CHM
2045C Name: _________________
Module Eight Homework Packet-Jespersen
Module Eight:
Solutions/Oxidation & Reduction (McMurry Chapter 4) ppt
A. _____(02) Solution Properties-From
Lecture(4.1, 4.2) Answers
B. _____(02) Factors Affecting Rate
of Dissolving-From Lecture(11.4) Answers
C. _____(02) Writing Ionzation Reactions acids/salts Sect 4.3, 4.4 Answers
D. _____(03) Solution Preparation Problems-Section 4.7 Answers
E. _____(03) Solution Dilution Problems-Section
4.7 Answers
F. _____(05) Solution Reaction Problems-Section
4.8, 4.9, 5.6 Answers
G. _____(05) Rewrite Equations Ionically –Section 4.3 Answers
H. _____(06) Redox Equations-Sections-Section 5.1, 5.2
Answers
I. _____ (02 Solution Discussion Question-Chapter 4, 5 Answers
K. ____
(02) Solution Definitions-Chapters 4, 5, From Lecture Answers
L. _____(03) pH calculations-Section 16.1 Answers
______(35)
Total = ______%
M-8
Required Homework:
______(35) M-8
Pretest Hardcopy Homework Packet
______(10/20) M-8 Multiple Choice (MC) Homework/Exam (Blackboard Online)
G1. ______(20) Hard Copy (Acid/base Media) Writing Net Ionic Reactions Homework (Sect 4.3)
H1.______(20) REDOX Post Lab
Part A: Solution Properties
Back in Module 1 a solution was
introduced as a homogeneous mixture of a solute and solvent and you included
them in your matter chart for Part A of Module 1. A solution is defined
as a homogeneous mixture of two or more substances. The two new words are
introduced, solute and solvent. Although
we usually think of solutions to be liquids, Table 13.1 (14.1 Corwin 6th )lists some common examples of solutions whose
physical state correspond to that of a solvent.
Module 8 we start with six properties
of a true solution which are not listed in chapter 4 as a separate list. You will
write five of these six properties for Part A:
1. It is
homogeneous mixture of two or more components, solute and
solvent
2. It
has variable composition, that is, the ratio of
solute and solvent may be varied.
3. The
dissolved solute is molecular or ionic in size
4. It may be
colored or colorless but it is usually transparent
5. The solute
remains uniformly distributed throughout the solution
and will not settle out with time (every drop
has exactly the same concentration)
6. The solute
generally can be separated from the solvent by purely
physical means (for example evaporation or distillation)
Part A: Solution Properties 02 points
List five of the six properties of a true solution:
(1)
(2)
(3)
(4)
(5)
(6)
Module 8: Solutions Chapter 4/5
Part B: Factors Affecting
Rate of Dissolving
Part B covers the dissolving process. A
discussion of solubility and temperature plus solubility and pressure of a gas
in a solution is as follows.
The following answers the first question:
State two factors greatly affecting the solubility of a gas in a
liquid:
(1) Temperature (increased temperature of
a solvent also generally increases the kinetic energy of the solute and the gas solute acquire more of a tendency
to escape from the solvent. Therefore, Cooling
the solvent increased the solubility of a gas in a liquid solvent.)
(2) Pressure (increasing the pressure
(partial pressure) of a gas solute increases the solubility proportionally of
that solute in the liquid (Henry’s Law)
The
properties of liquids dissolved in liquids focus on the main property which is
polarity. Now is the time to review polar covalent bonds chapter 7. Polar covalent bonds depend on the electronegativities of the two elements.
If two
elements differ in electronegativity between 0.4 and 1.7, then polar covalent
bonds are formed. If the difference between the two atoms is greater than 1.7
then ionization takes place. A solution containing ions must be dissolved by
polar molecules as a solvent.
Next you must
understand the three dimensional geometry of the molecules to determine if a
molecule is polar. Molecules with polar covalent bonds have dipoles, which are
vectors, created by the polar covalent bonds When they
are summed, if there is a net moment of force the molecule is polar. However,
it is possible for a compound to have polar covalent bonds and be nonpolar when the net summation of the vectors total zero .
