Transcript Chapter 10 - Solutions

Solutions & Their Properties
(Colligative Properties)
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Molarity (M):
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Dilution Formula:
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Moles of solute (mol) per liters (L) of solution:
Used when preparing diluted solutions from concentrated
ones.
Mole Fraction:
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Used previously for gas problems.
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XA = moles A = nA
total moles ntot
Mole fraction of a component of the solution will equal
moles of that component divided by the total moles present.
The sum of the mole fractions should equal 1.
Mass Percent (%)
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Mass percent is determined by the mass of the
solute divided by the total mass of the solution, then
multiplied by 100.
Example: If a solution is prepared by dissolving 24 g
of NaCl in 152 g of water, calculate the percent, by
mass, of NaCl.
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Sometimes when the amounts are so small (trace amounts)
we use parts per million (ppm) or parts per billion (ppb). For
example, if we have 5 x 10-8 grams of arsenic in 1.00 grams
of water that will be equivalent to 0.05 ppm of Arsenic in
water.
For ppm we multiply by a factor of 106 and for ppb we
multiply by a factor of 109.
Molality (m)
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Molality is determined my moles of solute per
kilograms of solvent.
m = moles of solute
kilograms of solvent
Conversions between concentration
units: Example 1
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The electrolyte in automobile lead storage
batteries is a 3.75 M sulfuric acid solution
that has a density of 1.23 g/mL. Calculate the
mass percent and molality of the sulfuric
acid.
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Ans: 29.9% H2SO4 and 4.35m H2SO4
Principles of Solubility
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Like dissolves like…..meaning polar generally dissolves polar
and nonpolar dissolves nonpolar.
Effects of Temperature:
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Solids – the higher the temperature, generally the higher the
solubility. An increase in temperature always shifts the position of
an equilibrium to favor an endothermic process, which means H
>0.
Gases – become less soluble as the temperature rises. Why?
Effects of Pressure:
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Pressure only has an effect on gas-liquid solutions. Do pressure
and the solubility of a gas have an inverse or direct relationship?
Henry’s Law:
Cg
=
kPg
Cg = concentration of the gas (M)
k = constant for the gas-liquid system – this will vary (M/atm)
Pg = Pressure of the gas (atm)
Henry’s Law: Example 2
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A certain soft drink is bottled so that a bottle at 25oC
contains CO2 gas at a pressure of 5.0 atm over the
liquid. Assuming that the partial pressure of CO2 in
the atmosphere is 4.0 x 10-4 atm, calculate the
quilibrium concentrations of CO2 in the soda both
before and after the bottle is opened. The Henry’s
law constant for CO2 in aqueous solution is 3.1 x 10-2
mol/L atm at 25oC.
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Ans: before 0.16M, after 0.000012M
Colligative Properties of Nonelectrolytes
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Properties of solutions differ from the
properties of the pure solvent.
The solution properties depend more so on
the concentration of the solute particles.
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Vapor Pressure Lowering
Boiling Point Elevation
Freezing Point Lowering
Vapor Pressure Lowering
Concentrated solutions evaporate more slowly
than pure water, meaning they have a lower vapor
pressure.
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With an increase in concentration of solute the vapor
pressure lowers – this is Raoult’s Law:
ΔP = X2 P10
or
P = X1 P10
ΔP = change in pressure (P10 - P1) in mmHg
X2 = mole fraction of solute
P10 = vapor pressure of the pure solvent in mmHg
P = vapor pressure of the solution
X1 = mole fraction of solvent
Vapor Pressure Lowering: Example 3
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Calculate the expected vapor pressure at
25oC for a solution prepared by dissolving
158.0 g of common table sugar (sucrose, M
= 342.3 g/mol) in 643.5 cm3 of water. At
25oC, the density of water is 0.9971 g/cm3
and the vapor pressure is 23.76 torr.
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ans: 23.46 torr
Boling Point Elevation
A solution does not begin to boil until the temperature
exceeds that of the solvent. The greater the
concentration of the solute the higher the temperature
needed to boil the solution:
ΔTb = kb x m
m = molality (m)
kb = Molal Boiling Point Constant (0C/m)
ΔTb = change in temp at which solution boils (Tb - Tb0)
-Tb = the temp at which the solution boils
-Tb0 = the temp at which the pure solvent boils
Boiling Point Elevation: Example 4
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A solution was prepared by dissolving 18.00
g of glucose in 150.0 g of water. The
resulting solution was found to have a boiling
point of 100.34 oC. The molal boiling-point
elevation constant for water is 0.51 oC
kg/mol. Calculate the molar mass of glucose.
