Download Answer Key, Problem Set 6 – complete, with explanations

Chemistry 121
Mines, Fall, 2016
Answer Key, Problem Set 6 – complete, with explanations
1. MP (4.62); 2. MP (4.64); 3. MP (4.65); 4. WO1; 5. WO2; 6. WO3; 7. MP (4.71); 8. MP (4.74); 9. MP (4.73); 10. WO4; 11. WO5;
12. WO6; 13. WO7; 14. MP (4.90); 15. MP (4.94); 16. MP (4.96); 17. WO8; 18. MP (4.98); 19. MP (4.97); 20. WO9; 21. WO10; 22.
MP (a Mastering Problem not in text)
---------------------------------------------------------------------------------------------------------------------------------------Molarity [Dilution], Stoichiometry w/ Molarity
1. 4.62.
If 3.5 L of a 4.8 M SrCl2 solution is diluted to 45 L, what is the molarity of the diluted solution?
Answer: 0.37 M SrCl2
Strategy: Remember that during a dilution, the moles of solute do not change—only the volume changes.
V
Hence M2V2 = M1V1  M2 = M1 x 1 (Note that V1/V2 is the inverse of the “dilution factor”)
V2
M 2  4.8 M x
3.5 L
 0.373...  0.37 M SrCl 2
45 L
2. 4.64. Answer not in key yet. Use same dilution ideas and relationship as shown in the prior problem’s
solution. You just have a different “given” and a different “unknown” variable here.
3. 4.65.
Consider the precipitation reaction (represented by):
2 Na3PO4(aq) + 3 CuCl2(aq) → Cu3(PO4)2(s) + 6 NaCl(aq)
What volume of 0.175 M Na3PO4(aq) solution is necessary to completely react with 95.4 mL of 0.102 M CuCl2(aq) ?
Answer: 37.1 mL (or 0.0371 L)
Strategy: 1) Use V (converted to L) and M (mol/L) of CuCl2(aq) to calculate moles of CuCl2.
2) Use the balanced equation / stoichiometry to calculate the moles of Na3PO4 needed.
3) Use moles and M (mol/L) of Na3PO4(aq) to calculate VNa3PO4 (in L).
Execution of Strategy:
VNa PO  95.4 mL CuCl 2 (aq) x
2
4
0.006487.. mol Na 3 PO 4 x
0.102 mol CuCl 2 2 mol Na 3 PO 4
1L
x
x
 0.006487.. mol Na 3 PO 4
1000 mL
1 L CuCl 2 (aq)
3 mol CuCl 2
(2)
(1)
1L
 0.03706..  0.0371 L ( 37.1 mL)
0.175 mol Na 3 PO 4
PS6-1
Answer Key, Problem Set 6
Nanoscopic pics of ionic compounds, Molarity of Ions (in Strong Electrolyte solutions)
4. WO1. (a) Draw a picture of a sample of solid sodium carbonate containing four formula units of compound. (b)
Draw a snapshot of your sample of compound after it has dissolved in a beaker of water. Do not draw the water
molecules—just represent how the sample of compound would look in the solution. Make sure that your second picture
has the same amount of compound in it as is depicted in the solid piece that you drew in part (a).
Na atom (or ion) =
CO32formula = Na2CO3  2 Na+'s per CO32-4 F.U. means
8 Na+'s and
4 CO32 's
C atom =
O atom =
ions separate from each other in solution
4 formula units of sodium carbonate dissolved in water
solid sodium carbonate(4 formula units)
Obviously, these are only 2D representations. The picture of the lattice is obviously not “correct”; note
that I have put the two positive ions next to one another, which wouldn’t happen in a real ionic lattice
(solid). My goal here is to represent certain aspects of the lattice and resulting solution:
1)
The sodium ions are not associated with any particular carbonate ion in the lattice; they are equally spaced
between multiple anions. This would be true in a “real” lattice; there are no molecules in an ionic compound!
There are only alternating positive and negative ions.
2)
The carbonate ions do have their atoms connected to one another because carbonate is a “polyatomic
anion”. They look like molecules in these pictures, but they are not molecules because each “three-oxygensand-a-carbon” grouping has an overall 2- charge.
3)
There are two Na+ ions per each CO32- ion, as indicated by the formula.
