Download Discussion 9, Mahaffy et al., Chapter 15

Chemistry 102 2017 Discussion #9, Chapter 15
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Things you should know when you leave Discussion today:
http://quantum.bu.edu/courses/ch102-spring-2017/notes/AcidBaseOverview.pdf
Solubility is the number of moles of the solid that dissolves in 1L of solution to form
saturated solution. Keq = Ksp – Solubility product constant
Solubility product: MX2 (s) <=> M2+ (aq)+ 2X-(aq)
Ksp= [M2+ (aq)][ X‾ (aq)]2
Oxidation Reduction Reactions
1. The Ksp of CaF2 is 4.0∙10-11, what is the molar solubility “s” for this compound in water?
a. Chemical equilibrium reaction:
b. Construct the ICE table:
c. Ksp=
d. s=
e. What is [Ca2+] =
and
[F-]=
2. What is the molar solubility “s” of CaF2 in an aqueous solution of 0.10 M NaF?
(Answers: 4.0∙10-9, 0.10)
a. Chemical equilibrium reaction:
b. Construct an ICE table:
c. Ksp=
d. s=
[Ca2+] =
[F-]=
1
3. What is the molar solubility “s” of CaF2 in an aqueous solution of 0.10 M CaCl2?
(Answers: 1.0∙10-5 , 2.0∙10-5 , 0.10)
a. Chemical equilibrium reaction:
b. Construct an ICE table:
c. Ksp=
d. s=
[Ca2+] =
and
[F-]=
4. If you mix 0.5L of 0.1M NaF with 0.5L of 0.1M CaCl2, there is the possibility of CaF2(s)
forming.
a. With regards to CaF2, circle the correct choice:
Qsp < Ksp
Qsp = Ksp
Qsp > Ksp
b. With regards to CaF2, circle the correct choice
 Precipitation will occur
 Precipitation will not occur
 More information needed
5. The number of moles of a solid that dissolves in 1 liter of water is called the molar solubility. For
Ag2SO4, the molar solubility is 1.4∙10-2 M. What is the Ksp for this compound? (Answers:1.1∙10-5)
a. Chemical equilibrium reaction:
b. Construct an ICE table:
c. Ksp=
2
6. A solution is made by adding 1 mmol of NaCl and 1 mmol of AgNO3 to 1 L of water. Will a
precipitate form? How many moles of AgCl will precipitate ? Ksp(AgCl)=1.6∙10–10
(Answer:1mmol)
7. A solution is prepared from 0.02 mol of MgCl2 and 0.004 mol of NaOH in 1L of water. What is
the pH of the solution? The Ksp of Mg(OH)2 is 6 x10−12. (Answer:9.3)
8. An 80. mL sample of 0.040 M lead nitrate, Pb(NO3)2, is titrated with 0.010 M sodium fluoride,
NaF, until the first appearance of precipitate. The precipitate first appears after 20. mL of sodium
fluoride are added. You can assume that the total final volume is 100. mL. What is the value of
Ksp for PbF2? (Answer: 1.3*10-7)
9. The molar solubility of silver chloride is 4.0x10−5. Calculate the Ksp of silver chloride.(Answer:
16*10-10)
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Things you should know and review:
Oxidation Reduction Reactions
a. Oxidation is loss of electrons (acts as a reducing agent)
b.Reduction is gain of electrons (acts as a oxidizing agent)
Assigning Oxidation numbers
c. Oxidation number is 0 for atoms in an element.
d.The sum of all oxidation numbers in a molecule or ion must add up to the total
charge.
e. In compounds, alkalis (group 1) have oxidation number +1; alkaline earths
(group 2) have oxidation number +2.
f. In compounds, fluorine (F) always has oxidation number -1. Other halogens
(Cl ext.) have oxidation number -1, except when bonded to elements that are
more electronegative such as fluorine or oxygen, where they can have positive
oxidation numbers.
g.In compounds, hydrogen has oxidation number +1 (when bonded to elements
with higher electronegativity). When bonded to metals hydrogen has oxidation
number -1 (when bonded to elements with lower elecronegativity).
h.In compounds, oxygen has oxidation number -2, except when bonded to
fluorine, where fluorine's rule takes precedence (in OF2 oxygen has oxidation
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number +2), and if there are O-O (peroxide) bonds. (in H2O2 or Na2O2 oxygen
has oxidation number -1).