The “Like
Dissolves Like Rule” depends on the polarity of the molecules of the solute and
the solvent (Table14.3 Corwin 6th is the same as Table 13.3 on page
378..
What is the main factor affecting the solubility of a liquid in
a liquid:
(3) Nature of the solute and solvent: the like dissolves like rule. The general principle that solubility is greatest when the
polarity of the solute is similar to that of the solvent
The discussion
of the dissolving leads to the rate of dissolving, which is the next question
in Section B of Module 8(They do not list #3 below.).
State four factors which governs the
rate of dissolving a solid in a liquid:
1. Particle
Size (increased surface area increases rate of solution i.e
powders have greater surface area than crystals
and will dissolve faster)
2. Temperature (increased
temperature of solvent generally increases rate of solution, except
gases in liquids is opposite)
3. Concentration of Solution- when
the solute and solvent are first mixed the rate of dissolving is at a maximum,
as saturation approaches the rate of dissolving slows
4. Agitation or stirring-the
effect of agitation is kinetic which increases the
rate of solution.
Part B: Factors Affecting Rate of Dissolving 02 points
State two factors greatly affecting the
solubility of a gas in a liquid and explain:
(1)
(2)
What is the main factor affecting the
solubility of a liquid in a liquid and explain the rule:
(3)
State four
factors which governs the Rate of dissolving a solid in a liquid:
(1)
(2)
(3)
(4)
Part C:
Writing Ionization Reactions
There is a discussion of the solubility of compounds in water
which produce ionic solutions. Strong Acids and Weak Acids are shown in table 14.2 below:
The
convention for writing the iodination reaction is wrong as the weak acid should
be a double arrow (reversible reaction), but the reversible reaction is not
discussed until 2046, therefore he shows a single arrow (which will be counted
wrong on the test).
Likewise, Strong
and Weak Bases are show:
And the
ionization reaction of a weak base should also be a reversible arrow or double
arrow
Soluble
Salts are show as a single direction arrow, while Insoluble salts have a double
arrow for a reversible reaction. Strong (Soluble) Electrolytes a and Weak
Electrolytes (Insoluble in water) are later in the text
These
sections lead to section 14.11 Writing Net Ionic Reactions from Corwin.
From the textbook (Corwin 6th ):
Part C: Writing Ionization Reactions 02 points
(Strong Acids/Weak Acids/Strong Bases/Weak Bases/
Soluble Salts and Oxides of Nonmetals and Metals)
Write the ionization reactions for the following:
(1) Strong Acid:
HCl (aq)
(2) Weak Acid:
HC2H3O2 (aq)
(3) Strong Base: NaOH (aq)
(4)
Weak Base: NH3 (aq)
(5)
Soluble Salts: NaCl (aq)
(6) Oxides of Nonmetals: CO2 (g)
(7) Oxides of Metals: CaO
(s)
Part D: Solution Preparation
Problems
There are three measurements of solutions in preparation problems of
which two will be given and the third will be asked in Part D for
preparing a solution in a laboratory. The three are: mass
of solute, volume of solution (not volume
of solvent-you should know the difference), and the concentration
of the solution.
There are six methods of measuring the concentration of a solution: Molarity, Weight (Mass) Percent, Volume Percent, Molality, Parts Per Million, and
Normality. Problems for Part D will focus mainly on Molarity, but Weight percent is also fair game. The
other four methods of measuring concentration will not be asked in Part D.
If the problem states the mass of the solute and the volume of the
solution prepared is given, then the Molarity is
unknown for one problem type. The other common problem is how to make a known
volume of a known concentration of a solution and you have to find the mass.
Part D: Solution Preparation Problems 03 points
1. How many grams of solute are needed to prepare 250 ml of a 0.0100 M KMnO4?
2. 20 grams of AgNO3 were placed in a 250 ml volumetric flask, calculate the Molarity of the Solution.
