kg of a very dilute aqueous solution will
have a volume very close to 1 L So, 1 ppm also means 1 mg/L of solution
ppm often used to express maximum concentration of toxic or carcinogenic substances in the environment In US, maximum arsenic in water is 0.010
ppm
ppb A solution that is VERY dilute is
often measured in terms of parts per billion (ppb) 1 ppb means 1 g of solute per billion
(109) g of solution Or 1 microgram (μg) per liter of
solution So the allowable concentration of
arsenic in water could also be 10 ppb.
Mole Fraction Concentration can also be expressed by
the number of moles of components of the solution.
Mole fraction we’ve talked about
No units, just a ratio The sum of all of the mole fractions of
the components adds up to 1
components all of moles total
component of moles component offraction mole X
Molarity The big one Molarity (M) relates moles of solute to
volume of solution
Very useful when doing stoichiometry, because of the easy mole relationship
solution liters
solute moles Molarity
Molality Molality (m) is the concentration unit that
looks at the number of moles of solute per kilogram of solvent
Don’t get it confused with molarity Molarity is moles/L Molality is moles/kg
Molality doesn’t change with temperature, while molarity does
Molality often used when the solution is used over a range of temperatures
solvent of kilograms
solute of moles Molality
Conversion of Concentration Units Sometimes the concentration
needs to be known in several different units
It is possible to convert concentration units
Example An aqueous solution of
hydrochloric acid contains 36% HCl by mass. Find the mole fraction of HCl in the solution. Find the molality of HCl in the solution
Working Mole fraction When dealing with percentages, assume
100g of the substance So 36% = 36g per 100 g of the substance So if HCl is 36g, that means that water
would be 64g To do mole fraction, we now convert the
g of each of the components to moles
HCl mol 99.0HCl g 36.5
HCl mol 1HCl g 36HCl Moles
OH mol 6.3OH g 18
OH mol 1OH g 64OH Moles 2
2
222
22.099.06.3
99.0
HCl moles OH moles
HCl molesX
2HCl
Converting to molality Need moles of HCl and mass of water Luckily, we got those from doing the mole
fraction 0.99 mol HCl and 0.064 kg H2O
m 15H kg 0.064
HCl mol 0.99 HCl ofMolality
2
O
Molality ↔ Molarity To convert between molality and
molarity we need to know the density of the solution
Example A solution contains 5.0g of toluene
(C7H8) and 225 g of benzene and has a density of 0.876 g/mL. Calculate the molarity of the solution.
To find molarity, we need mol of toluene and volume of solution
We find volume by using the density of the solution But first, we need total mass of solution
mol 054.0HC g 92
HC mol 1HC g 0.5HC Moles
87
878787
To find volume, we use density Note: We actually have 230 g of substance
225 g benzene and 5 g of toluene
Finally, to find molarity, we just divide moles of toluene by liters of solution
mL263g 0.876
mL 1g 230 Volume
Final Step – Almost There!
Note: To solve this problem we needed the following Density (allows us to move from mass of
solution to volume of solution) Mass of solute (to find moles of solute)
M21.0solution L 1
mL 1000
solution mL 263
HC mol 0.054
solutionliter
HC moles Molarity 8787
13.5 – Colligative Properties The physical properties of many
solutions differ from those of the pure solvent We add ethylene glycol to water in
radiators to lower the freezing point (antifreeze) and to raise the boiling point (so that the water won’t boil in the radiator)
This lowering of the freezing point and raising of the boiling point are physical properties These properties (in a solution) depend on
the quantity (concentration) but not the identity of the solute
These types of properties are called colligative properties.
Colligative means “depending on the collection” So these properties depend on the collective
effect of the number of solute particles
Common Colligative Properties The common (the one’s we’re going to
look at) colligative properties include: Lowering freezing point Raising boiling point Vapor-pressure reduction Osmotic pressure
For each of these, notice how the concentration of the solute affects the property we’re looking at (relative to the pure substance)
Lowering the Vapor Pressure We know that a liquid in a closed
container will create a dynamic equilibrium with its vapor At equilibrium, the pressure exerted by
the vapor is called the vapor pressure Remember, a substance that has no
measurable vapor pressure is called nonvolatile
And a substance that has a vapor pressure is called volatile
We find that the vapor pressure of solutions (if we have added a nonvolatile solute) will be lower than the vapor pressure of the solvent The amount the vapor pressure is
lowered is directly proportional to the concentration of the solute
This relationship is given by Raoult’s law
Raoult’s Law PA = partial pressure of the vapor from
the solution In other words: vapor pressure of the
solution XA= mole fraction of the solvent PºA= vapor pressure of the pure solvent
AAA PXP
Example The vapor pressure of water is 17.5
torr at 20ºC. If we add glucose to the water, so that the mole fraction of water is 0.800, then the new vapor pressure of water will be PH2O= (0.800)(17.5 torr) = 14.0 torr Doesn’t matter which pressure units
we use! Just be consistent
Raoult’s law assumes that the solutes are non-volatile, molecular compounds. Not always the case Will deal with ionic compounds (and
their effects) when we discussion freezing and boiling points
Deviation from Prediction An ideal solution obeys Raoult’s
law. Real solutions best approximate ideal
solutions when solute concentration is low and when solute and solvent have similar molecular sizes and types of intermolecular attractions Understand that this occurs, but assume
ideal solution unless told otherwise
Boiling Point Elevation Remember, adding solute lowers vapor
pressure Also remember, boiling occurs when
vapor pressure = atmospheric pressure So we need a higher temperature to
give us more vapor pressure when we add a solute
So, the boiling point of a solution is higher than that of a pure liquid
The increase in boiling point (relative to the pure solvent) is given by the term ΔTb
Is directly proportional to the number of solute particles per mole of solvent molecules
Molality is the number of moles of solute per 1000 g of solvent, which shows a fixed number of moles of solvent.