4)
The cations separate from the anions when any strong electrolyte is dissolved in water. Note that the
atoms of the polyatomic ion do not separate from each other (that is, the individual carbonate ions
themselves remain intact).
5)
The ions are freely moving about in the solution; I try to show this by showing them randomly distributed
throughout. A solution is a homogeneous mixture. Even though there are two different types of ions (as well
as water molecules, which are not shown), if you had a macroscopic sample of the solution represented
here, you could not pick out a drop of solution that would have a different composition than another drop
taken from elsewhere in the same solution.
5. WO2.
In each of the following solutions, calculate the concentration (in M) of i) the solute (i.e., the compound
dissolved), ii) the cation, and iii) the anion (E.g., in the first one, calculate [Na3PO4], [Na+], and [PO43-] in the solution
described)
Answers: (a) 2.00 M Na3PO4, 6.00 M Na+, 2.00 M PO43-; (b) 0.666 M (NH4)2SO4, 1.33 M NH4+, 0.666 M SO42(a) 0.0200 mol of sodium phosphate dissolved in 10.0 mL of solution
i) [Na3PO4] 
0.0200 mol Na 3 PO 4
-3
10.0 mL x
10 L
solution
1 mL

0.0200 mol Na 3 PO 4
 2.00 M Na 3 PO 4
0.0100 L solution
ii) [Na+] = 3 x [Na3PO4] (because there are “3 mol Na+ ions per mol of Na3PO4” (subscript is “3”)):
[Na+] 
2.00 mol Na 3 PO 4
3 mol Na 
x
 6.00 M Na 
L solution
1 mol Na 3 PO 4
(Na3PO4(aq)  3 Na+(aq)
iii) [PO43-] = [Na3PO4] (because there is 1 mol PO43- per mol Na3PO4 (subscript is “1”)
PS6-2
+ PO43-(aq))
Answer Key, Problem Set 6
3
[PO43-] 
2.00 mol Na 3 PO 4
1 mol PO 4
3
x
 2.00 M PO 4
L solution
1 mol Na 3 PO 4
(d) 132 g of ammonium sulfate dissolved in 1.50 L of solution
Molar mass (NH4)2SO4 = 2(14.01) + 8(1.01) + 32.07 + 4(16.00) = 132.17 g/mol
132 g
 0.9987..mol (NH4 )2SO4 ;
132.17 g/mol
i) [(NH4)2SO4] 
Subscripts indicate: (NH4)2SO4(aq)  2 NH4+(aq) + SO42-(aq)
0.9987.. mol (NH 4 ) 2 SO 4
 0.6658  0.666 M (NH 4 ) 2 SO 4
1.50 L solution

ii) [NH4+] 
0.6658 mol (NH 4 ) 2 SO 4
2 mol NH 4

x
 1.3316 M  1.33 M NH 4
L solution
1 mol (NH 4 ) 2 SO 4
iii) [SO42-] 
0.6658 mol (NH 4 ) 2 SO 4
1 mol SO 4
2
x
 0.666 M SO 4
L solution
1 mol (NH 4 ) 2 SO 4
2
6. WO3.
Calculate the sodium ion concentration when 70.0 mL of 3.0 M sodium carbonate is added to 30.0 mL of 1.0 M
sodium hydrogen carbonate.
Answer: 4.5 M Na+
Strategy:
1) Recognize that the final [Na+] is just equal to
total moles Na+
total volume of solution
2) Calculate the moles of Na+ in each of the two solutions using volume and molarity, and by
looking at their chemical formulas and recognizing that they are strong electrolytes.
3) Sum the two volumes (70.0 mL and 30.0 mL) and convert to liters to get the final volume, in L.
4) Divide the total moles of Na+ by the total volume.
Execution:
70.0 mL x
3.0 moles Na 2CO3
1L
2 moles Na+
x
x
 0.42 moles Na+ (from Na 2CO3 )
1000 mL
1L
1 mol Na 2CO3
Converts mL to L
30.0 mL x
Na2CO3(s)  2 Na+(aq) + CO32-(aq)
1.0 moles NaHCO3
1L
1 moles Na+
x
x
 0.030 moles Na+ (from NaHCO3 )
1000 mL
1L
1 mol NaHCO3
NaHCO3(s)  Na+(aq) + HCO3-(aq)
[Na+] =
total moles Na+
0.42 + 0.030 mol Na+

 4.5 M Na+
total volume of solution 100.0 mL
1000 mL/L
Electrolytes, Solubility Rules, Ions in SE solutions
7. 4.71.
