Download Chapter 4 Student Notes

Chemistry 20H
Chemistry: The Central Science
Chapter 4 Aqueous Reactions and Solution Stoichiometry
Objectives
1.
Discuss the general properties of aqueous solutions. (4.1)
2.
Be able to predict whether a precipitate will form when two solutions are mixed (double replacement,
exchange, or metathesis reactions). (4.2)
3.
Write complete ionic equations and net ionic equations for precipitation reactions. (4.2)
4.
Discuss the nature of acids and bases. (4.3)
5.
Distinguish between strong and weak acids and bases. (4.3)
6.
Be able to list several strong acids and bases. (4.3)
7.
Be able to predict whether a compound is a strong, weak, or non-electrolyte. (4.3)
8.
Be able to identify neutralization reactions and predict the products of such reactions. (4.3)
9.
Be able to predict which neutralization reactions will form gases. (4.3)
10.
Discuss the process of oxidation and reduction. (4.4)
11.
Be able to determine the oxidation number of each atom in any element or compound. (4.4)
12.
Be able to identify displacement reactions (single replacement). (4.4)
13.
Use activity series to determine whether a displacement reaction will proceed. (4.4)
14.
Be able to calculate the concentration of solutions and incorporate what we know about moles. (4.5)
15.
Be able to perform dilution calculations. (4.5)
16.
Be able to solve problems involving solution stoichiometry. (4.6)
17.
Use titration to collect data for solution stoichiometry. (4.6)
Vocabulary
See the Summary and Key Terms on pages 156 & 157 of the text.
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4.1
General Properties of Aqueous Solutions
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Electrolytic Properties
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Ionic Compounds in Water
Dissociation
o
o
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
o
Electrolytes:
Nonelectrolytes:
Strong electrolyte:
Weak electrolyte:
o
Solvation equations
o

o
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to write a dissociation equation for any ionic compound identify of the cation and anion, then write the balanced
equation:
Write dissociation equations for the ionic substances in questions 4.15 and 4.16
4.2 Precipitation Reactions
o
Reactions that result in the formation of an insoluble product are known as precipitation reactions.
Solubility Guidelines for Ionic Compounds
o
o
o
o
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Solubility rules for common ionic compounds in water:
o
Compounds containing alkali metal ions or ammonium ions are soluble.
o
Compounds containing NO31- or C2H3O21- are soluble.
 An exception is Ag C2H3O2
o
Compounds containing Cl1-, Br1- or I1- are soluble.
 Exceptions are the compounds of Ag1+, Hg22+, and Pb2+.
o
Compounds containing SO42- are soluble.
 Exceptions are the compounds of Sr2+, Ba2+, Hg22+, and Pb2+.
o
Compounds containing S2- are insoluble.
 Exceptions are the compounds of NH41+, the alkali metal cations, and Ca2+, Sr2+, and Ba2+.
o
Compounds of CO32- or PO43- are insoluble.
 Exceptions are the compounds of NH41+ and the alkali metal cations.
o
Compounds of OH1- are insoluble.
 Exceptions are the compounds of NH41+, the alkali metal cations, and Ca2+, Sr2+, and Ba2+.
Copy examples in class:
Complete questions 4.19 and 4.20
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Exchange (Metathesis, Double Replacement) Reactions
o
o
o
o
Example:
o
Chemical Equations of Metathesis Reactions
Molecular Equations –
Complete Ionic Equations –
Net Ionic Equations –
Writing Net Ionic Equations
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For the following write:
1. dissociation equations for the reactants
2. predict the precipitate
3. a balanced molecular equation
4. complete ionic equation
5. net ionic equation

a)
FeCl2 (aq) + K2S
b)
AlBr3 (aq) + NaOH
c)
(NH4)3PO4 (aq) + Ca(NO3)2 (aq) 
d)
Aqueous solutions of silver nitrate and sodium carbonate react.
e)
Aqueous solutions of barium chloride and potassium sulphate react.
(aq)
(aq)