Thus, ΔTb is directly proportional to molality
ΔTb = Kbm
The magnitude of Kb (called the molal boiling-point-elevation constant) depends on the solvent in question Typical solvents listed on pg. 551 For water, Kb = 0.51ºC/m So a 1 m aqueous solution of sucrose
(sugar) would boil at 0.51ºC higher than pure water
The boiling-point elevation is proportional to the concentration of the solute particles Doesn’t matter if they are ions or molecules Note: When an ionic compound dissolves, total
molality increases When 1 mol of NaCl dissolves, 2 moles of solute form
(1 mol of Na+ and 1 mol of Cl-) So if we have a 1 m aqueous solution of NaCl, we
actually have a total of 2 m in solute particles So double the effect of boiling point increase
Freezing-Point Depression The freezing point of a solution is
the temperature at which the first crystals of pure solvent begin to form In equilibrium with the solution Lower vapor pressure effects freezing
point too Lower vapor pressure (we’ve added
solute) means the freezing point is lower than that of a pure solvent
ΔTf = Kfm Like the boiling-point elevation, the
decrease in freezing point, ΔTf is directly proportional to the molality of the solute
The values of Kf, the molal freezing-point-depression constant for several common solvents are also found on pg. 551
For water, Kf = 1.86ºC/m So 1m aqueous solution of sucrose will freeze
1.86ºC lower than pure water While the same rule for ionic compounds
applies here A 1m aqueous solution of NaCl will freeze 3.72ºC
below the freezing point of the pure solvent
Example Calculate the freezing point of a
solution containing 0.600 kg of CHCl3 and 42.0g of eucalyptol (C10H18O), a fragrant substance found in the leaves of eucalyptus trees.
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Kf = 4.68ºC/m normal freezing point = -63.5ºC Find molality!
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Example 2 Which of the following solutes will
produce the largest increase in boiling point upon addition to 1 kg of water? 1 mol of Co(NO3)2
2 mol of KCl 3 mol of ethylene glycol (C2H6O2)
Osmosis Certain materials are
semipermeable Often in biological systems and
synthetic substances When in contact with a solution,
allows some molecules to pass through them, but not others
Generally allow small solvent molecules (like water) to pas through
But blocks larger solute molecules or ions
Consequences of Osmosis Consider a membrane that only allows
solvent particles to pass If placed between two solutions of different
concentration, solvent molecules move in both directions through the membrane
However, this process is called osmosis, and the net movement of solvent is always toward the solution with the higher SOLUTE concentration
So the less concentrated solution loses solvent, and more concentrated solution gains solvent
What does this mean?
If we wanted to stop this from happening, we apply pressure on the side of the more concentrated solution
The pressure required to prevent osmosis by pure solvent is the osmotic pressure, π
The osmotic pressure obeys a law similar in form to the ideal gas law
πV=nRT V is the volume of the solution n is the number of moles of solute R is the ideal-gas constant T is temperature in Kelvin scale
From this equation, we can write
Where M is the molarity of the solution R value determines which units your
pressure will be in
MRTRTV
n
If two solutions of identical osmotic pressure are separated by a semi-permeable membrane, no osmosis will occur The two solutions are called isotonic
If one solution is of lower osmotic pressure, that solution is hypotonic with respect to the more concentrated solution The more concentrated solution is hypertonic
with respect to the dilute solution
Biological Systems See! I know biology! The membranes of blood cells are
semipermeable Placing a red blood cell in solution that is
hypertonic relative to the solution inside the cell causes water to come out of the cell
Causes cell to shrivel (crenation) Placing a red blood cell in a solution that is
hyptotonic relative to the solution inside the cell causes water to move into the cell
Causes cell to rupture (hemolysis) When giving IV infusions to people, IV solution
must be isotonic with the solution inside the cell
Example What is the osmotic pressure at
20ºC of a 0.0020 M sucrose (C12H22O11) solution.
Determination of Molar Mass Any of the colligative properties
can be used to determine the molar mass of a solute.
Molar Mass from Boiling Point Elevation A solution of unknown nonvolatile
non-ionic solute was prepared by dissolving 0.250 g of the substance in 40.0 g of CCl4. The boiling point of the solution was 0.357ºC higher than that of the pure solvent. What is the molar mass of the solute?
To find molar mass, we need grams of solute and moles of solute
Step 1: Solve for molality Step 2: Use molality to solve for mol
solute Step 3: Use given data and mol
solute to find molar mass
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Molar Mass from Osmotic Pressure A solution contains 3.50 mg of
protein dissolved in enough water to form 5.00 mL of solution. The osmotic pressure of the solution at 25ºC was found to be 1.54 torr. Calculate the molar mass of the protein.
Step 1: Solve for molarity Step 2: Using molarity, solve for
number of moles Step 3: Divide number of grams
given by moles
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