For each compound (all water soluble), would you expect the resulting aqueous solution to conduct electrical
current?
NOTE: This problem is essentially asking you if the compounds are strong electrolytes (technically, they should have
clarified this, since a solution with a substantial concentration of a weak electrolyte can also conduct electricity pretty well. But none of
the choices are weak electrolytes here—they clearly were looking for strong vs. non). The classes of compounds that are
strong electrolytes are: 1) soluble ionic compounds and 2) the six strong acids. Since the problem
PS6-3
Answer Key, Problem Set 6
STATES that all of the compounds are soluble, this question further reduces to the question: “Are the
following substances ionic or one of the six strong acids?”!
(a) CsCl
YES (this is an ionic compound [and it’s water soluble]  strong electrolyte / conductive sol’n)
(b) CH3OH NO (this is a molecular compound (an “alcohol”) that is not—don’t let the “OH” fool you
into thinking this is a hydroxide compound—it isn’t! There is no metal or “NH4” listed
first!! So this is a nonelectrolyte)
(c) Ca(NO2)2 YES (ionic compound and water soluble  strong electrolyte / conductive sol’n)
(d) C6H12O5 NO (this is a molecular compound that is not one of the six strong acids; nonelectrolyte)
8. 4.74.
Determine whether each compound is soluble or insoluble. For the soluble compounds, list the ions present in
solution.
Strategy: Use Table 4.1 (or the solubility “rules” that were formatted slightly differently in my
PowerPoint presentation). Make sure you understand how to apply the rules.
(a) AgI INSOLUBLE. Cl-, Br-, & I- compounds are soluble, unless the cation is Ag+, Hg22+, or Pb2+
(b) Cu3(PO4)2 INSOLUBLE. PO43- compounds are generally insoluble. (The cation here is not a
Group I metal ion or NH4+, which would “make” the phosphate compound soluble.)
(c) CoCO3 INSOLUBLE. CO32- compounds are generally insoluble. (The cation here is not a Group
I metal ion or NH4+.)
(d) K3PO4. SOLUBLE. Although phosphates are generally insoluble, the cation here is K+ , which is
a Group I metal ion. Thus, soluble.
9. 4.73 Answer not in key yet. Problem is analogous to the prior problem (4.70); see above for strategy.
Predicting the products in a potential exchange [precipitation] reaction
10. WO4. Tro, 4.78 (w/ part (e) added) Write a molecular equation for the precipitation reaction that occurs (if any)
when each pair of aqueous solutions is mixed. If no reaction occurs, write NO REACTION.
Answers:
(a) 2 NaCl(aq) + Pb(C2H3O2)2(aq) → 2 NaC2H3O2(aq) + PbCl2(s)
(b) K2SO4(aq) + SrI2(aq) → 2 KI(aq) + SrSO4(s)
(c) NO REACTION (2 CsCl(aq) + CaS(aq) → Cs2S(aq)
(d) Cr(NO3)3(aq) + Na3PO4(aq)
(e)
+ CaCl2(aq) ) Just a bunch of ions in solution!
→ CrPO4(s) + 3 NaNO3(aq)
2 Fe(NO3)3(aq) + 3 Na2S(aq)  Fe2S3(s) + 6 NaNO3(aq)
Strategy: 1)
Swap the anions to get possible products.
2) Use the two solubility rules that you must memorize, along with the other rules (Table 4.1)
to decide if either of the potential product compounds is insoluble.
a) If “yes”, then that insoluble ionic compound is the precipitate (and a precipitation
reaction occurs)
b) If “no”, then no precipitate forms.
3) Use the neutrality principle, along with your knowledge of the cation / anion formulas and
charges to create the proper formulas of each product and reactant compound. Review
PS3 key for how to do this.
4) Express as a molecular equation and then balance it.