Steps for Balancing Redox reactions in Acidic or Basic media (always start with
acidic).
Step 1. Write the two half reactions.
Step 2. Balance all the elements except for O and H.
Step 3. Balance O by adding H2O.
Step 4. Balance H by adding H+.
Step 5. Balance charge by adding electrons to the side of the half reaction that is
most positive.
Step 6. Balance electrons lost in one half reactions with electrons gained in the
other half reaction.
Step 7. Add half reactions together canceling what can be canceled.
Step 8. Determine if your answer should be in acidic media or basic media
If in acidic media: Add water to both sides in order to convert all H+ to H3O+
a) Combine H+ and H2O to form H3O+
b) Simplify the reaction by canceling the H2O that appear on both sides
If in basic media: Add enough OH- to neutralize H+ to both sides of the reaction.
a) Combine OH- and H+ to form H2O.
b) Simplify the reaction by canceling H2O that appear on both sides.
http://quantum.bu.edu/courses/ch102-spring-2016/notes/BalancingRedoxEquations.pdf
1. Balance the following reaction: MnO4-(aq) + C2O42- (aq)  MnO2 (aq) + CO2 (g).
Oxidation half RXN:
Reduction half RXN:
Net RXN in Acidic Media:
Net RXN in Basic Media:
4
2. Balance the following reaction: IO-3(aq) + I-(aq)  I2(aq).
Oxidation half RXN:
Reduction half RXN:
Net RXN in Acidic Media:
Net RXN in Basic Media:
3. Balance the redox reaction of H2 (g) reducing Au3+ (aq) to produce solid gold in an acidic
solution.
4. Balance the following reaction in basic media: Fe2+ (aq) + MnO4- (aq) → Fe3+ (aq) + Mn2+ (aq)
5. Use the steps provided to balance following oxidation reduction reactions in acidic and basic
media.
a. Cu(s) + NO3- (aq)  Cu2+(aq) + NO2(g)
b. S2-(aq) + I2(aq)  SO42-(aq) + I- (aq)
6. Write down the balanced oxidation half-reaction for the complete combustion of glucose,
C6H12O6, in aqueous acidic solution.
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In preparation for the weeks ahead:
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Potential E°cell at standard conditions.(all concentrations are at 1M)
Cell potential E° [J/c = J/charge] is a measure of electrical potential difference.
E °cell = E °red(cathode) - E °red(anode) = E °red + E °ox
0.05912V
E° =
logK (at 25°C)
ne
a. http://quantum.bu.edu/courses/ch102-spring-2016/notes/StandardRedcutionPotentials.pdf
b. Pages 619-643, Chemistry: Human Activity, Chemical reactivity, Mahaffy et al., 2e
1. Choose all that must be true about standard reduction potential E°:
i. Most positive reduction potential represents the strongest Oxidizing Agent (an
oxidizing agent is an electron acceptor and is most likely to get reduced)
ii. Most negative reduction potential represents strongest Reducing Agent (most
likely to get oxidized)
iii. E° is an Intensive quantity as long as concentration does not change it does not
depend on the size of the cell
iv. E° does not depend on the amount of charge transferred.
v. E° does not depend on stoichiometry. (doubling the stoichiometry of all the
species in the reaction has no effect on the E°)
vi. Sign of the cell potential tells the direction of electron flow.
2. Choose all that must be true:
i. Anode is an electrode where oxidation takes place
ii. Cathode is an electrode where reduction takes place
Exam 2 Answers:
1.
2.
3.
4.
5.
8. 1.3∙10-8
9. 1.40∙105g/mol
10. 6.6M
11. -5.24ºC
12. 0.0197mol
13. 6∙10-5
14. 9.48∙10-14
0.309
1.2%
Kb(NH3) < Ka(HB)
0.101g
a. < 0.57 bar
b. < 0.57 bar
c. > 0.57 bar
6. 3Mg2+(g) + 2PO43-(g) 3Mg2+(aq) +
2PO43-(aq)
7. NaCl, BeO
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