Part E: Solution Dilution/Concentration
Problems
In Section 13.10 of Chapter 13 there is a
second type of method for preparing a solution. On page 391 Dilution of a
Solution is discussed. On page 392 is Example 13.10 of diluting a more
concentrated solution. At the end of the chapter are two additional
exercises #65 and #68 on page 401-402.
Part E: Solution Dilution/Concentration Problems 03 points
1. How many milliliters of a 12 M HCl (concentrated) are required to make 2.00 L of a 1.00 M HCl?
2. A stock bottle of concentrated nitric acid indicates that the solution is 67.0 HNO3 by mass and has a density of 1.40 g/ml. Calculate the molarity of concentrated nitric acid.
Part F: Solution Reaction Problems
Part F: Solution Reaction Problems 5 points
Acid-Base Neutralization (2.5 Points):
1. Calculate the molarity of a Calcium hydroxide solution if 18.50 mL of it requires 28.27 mL of a 0.0125 M HCl to reach its neutralization point.
REDOX Titration: (2.5 Points)(Ask
for Balanced)
2. A KMnO4 (aq) solution is to be standardized by titration against As2O3(s), A 0.1156 g sample of As2O3 requires 27.08 mL of KMnO4. What is the molarity of the KMnO4?
As2O3 + MnO4
1- + H2O +
H1+ -------> H3AsO4 +
Mn 2+
Module
Eight: Part G Rewriting Equations Ionically
In Section 14.11 on page 427 is the
discussion of Net Ionic Equations. To do Part H, REOX Equations, you need to be good
at writing equations in aqueous solutions ionically. Example Exercise 15.15 (14.15 Corwin 7th ) on page 428 and Exercise 15.16 (14.16
Corwin 7th )on page 429 demonstrate the process step by step. You
should work the Practice Exercise on page 429 plus the end of the chapter
exercises #71-76 on page 436.
The following Rules are written at the top
of the Part G test:
Show as ions: soluble salts (aq) and strong acids (aq); leave
as molecules/formula units insoluble salts (s), weak acids (aq),
covalent molecules. (Strong hydroxide bases are hydroxides which are also
soluble salts are written ionically.)
Strong Acids are: Perchloric Acid; Hydrochloric Acid; Nitric Acid; Sulfuric Acid; Hydrobromic Acid; Hydroiodic
Acid.
Strong Bases are: Sodium hydroxide, Potassium hydroxide, Calcium hydroxide, Barium hydroxide,
Strontium hydroxide
Here are a
set of worked examples:
1. KOH (aq) + HNO3(aq) à KNO3(aq) + HOH(l)
Decide
which compounds should be split into ions:
K|OH (aq) +
H|NO3(aq) à K|NO3(aq) + HOH(l)
Strong
base strong acid soluble salt covalent molecule
Rewrite
the ionic species as ions, and leave the rest as compound formulas:
K1+(aq)+OH1-(aq)+H1+(aq)+NO31-(aq)àK1+(aq)+NO31-(aq)+HOH(l)
Cancel the
Spectator Ions:
K1+(aq)+OH1-(aq)+H1+(aq)+NO31-(aq)àK1+(aq)+NO31-(aq)+HOH(l)
Final
Net Ionic Answer:
1. KOH (aq) + HNO3(aq) à KNO3(aq) + HOH(l)
OH 1-(aq) + H 1+(aq) à
HOH (l)
------------------------------------------------------------------------------
2.