PS6-4
Answer Key, Problem Set 6
Execution of Strategy (for each)
(a) sodium chloride and lead(II) acetate
swap anions to get
sodium acetate and lead(II) chloride
soluble (Na+)
Formulas of ions:

Na+
Cl-
Pb2+
insoluble (Cl- exception)
C2H3O2-
NaCl + Pb(C2H3O2)2 → NaC2H3O2 + PbCl2(s) (Balancing and adding (aq) state designations
leads to answer shown in box above)
swap anions to get
(b) potassium sulfate and strontium iodide
potassium iodide and strontium sulfate
soluble (K+)
Formulas of ions:

K+
I-
Sr2+
insoluble (SO42- exception)
SO42-
K2SO4 + SrI2 → KI + SrSO4(s) (Balancing and adding (aq) state designations leads to answer
shown in box above)
(c) cesium chloride and calcium sulfide
swap anions to get
cesium sulfide and calcium chloride
soluble (Cs+)
soluble (Cl-
)
NO REACTION (because both reactants and both “products” are strong electrolytes;
all ions
“before reaction” and all ions “after reaction”  nothing actually happened!)
swap anions to get
(d) chromium(III) nitrate and sodium phosphate
chromium(III) phosphate and sodium nitrate
insoluble (PO43-)
Formulas of ions:

(e)
Cr3+
NO3-
Na+
Cr(NO3)3 + Na3PO4 → CrPO4(s) + NaNO3 (Balancing and adding (aq) state designations
leads to answer shown in box above)
2) & 3) swap anions to get
Fe(NO3)3(aq) + Na2S(aq)
1) “separate” into ions
(just to create new formulas)
Ions:

soluble (Na+)
PO43-
Fe3+
NO3-
Na+
Fe2S3
and
insoluble (S2-)
S2-
(Balancing and adding all state designations leads to answer shown in box above)
PS6-5
NaNO3
soluble (Na+, NO3-)
Answer Key, Problem Set 6
Nanoscopic representation of precipitation reaction, Net Ionic Equations (NIE’s)
11. WO5.
Given the following equation: 2 CsF(aq) + CaCl2(aq)  CaF2(s) + 2 CsCl (aq),
Draw a representation of what a solution with 8 formula units of CsF and a solution with 5 formula units of CaCl2
would look like, and then draw what a beaker would look like if the two solutions were mixed and the reaction was
allowed to proceed to completion. Write the correct number and proper depiction of all species. Use the symbols
below in your pictures.
(b) Write the net ionic equation representing the reaction that occurred, and then describe in words what actually
(a)
occurs (changes) during this reaction.
Cs+ =
F =
Ca2+ =
Cl =
Answers:
mix
8 F.U. CsF(aq)
(b) Net ionic equation:
5 F.U. CaCl2(aq)
after mixing and reaction is complete
Ca2+(aq) + 2 F-(aq)  CaF2(s)
(the order of Ca2+ and F- on the left side could be reversed)
Reasoning / Strategy:
For (a):
1) First recognize that 8 FU of CsF is composed of 8 Cs+ ions and 8 F- ions, and 5 FU of CaCl2 is composed of 5
Ca2+ ions and 2x 5 = 10 FU of Cl- ions (i.e., interpret the formulas correctly). Draw those ions in the two boxes at
the left, recognizing that they should be shown as homogeneous mixtures (i.e., ions spread out in the solution,
and randomly “mixed up”).
2) To figure out the box on the right, you could use the balanced equation as a “recipe” and circle “equation units’”
worth of reaction that occur, one at a time, to see a) how many “cycles” of reaction occur (and thus how many
“products” form) as well as whether or not one reactant runs out before another (leaving “leftover reactant ions”)
(analogously to WO1 on PS5 and worksheet problem done in class on the board):
mix
Since four “equation units” of reaction
occur before the CsF FUs run out (four
sets of “circles”), you should end up
with four “equation units” of products
formed (which equals 4 CaF2(s) and 8
CsCl(aq). Also, note that there is one
FU of CaCl2 left over after the reaction,
which just remains in the solution (and
thus must be shown in the “after” box.
You could also have treated this like a limiting reactant type of problem (PS5) without doing the “circling”.