Complete questions 4.22 and 4.24. In each reaction with a precipitate do the 4 steps above.
4.3 Acid-Base Reactions
Acids
Bases
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Weak acids and weak bases
Identifying Strong and Weak Electrolytes
Complete questions 4.35 to 4.38
Neutralization Reactions and Salts
o
A neutralization reaction occurs when an acid and a base react:
Molecular equation:
Complete ionic equation:
Net Ionic Equation:
Another Example:
Mg(OH)2 (s) + 2 HCl(aq)
Molecular Equation:
Complete Ionic Equation:
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Net Ionic Equation:
o
Note that the magnesium hydroxide is an insoluble solid; it appears in the net ionic equation.
Complete questions 4.39 and 4.40
Acid-Base Reactions with Gas Formation
o
There are many bases besides OH1- that react with H1+ to form molecular compounds:
o
Reaction of sulfides with acid gives rise to H 2S(g).
o
Reaction of sulfites with acid gives rise to SO 2 (g).
o
Carbonates and hydrogen carbonates (or bicarbonates) will form CO 2 (g) when
treated with an acid:
o
Bases with ammonium compounds gives ammonia gas:
Complete question 4.43
4.4 Oxidation-Reduction Reactions
Oxidation and Reduction
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o
o
o
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Oxidation Numbers
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Oxidation numbers are assigned to atoms using specific rules:
1.
2.
3.
4.
5.
6.
o
Elements have an oxidation number of 0.
The oxidation number of a monatomic ion is the same as its charge.
Oxygen has an oxidation number of 2- (except peroxides, O22-)
Hydrogen has an oxidation number of 1+ (except hydrides, H1-)
The sum of the oxidation numbers in a neutral compound is 0.
The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.
The oxidation of an element is evidenced by its increase in oxidation number; reduction is accompanied
by a decrease in oxidation number.
Copy Examples in class:
Complete questions 4.49 and 4.50
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Recognizing Oxidation and Reduction in Chemical Reaction Equations
Consider the following balanced chemical equation:
3 Cu2+ + 2 Al  3 Cu + 2 Al3+
Let’s apply what we know about oxidation numbers to this equation:
O.N.
o
o
3 Cu2+
+ 2 Al 
3 Cu
2+
0
0
+
2 Al3+
3+
the copper goes from 2+ on the reactant side to 0 on the product side
o it’s charge got lower
o it is reduced (it gained electrons).
The aluminum goes from 0 to 3+
o it’s charge got greater
o it is oxidized (it lost electrons):
reduction (gained 2 electrons each)
┌───────────┐
3 Cu2+ + 2 Al  3 Cu + 2 Al3+
O.N.
2+
0
0
3+
└─────────┘
oxidation (lost 3 electrons each)
notice there are 3 coppers in the balanced equation; the total electrons gained is 3(2 e 1-) = 6 electrons.
there are 2 aluminums in the equation; the total electrons lost is 2(3 e 1-) = 6 electrons.
A redox equation is balanced if:
it is balanced for atoms on each side.
the total electrons lost and gained are equal.
the reducing agent:
the oxidizing agent:
Our equation now looks like this:
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Copy examples in class:
1. CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l)
2. 2 PbS(s) + 3 O2
(g)
 2 PbO
(s)
+ 2 SO2 (g)
Complete questions 4.51 and 4.52
Oxidation of Metals by Acids and Salts
o
o
o
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o
Example: metals produce hydrogen gas with acids
o
Example: It is possible for metals to be oxidized in the presence of a salt:
Complete questions 4.53 and 4.54
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The Activity Series
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This list is an activity series:
o
o
o
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o
or
Complete questions 4.56 to 4.58
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4.5 Concentrations of Solutions
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Expressing the Concentration of an Electrolyte
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o


Write the equation for the dissociation of the following ionic substances. Calculate the
concentration of each ion in solution:
a)
0.35 mol/L NaOH
b)
1.12 mol/L (NH4)2CO3
c)
0.056 mol/L V3(PO4)5
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Interconverting Molarity, Moles, and Volume
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o
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o
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Examples:
1)
Calculate the molar concentration when 0.256 mol of CuSO4 is dissolved in 1.1 L of solution.
2)
Calculate the molar concentration when 28.2 g of KNO 3 is dissolved in 750. mL of solution.
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3)
Calculate the volume of solution made with 0.455 mol of sodium phosphate in a 2.00 M solution.
4)
How many moles are present in 4.50 L of a 0.025 mol/L solution of magnesium nitrate?
Complete questions 4.61 to 4.69, odd
Other Measures of Concentration
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Parts per hundred
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Parts per thousand
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Parts per million
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Parts per billion
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Parts per trillion
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Parts per quadrillion
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Dilution
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Example
•
•
Have: 12.4 mol/L stock solution HCl
Want: 2.00 L of 0.100 mol/L solution.
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o
There is an easier way to do this:
Complete examples in class
•
What volume of concentrated sulfuric acid (18 M) is needed to make 500. mL of 2.00 M
dilute solution?
•
What is the concentration of dilute solution if 25.0 mL of glacial acetic acid (24.0 mol/L) is
added to make 2.00 L of dilute solution?
•
What volume of a 1.00 mol/L stock solution of aluminum chloride is needed to make 500.
mL of 3.0 x 10-4 mol/L solution?
Complete questions 4.71 to 4.74
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Mixing Ionic Solutions
Consider these solutions:
•
•
CaCl2 (aq) →
0.25 mol/L
Ca2+(aq) +
2 Cl1-(aq)
1(0.25 mol/L)
2(0.25 mol/L)
= 0.25 mol/L
= 0.50 mol/L
Mg3(PO4)2 (aq)
0.011 mol/L
→
3 Mg2+(aq)
+
2 PO43-(aq)
3(0.011 mol/L)
2(0.011 mol/L)
= 0.033 mol/L
= 0.022 mol/L
take 300. mL of the CaCl2 and 200. mL of the Mg3(PO4)2 and we mix them.
Assuming that no precipitate is formed, what will the concentration of each ion in solution be after
mixing ?
Complete question 4.70
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4.6 Solution Stoichiometry and Chemical Analysis
o
o
We can apply what we know to stoichiometry problems.
Consider this problem:
o 150. mL of 0.500 mol/L lead (II) nitrate is added to a solution containing sufficient potassium
iodide. What mass of lead (II) iodide will precipitate ?
o
Step 1. Write and balance the chemical equation. Write down the knowns and unknowns.
o
Step 2. Convert given information to moles
o
Step 3. Use the mole ratio to determine the moles of PbI 2
o
Step 4. Convert moles of unknown to desired units
o
Step 5. Make a concluding statement
Complete the following:
Aqueous cadmium chloride reacts with sodium sulfide to produce bright-yellow cadmium sulfide. What mass of
cadmium sulfide would be produced if 50.00 mL of a 3.91 M solution of sodium sulfide reacts with sufficient
cadmium chloride?
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Solutions of calcium nitrate and sodium phosphate react. If 256 mL of a 0.356 M solution of calcium nitrate
reacts with sufficient sodium phosphate, what mass of precipitate would result?
These problems are often connected to a lab technique called titration.
Titrations
o
o
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o
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Example Calculations:
•
What is the concentration of HCl if 50.00 mL of HCl solution is neutralized by 16.15 mL of
0.100 M NaOH ?

What volume of 0.250 mol/L KOH will be neutralized by 160. mL of 0.400 mol/L HNO 3 ?
Indicators
o
o
o
o
o
Complete questions 4.77 to 4.86