CuSO4
(aq) + Na2CO3
(aq) à CuCO3 (s) +
Na2SO4
(aq)
Decide
which compounds should be split into ions:
Cu|SO4 (aq) +
Na2|CO3
(aq) à CuCO3 (s)
+ Na2|SO4
(aq)
Soluble Salt Soluble Salt Insoluble salt Soluble Salt
Rewrite
the ionic species as ions, and leave the rest as compound formulas
Cu2+(aq)+SO42-(aq)+2Na1+(aq)+CO32-(aq)à
CuCO3(s) + 2Na1+(aq)+SO42-(aq)
Cancel the
Spectator Ions:
Cu2+(aq)+SO42-(aq)+2Na1+(aq)+CO32-(aq)à
CuCO3(s) + 2Na1+(aq)+SO42-(aq)
Final Net Ionic Answer:
2. CuSO4
(aq) + Na2CO3
(aq) à CuCO3 (s)
+ Na2SO4
(aq)
Cu 2+ + CO3 2- à CuCO3 (s)
--------------------------------------------------------------------------------------------
3. NaOH (aq) + NH4NO3
(aq) à NaNO3
(aq) + NH3
(g) + HOH
(l)
Decide
which compounds should be split into ions:
Na|OH (aq) + NH4|NO3
(aq) à Na|NO3
(aq) + NH3 (g) + HOH (l)
Strong Base Soluble Salt Soluble Salt molecular molecular
Rewrite
the ionic species as ions, and leave the rest as compound formulas:
Na1+(aq)+OH1-(aq)+NH41+(aq)+NO31-(aq)à
Na1+(aq)+NO31-(aq) +
NH3 (g)+HOH(l)
Cancel the
Spectator Ions:
Na1+(aq)+OH1-(aq)+NH41+(aq)+NO31-(aq)à
Na1+(aq)+NO31-(aq) +
NH3
(g)+HOH(l)
Final Net Ionic Answer:
3. NaOH
(aq) + NH4NO3
(aq) à NaNO3
(aq) + NH3
(g) +
HOH (l)
OH 1- + NH4 1+ à NH3 (g) + H2O (l)
------------------------------------------------------------------------------------------
4.
BaBr2
(aq) + ZnSO4
(aq) à BaSO4
(s) + ZnBr2 (aq)
Decide
which compounds should be split into ions:
Ba|Br2
(aq) + Zn|SO4
(aq) à BaSO4 (s)
+ Zn|Br2 (aq)
Soluble Salt Soluble Salt Insoluble salt Soluble Salt
Rewrite
the ionic species as ions, and leave the rest as compound formulas
Ba2+(aq)+ 2Br1-(aq) + Zn2+(aq) + SO42-(aq)à
BaSO4(s) + Zn2+aq) + 2Br1-(aq)
Cancel the
Spectator Ions:
Ba2+(aq) + 2Br1-(aq) + Zn2+(aq) + SO42-(aq)à
BaSO4(s) + Zn2+(aq) + 2Br1-(aq)
Final Net Ionic Answer:
4. BaBr2
(aq) + ZnSO4
(aq) à BaSO4
(s) + ZnBr2 (aq)
Ba 2+ + SO4 2- à BaSO4 (s)
---------------------------------------------------------------------
5.
Cr(OH)2
(s) +
HCl (aq) à
First
Complete the Product:
Cr(OH)2
(s) +
2HCl (aq) à
CrCl2 (aq) +
HOH (l)
Decide
which compounds should be split into ions:
Cr(OH)2 (s)
+ 2 H|Cl (aq) à
Cr|Cl2
(aq) + HOH (l)
Insoluble Salt Strong Acid Soluble Salt Molecular
Rewrite
the ionic species as ions, and leave the rest as compound formulas:
Cr(OH)2 (s) +2H1+(aq) + 2Cl1-(aq)àCr2+(aq)+ 2Cl1-(aq) + HOH(l)
Cancel the
Spectator Ions:
Cr(OH)2 (s)+ 2H1+(aq)+ 2Cl1-(aq)à Cr2+(aq)+ 2Cl1-(aq) + HOH(l)
Final Net Ionic Answer:
5. Cr(OH)2
(s) +
2 HCl
(aq)
à CrCl2 (aq) +
H2O
(l)
Cr(OH)2 (s) + 2 H 1+ (aq) à Cr 2+
(aq) + 2 H2O (l)
From the Corwin Textbook:
Video:
http://www.brightstorm.com/science/chemistry/chemical-reactions/net-ionic-equation/
Module
Eight: Part G Rewriting Equations Ionically 5 points
Rewrite the
following (unbalanced) equations ionically, cancel
spectator ions and then balance the net ionic reactions. Show as ions: soluble
salts and strong acids and strong bases;
leave as molecules/formula units insoluble salts, weak acids, covalent molecules.