3) “Translate” the remaining information into “pictures” by (carefully) representing each type of formula unit properly
in the “after mixing” box:
i) Draw the 8 FU of CsCl(aq) as separated ions, roughly evenly spread out in it, because CsCl is a strong
electrolye.
PS6-6
Answer Key, Problem Set 6
ii) Draw the 4 FU of CaF2(s) (the precipitate / insoluble ionic compound) as one single crystal of the solid (since
there are so few FU’s of it here), much like the depiction in WO1 above (except since there are no polyatomic
ions, I have shown the ions as “touching” here—you could have shown them with a bit of space in between
them as well, as long as the amount of space in between was roughly “equal” for all adjacent ions). To further
stress this point:
Note that you cannot tell to which anions the cations belong (or vice versa) in the representation of the CaF2
precipitate. That’s because they DON’T “belong” to any one cation! The atoms in molecules are bound strongly
to one another but are only weakly attracted to atoms in other molecules; anions are attracted equally to all
cations around them and vice versa! So you must draw ionic solids in a way that does not make the formula
units “look like molecules”!
iii) Don’t forget to add the leftover (unreacted) FU of CaCl2, which should be shown as separated ions (because
CaCl2 is a strong electrolyte.
iv) Count up all your ions on both sides to make sure you haven’t “created or destroyed any atoms”! 
There are 8 Cs+, 8 F-, 5 Ca2+, and 10 Cl- ions in the left beakers, and the same amount in the mixed beaker.
Note : Macroscopic amounts of the two solutions on the left would be “clear” in real life (you could see through
them; there would be no opaqueness or cloudiness). Upon mixing, a precipitate would form, which is to say
that the solution would turn cloudy or solid clumps would be seen to appear and, with time, likely settle to the
bottom of the container (right hand picture above).
For (b):
Here, you could do this “visually” rather than from the molecular equation! Hopefully you can see
that (from the pictures), the only things that change are the white circles (F- ions) and the dark ones
with the x’s in them (the Ca2+ ions)—they come together to form an insoluble compound with
formula CaF2. Hence, the next ionic equation is just: Ca2+(aq) + 2 F-(aq)  CaF2(s)
Of course, you could also get to the net ionic equation the “traditional” way:
2 CsF(aq) + CaCl2(aq)  CaF2(s) + 2 CsCl (aq)
2
Cs+(aq)
12. WO6 4.79(d) & 4.80(d).
+ 2
F-(aq)
+ Ca2+(aq) + 2 Cl-(aq)  CaF2(s) + 2 Cs+(aq) + 2 Cl- (aq)
Write balanced complete ionic and net ionic equations for each reaction.
**Remember: Only write STRONG electrolytes as separated ions in a complete or net ionic equation.
WEAK ACIDS are not strong electrolytes. Nor are insoluble ionic solids. You must be careful!!**
Answers:
4.79(d)
6 Na+(aq) + 2 PO43-(aq) + 3 Ni2+(aq) + 6 Cl-(aq) → Ni3(PO4)2(s) + 6 Na+(aq) + 6 Cl-(aq)
3 Ni2+(aq) + 2 PO43-(aq)
4.80(d)
2 HC2H3O2(aq) + 2
→ Ni3(PO4)2(s)
K+(aq)
+ CO32-(aq)
2 HC2H3O2(aq) + CO32-(aq)
→ H2O(l) + CO2(g) + 2 K+(aq) + 2 C2H3O2-(aq)
→ H2O(l) + CO2(g) + 2 C2H3O2-(aq)
Reasoning:
4.79(d) 2 Na3PO4(aq) + 3 NiCl2(aq)
→ Ni3(PO4)2(s) + 6 NaCl(aq) (balance the equation first)
The three ionic compounds that are soluble (here you are told this with the state designations)
are strong electrolytes. Separate each of them, stoichiometrically and with proper charges, to
obtain the answer shown above.
4.80(d) 2 HC2H3O2(aq) + K2CO3(aq) → H2O(l) + CO2(g) + 2 KC2H3O2(aq) (balance the equation first)
PS6-7
Answer Key, Problem Set 6
HC2H3O2 is an acid (H listed in front), but it is a weak acid (not one of the six strong ones you
should memorize). So it is NOT a strong electrolyte (so leave it unchanged).
