chm101 chapter15 re
TRANSCRIPT
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Chapter 15
Solutions
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15.1 Solubility
15.2 Solution Composition: An Introduction
15.3 Solution Composition: Mass Percent
15.4 Solution Composition: Molarity
15.5 Dilution15.6 Stoichiometry of Solution Reactions
15.7 Neutralization Reactions
15.8 Solution Composition: Normality
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What is a Solution?
Solution homogeneous mixture
Solventsubstance present in largest
amount
Solutesother substances in the solution
Aqueous solutionsolution with water as the
solvent
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Various Types of Solutions
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Ionic substances breakup into individual
cations and anions.
Solubility of Ionic Substances
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Polar water molecules interact with the positive
and negative ions of a salt.
Solubility of Ionic Substances
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Ethanol is soluble in water because of the
polar OH bond.
Solubility of Polar Substances
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Why is solid sugar soluble in water?
Solubility of Polar Substances
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Nonpolar oil does not interact with polar water.
Water-water hydrogen bonds keep the water
from mixing with the nonpolar molecules.
Substances Insoluble in Water
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A hole must be made in the water structure
for each solute particle.
The lost water-water interactions must be
replaced by water-solute interactions.
like dissolves like
How Substances Dissolve
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Concept Check
Which of the following solutes will generally not
dissolve in the specified solvent? Choose the best
answer. (Assume all of the compounds are in theliquid state.)
a) CCl4 mixed with water (H2O)
b) NH3 mixed with water (H2O)c) CH3OH mixed with water (H2O)
d) N2 mixed with methane (CH4)
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The solubility of a solute is limited.
Saturated solution contains as much
solute as will dissolve at that
temperature.
Unsaturated solution has not
reached the limit of solute that will
dissolve.
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Supersaturated solution occurs when a
solution is saturated at an elevated
temperature and then allowed to cool but
all of the solid remains dissolved. Contains more dissolved solid than a
saturated solution at that temperature.
Unstable
adding a crystal causesprecipitation.
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Solutions are mixtures.
Amounts of substances can vary in different
solutions.
Specify the amounts of solvent and
solutes.
Qualitative measures of concentration
concentrated relatively large amountof solute
dilute relatively small amount of solute
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mass of soluteMass percent = 100%
mass of solution
grams of soluteMass percent = 100%
grams of solute + grams of solvent
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Exercise
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%
[5.5 g / (5.5 g + 78.2 g)] 100 = 6.6% glucose
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Molarity (M) = moles of solute per
volume of solution in liters:
moles of solute= Molarity =liters of solution
M
6 moles of HCl3 HCl =2 liters of solution
M
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Exercise
You have 1.00 mol of sugar in 125.0 mL of
solution. Calculate the concentration in units
ofmolarity.
8.00 M
1.00 mol / (125.0 / 1000) = 8.00 M
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Exercise
A 500.0-g sample of potassium phosphate
is dissolved in enough water to make 1.50 L
of solution. What is the molarity of thesolution?
1.57 M
500.0 g is equivalent to 2.355 mol K3PO4 (500.0 g /
212.27 g/mol). The molarity is therefore 1.57 M
(2.355 mol/1.50 L).
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Exercise
You have a 10.0 Msugar solution. What
volume of this solution do you need to have
2.00 mol of sugar?
0.200 L
2.00 mol / 10.0 M= 0.200 L
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Exercise
Consider separate solutions of NaOH and KCl
made by dissolving 100.0 g of each solute in
250.0 mL of solution. Calculate theconcentration of each solution in units of
molarity.
10.0 MNaOH[100.0 g NaOH / 39.998 g/mol] / [250.0 / 1000] = 10.0 MNaOH
5.37 MKCl[100.0 g KCl / 74.55 g/mol] / [250.0 / 1000] = 5.37 MKCl
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Concept Check
You have two HCl solutions, labeled Solution A andSolution B. Solution A has a greater concentration thanSolution B. Which of the following statements are true?
a) If you have equal volumes of both solutions,Solution B must contain more moles of HCl.
b) If you have equal moles of HCl in both solutions,Solution B must have a greater volume.
c) To obtain equal concentrations of both solutions,you must add a certain amount of water toSolution B.
d) Adding more moles of HCl to both solutions will makethem less concentrated.
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For a 0.25 MCaCl2 solution:
CaCl2 Ca2+ + 2Cl
Ca2+: 1 0.25 M= 0.25 MCa2+
Cl: 2 0.25 M= 0.50 MCl.
Concentration of Ions
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Concept Check
Which of the following solutions contains
the greatest number of ions?
a) 400.0 mL of 0.10 M NaCl.
b) 300.0 mL of 0.10 M CaCl2.
c) 200.0 mL of 0.10 M FeCl3.d) 800.0 mL of 0.10 M sucrose.
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Where are we going?