Strong acids are: Perchloric
Acid; Hydrochloric Acid; Nitric Acid; Sulfuric Acid; Hydrobromic
Acid; Hydroiodic Acid.
Strong
bases are Sodium hydroxide, Potassium hydroxide, Calcium hydroxide, Barium
hydroxide and Strontium hydroxide
1. KOH
(aq)
+ HNO3(aq) à KNO3(aq) + HOH(l)
2. CuSO4 (aq) + Na2CO3 (aq) à CuCO3 (s)
+
Na2SO4 (aq)
3. NaOH (aq) +
NH4NO3 (aq) à NaNO3 (aq) + NH3 (g) +
HOH (l)
4. BaBr2 (aq) + ZnSO4 (aq) à BaSO4 (s) +
ZnBr2 (aq)
5. Cr(OH)2
(s) +
HCl (aq) à
Module Eight Homework Packet
Module
8 Part H: Redox
Equations 6 points
A separate Study
Guide will be distributed along with additional Homework.
Balance the
following redox equations written in net ionic form:
Acid Media: (3 points)
1. C2O4
2- (aq) +
MnO4 1- (aq) + H 1+ (aq)
→ Mn 2+ (aq) + CO2
(g) +
HOH (l)
half equation:
half equation:
Basic Media (3 points)
2. Bi2O3
(s) + OH 1- (aq) + OCl 1-
(aq) → BiO3
1- (aq) + Cl 1-
(aq) + HOH (l)
half equation:
half equation:
Module Eight Homework Packet
Part I: Solution Discussion Questions 2 points
Answer the following:
(1) Define oxidation and reduction in terms of electron transfer.
(2) In an oxidation reduction equation, what is the oxidizing agent?
What is the reducing agent?
(3) How can you recognize whether a reaction is a redox reaction or not?
(4) Name three types of reactions which are NOT oxidation reduction and give at least one example
(5) What is the equivalence point in a titration of an acid and a base? Is there a difference between the equivalence point and the end point of the titration when using an acid/base indicator to determine the end point?
Key Strokes:
Negative Sign then Log Key then [H3O1+]
then Exe = pH
Key Strokes:
2nd Sign then Log Key then Neg Sign then pH then Exe = [H3O1+]
Part L: pH Scale Calculations (Section 15.8-15.9) 3 points
What is the pH and pOH of the following solutions:
pOH = 14.00 –
pH Kw
= 1.00 x 10-14 = [H3O1+][OH1-]
(1) 0.0100 M HCl
(2) 0.0055 M Mg(OH)2
(3) The pH of vinegar is 2.85,
calculate the [H3O1+]:
(4) Easy Off Oven Cleaner has a pH of 11.70. What is the [H3O1+]
and [OH1-]?
Module Eight Homework Packet
Part K: Solution Definitions 2 points
Fill in the blank with word for the definition listed:
1.______________When dynamic equilibrium is established between undissolved solute in a
Solution, we say the solution is __.
2. ______________Any solution containing less solute than it could hold at equilibrium is said
To be a(n) _______________solution.
3. ______________That part of a solution which you are dissolving.
4. ______________That part of a solution which does the dissolving (or the media of the
solution).
5. ______________are dispersions in which that the dispersed matter has one or more
dimensions in the range of 1 to 1000 nm and has the unique ability to scatter light.
6. ______________of a solution is the number of moles of solute per Liter of solution.
7. ______________of a solution is the grams of solute per 100 grams of solution
8._______________of a solution is the milliliters of solute per 100 milliliters of solution.
9.______________homogeneous mixture of a solute and a solvent.
10._____________ .Any solution containing more solute than it could hold at equilibrium is said
To be a(n) _______________solution.
Define Normality; define equivalent.
What is the difference between a solution, colloidal solution, colloidal suspension, and a suspension