Species with liquid or gas designations (H2O and CO2) are not strong electrolytes / not separated.
The two ionic compounds in this problem are both soluble, so they ARE strong electrolytes. They
are the only ones whose formulas are “separated” in the complete ionic equation. Separate each
of them, stoichiometrically and with proper charges, to obtain the answer shown above.
Predicting the products in a potential exchange [acid-base] reaction, NIE’s
13. WO7.
Write the balanced (i) molecular, (ii) complete ionic, and (iii) net ionic equations for each of the following
[unbalanced] uncompleted equations representing acid-base reactions. If no reaction occurs, write NO REACTION.
(a)
(b)
(c)
(d)
(e)
HNO3(aq) + Al(OH)3(s) 
KC2H3O2(aq) + H2SO4(aq) 
NaNO3(aq) + HCl(aq) 
The equation in Tro, 4.81(c): H2SO4 + NaOH 
The equation in Tro, 4.82(b): HC2H3O2 + Ca(OH)2 
Answers:
(a)
(i)
3 HNO3(aq) + Al(OH)3(s)  Al(NO3)3(aq) + 3 H2O(l)
(ii)
3 H+(aq) + 3 NO3-(aq) + Al(OH)3(s)  Al3+(aq) + 3 NO3-(aq) + 3 H2O(l)
(iii) 3 H+(aq) + Al(OH)3(s)  Al3+(aq) + 3 H2O(l)
(b)
(i)
2 KC2H3O2(aq) + H2SO4(aq)  2 HC2H3O2(aq) + K2SO4(aq)
(ii)
2 K+(aq) + 2 C2H3O2-(aq) + 2 H+(aq) + SO42-(aq)  2 HC2H3O2(aq) + 2 K+(aq) + SO42-(aq)
(iii) H+(aq) + C2H3O2-(aq)  HC2H3O2(aq)
(c)
(d)
(e)
(i)
NaNO3(aq)
(ii)
Na+(aq)
(iii)
NO REACTION
(i)
H2SO4(aq)
H+(aq)
+
+
+

HCl(aq)
NO3-(aq)
+
+
H+(aq)
NaCl(aq)
+
2 NaOH(aq)
SO42-(aq)
+ 2
Na+(aq)
(ii)
2
(iii)
H+(aq) + OH-(aq)  H2O(l)
Cl-(aq)

+ 2
+
HNO3(aq)
 Na+(aq) + Cl-(aq) + H+(aq) + NO3-(aq)
Na2SO4(aq)
OH-(aq)
 2
+
2 H2O(l)
Na+(aq)
+ SO42-(aq) + 2 H2O(l)
(i)
2 HC2H3O2(aq) + Ca(OH)2(aq)  Ca(C2H3O2)2(aq) + 2 H2O(l)
(ii)
2 HC2H3O2(aq) + Ca2+(aq) + 2 OH-(aq)  Ca2+(aq) + 2 C2H3O2-(aq) + 2 H2O(l)
(iii) HC2H3O2(aq) + OH-(aq)  C2H3O2-(aq) + H2O(l)
Strategy: 1) Separate the reactants’ formula units into “ions” (where “H+” is treated like a cation in any
acid, and what’s left over is treated like an anion) just to figure out what the possible
products will be (next step).
2) Swap the “anions” to get possible products.
3) Use the neutrality principle, along with your knowledge of the cation / anion formulas and
charges to create the proper formulas of each product compound. Review PS3 key for
how to do this. Note that when “H+” and “OH-“ come together, they make H2O (although if
you write HOH, that’s fine; just realize that it is not an ionic compound!  ). When H+ comes
together with any anion besides OH-, an acid is formed (also not ionic).
4) Use the two solubility rules that you must memorize, along with the other rules (Table 4.1)
to decide if either of the potential product compounds is insoluble (Although none of the
PS6-8
Answer Key, Problem Set 6
examples in this problem have a precipitate, I’m putting this step in because you should
always check this [don’t assume it]).
a) If “yes”, then that insoluble ionic compound is the precipitate (and a precipitation reaction occurs)
b) If “no”, then no precipitate forms.
5) Express as a molecular equation and balance it.