Glucose is a molecular solid that is present in
individual molecules in solution.
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ans: 180.0 g/mol
Freezing Point Depression
When a solution is cooled it does not begin to freeze
until a temperature below the freezing point of the
solvent is reached. The greater the concentration the
lower the temperature required for freezing:
ΔTf = kf x m
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m = molality
kf = Molal freezing point constant
ΔTf = change in temp at which solution freezes (Tf0 - Tf)
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Tf = the temp at which the solution freezes
Tf0 = the temp at which the pure solvent freezes
Osmotic Pressure
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Osmosis is a process taking
place through a membrane
permeable to only the solvent is
called osmosis. Water moves
from a region where its vapor
pressure or mole fraction is high
to one in which it’s vapor
pressure or mole fraction is low.
Osmotic pressure is directly
proportional to the molarity (M)
– this is why it is a colligative
property:
π = nRT = MRT
V
Osmotic Pressure: Example 5
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To determine the molar mass of a certain
protein, 1.00 x 10-3 g of it was dissolved in
enough water to make 1.00 mL of solution.
The osmotic pressure of this solution was
found to be 1.12 torr at 25 oC. Calculate the
molar mass of the protein.
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ans: 1.66 x 104 g/mol
FRQ #1
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An unknown compound contains only the three elements C, H, and
O. A pure sample of the compound is analyzed and found to be
65.60 percent C and 9.44 percent H by mass.
(a) Determine the empirical formula of the compound.
(b) A solution of 1.570 grams of the compound in 16.08 grams of
camphor is observed to freeze at a temperature 15.2 Celsius
degrees below the normal freezing point of pure camphor.
Determine the molar mass and apparent molecular formula of the
compound. (The molal freezing-point depression constant, Kf, for
camphor is 40.0 kg•K•mol-1.)
(c) When 1.570 grams of the compound is vaporized at 300°C
and 1.00 atmosphere, the gas occupies a volume of 577 milliliters.
What is the molar mass of the compound based on this result?
(d) Briefly describe what occurs in solution that accounts for the
difference between the results obtained in parts (b) and (c).
FRQ #2
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Concentrated sulfuric acid (18.4-molar H2SO4) has a
density of 1.84 grams per milliliter. After dilution with
water to 5.20-molar, the solution has a density of 1.38
grams per milliliter and can be used as an electrolyte in
lead storage batteries for automobiles.
(a) Calculate the volume of concentrated acid required
to prepare 1.00 liter of 5.20-molar H2SO4.
(b) Determine the mass percent of H2SO4 in the original
concentrated solution.
(c) Calculate the volume of 5.20-molar H2SO4 that can
be completely neutralized with 10.5 grams of sodium
bicarbonate, NaHCO3.
(d) What is the molality of the 5.20-molar H2SO4?
FRQ #3
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Elemental analysis of an unknown pure substance indicated that the
percent composition by mass is as follows: carbon 49.02%, hydrogen
2.743%, chlorine 48.23%.
A solution that is prepared by dissolving 3.150 grams of the
substance in 25.00 grams of benzene, C6H6, has a freezing point of
1.12°C. (The normal freezing point of benzene is 5.50°C and the
molal freezing-point depression constant, Kf, for benzene is
5.12°C/molal.)
(a) Determine the empirical formula of the unknown substance.
(b) Using the data gathered from the freezing-point depression
method, calculate the molar mass of the unknown substance.
(c) Calculate the mole fraction of benzene in the solution described
above.
(d) The vapor pressure of pure benzene at 35°C is 150. millimeters
of Hg. Calculate the vapor pressure of benzene over the solution
described above at 35°C.
Equations Practice
(a) A solution of sodium hydroxide is added to a solution of lead(II)
nitrate.
(i) Balanced equation:
(ii) If 1.0 L volumes of 1.0 M solutions of sodium hydroxide and
lead(II) nitrate are mixed together, how many moles of product(s) will
be produced? Assume the reaction goes to completion.
(b) Excess nitric acid is added to solid calcium carbonate.
(i) Balanced equation:
(ii) Briefly explain why statues made of marble (calcium carbonate)
displayed outdoors in urban areas are deteriorating.
(c) A solution containing silver(I) ion (an oxidizing agent) is mixed with a
solution containing iron(II) ion (a reducing agent).
(i) Balanced equation:
(ii) If the contents of the reaction mixture described above are
filtered, what substance(s), if any, would remain on the filter paper.