To find the solution that contains the greatest
number of moles of ions.
How do we get there? Draw molecular level pictures showing each
solution. Think about relative numbers of ions.
How many moles of each ion are in each
solution?
Lets Think About It
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The solution with the greatest number of
ions is not necessarily the one in which:
the volume of the solution is the
largest.
the formula unit has the greatest
number of ions.
Notice
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A solution whose concentration is
accurately known.
Standard Solution
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Weigh out a sample of solute.
Transfer to a volumetric flask.
Add enough solvent to mark on flask.
To Make a Standard Solution
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The process of adding water to a
concentrated orstock solution to achieve
the molarity desired for a particular
solution. Dilution with water does not alter the
numbers of moles of solute present.
Moles of solute before dilution = moles ofsolute after dilution
M1V1 = M2V2
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Transfer a measured amount of original solution
to a flask containing some water.
Add water to the flask to the mark (with swirling)
and mix by inverting the flask.
Diluting a Solution
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Concept Check
A 0.50 Msolution of sodium chloride in an open
beakersits on a lab bench. Which of the
following would decrease the concentration of
the salt solution?
a) Add water to the solution.
b) Pour some of the solution down the sink drain.
c) Add more sodium chloride to the solution.d) Let the solution sit out in the open air for a
couple of days.
e) At least two of the above would decrease the
concentration of the salt solution.
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Exercise
What is the minimum volume of a 2.00 M
NaOH solution needed to make 150.0 mL of
a 0.800 MNaOH solution?
60.0 mL
M1V1 = M2V2
(2.00 M)(V1) = (0.800 M)(150.0 mL)
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1. Write the balanced equation for the reaction. For
reactions involving ions, it is best to write the net
ionic equation.
2. Calculate the moles of reactants.
3. Determine which reactant is limiting.
4. Calculate the moles of other reactants or
products, as required.
5. Convert to grams or other units, if required.
Steps for Solving Stoichiometric Problems Involving Solutions
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Concept Check (Part I)
10.0 mL of a 0.30 Msodium phosphate
solution reacts with 20.0 mL of a 0.20 M
lead(II) nitrate solution (assume no volumechange).
What precipitate will form?
lead(II) phosphate, Pb3(PO4)2 What mass of precipitate will form?
1.1 g Pb3(PO4)2
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Where are we going? To find the mass of solid Pb3(PO4)2 formed.
How do we get there?
What are the ions present in the combined solution? What is the balanced net ionic equation for the
reaction?
What are the moles of reactants present in the
solution?
Which reactant is limiting?
What moles of Pb3(PO4)2 will be formed?
What mass of Pb3(PO4)2 will be formed?
Lets Think About It
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Where are we going?
To find the concentration of nitrate ions left in
solution after the reaction is complete.
How do we get there? What are the moles of nitrate ions present in the
combined solution?
What is the total volume of the combined
solution?
Lets Think About It
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Concept Check (Part III)
10.0 mL of a 0.30 Msodium phosphate
solution reacts with 20.0 mL of a 0.20 M
lead(II) nitrate solution (assume no volumechange).
What is the concentration of phosphateions left in solution after the reaction is
complete?
0.011 M
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Where are we going? To find the concentration of phosphate ions left in
solution after the reaction is complete.
How do we get there? What are the moles of phosphate ions present in
the solution at the start of the reaction?
How many moles of phosphate ions were used
up in the reaction to make the solid Pb3(PO4)2? How many moles of phosphate ions are left over
after the reaction is complete?
What is the total volume of the combined
solution?
Lets Think About It
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An acid-base reaction is called a
neutralization reaction.
Steps to solve these problems are the same
as before. For a strong acid and base reaction:
H+(aq) + OH(aq) H2O(l)
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Concept Check
For the titration of sulfuric acid (H2SO4) with
sodium hydroxide (NaOH), how many moles
of sodium hydroxide would be required toreact with 1.00 L of 0.500 Msulfuric acid?
1.00 mol NaOH
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Where are we going?
To find the moles of NaOH required for the
reaction.
How do we get there? What are the ions present in the combined
solution? What is the reaction?
What is the balanced net ionic equation for the
reaction? What are the moles of H+ present in the solution?
How much OH is required to react with all of the
H+ present?
Lets Think About It
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One equivalent of acid amount of acid
that furnishes 1 mol of H+ ions.
One equivalent of base amount of base
that furnishes 1 mol of OH ions Equivalent weight mass in grams of 1
equivalent of acid or base.
Unit of Concentration
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To find number of equivalents:
number of equivalents equivalents equivNormality = = = =
1 liter of solution liter LN
equiv
V = L = equivLN
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Concept Check
If Ba(OH)2 is used as a base, how many
equivalents of Ba(OH)2 are there in 4 mol
Ba(OH)2?
a) 2
b) 4
c) 8d) 16