6) To get the complete ionic equation, make sure to assess whether or not each substance is
a strong electrolyte (SE) or not. ONLY SEPARATE THE STRONG ELECTROLYTES
7) To get the net ionic equation, omit any spectator ions.
Execution of Strategy:
(a)
2) & 3) swap anions to get
HNO3 + Al(OH)3
H2O
+
1) “separate” into ions
(just to create new formulas)
H+
NO3-
Al(NO3)3
soluble (NO3-)
Al3+ OH-
3 HNO3(aq) + Al(OH)3(s)  Al(NO3)3(aq) + 3 H2O(l) (add the 3’s to balance)
(i)
SE (strong acid)
(ii)
(insoluble)
SE (soluble ionic)
not a SE (not a strong acid)
3 H+(aq) + 3 NO3-(aq) + Al(OH)3(s)  Al3+(aq) + 3 NO3-(aq) + 3 H2O(l)
(iii) 3 H+(aq) + Al(OH)3(s)  Al3+(aq) + 3 H2O(l)
(b)
[only NO3- is a spectator ion]
2) & 3) swap anions to get
KC2H3O2(aq) + H2SO4(aq)
1) “separate” into ions
(just to create new formulas)
K+
C2H3O2-
K2SO4
+
HC2H3O2
soluble (NO3-)
H+ SO42-
2 KC2H3O2(aq) + H2SO4(aq)  2 HC2H3O2(aq) + K2SO4(aq) (add the 2’s to balance)
(i)
SE (soluble ionic)
(ii)
SE (a strong acid)
WE (a weak acid) (SE (soluble ionic)
2 K+(aq) + 2 C2H3O2-(aq) + 2 H+(aq) + SO42-(aq)  2 HC2H3O2(aq) + 2 K+(aq) + SO42-(aq)
[HC2H3O2 is a weak acid—do not separate it!]
(iii) 2
H+(aq)
+2
C2H3O2-(aq)
+
HCl(aq)
 2 HC2H3O2(aq) (which reduces to the 1:1:1 stoichiometry
shown above)
2) & 3) swap anions to get
(c)
NaNO3(aq)
NaCl + HNO3
1) “separate” into ions
(just to create new formulas)
Na+
(i)
NO3-
H+
NaNO3(aq)
SE (soluble ionic)
soluble (NO3-)
Cl+
HCl(aq)
SE (a strong acid)

NaCl(aq)
+
SE (soluble ionic)
+ H+(aq) + Cl-(aq)
HNO3(aq)
SE (a strong acid)
 Na+(aq) + Cl-(aq) + H+(aq) + NO3-(aq)
(ii)
Na+(aq) + NO3-(aq)
(iii)
All SE’s  NO REACTION (all ions on both sides; everything cancels, nothing “happens”)
H+(aq) + C2H3O2-(aq)  HC2H3O2(aq)
PS6-9
Answer Key, Problem Set 6
(d)
H2SO4(aq)
+
NaOH(aq)
2) & 3) swap anions to get
H2O + Na2SO4
1) “separate” into ions
(just to create new formulas)
H+
SO42-
(i)
Na+
H2SO4(aq)
+
SE (strong acid)
(e)
soluble (NO3-)
OH
2 NaOH(aq)
SE (soluble ionic)
Na2SO4(aq)
+
2 H2O(l) (added the 2’s to balance)
SE (soluble ionic)
not a SE
(ii)
2 H+(aq) + SO42-(aq) + 2 Na+(aq) + 2 OH-(aq)  2 Na+(aq) + SO42-(aq) + 2 H2O(l)
(iii)
2 H+(aq) + 2 OH-(aq)  2 H2O(l)
2 HC2H3O2(aq) + Ca(OH)2(aq)
(which reduces to a 1 : 1 : 1 stoichiometery shown in box above)
2) & 3) swap anions to get
2 H2O(l) + Ca(C2H3O2)2(aq)
1) “separate” into ions
(just to create new formulas)
H+
C2H3O2-
Ca2+
soluble (C2H3O2-)
OH-
2 HC2H3O2(aq) + Ca(OH)2(aq)  Ca(C2H3O2)2(aq) + 2 H2O(l) (added the 2’s to balance)
(i)
WE (weak acid)
SE ([slightly] soluble ionic)
SE (soluble ionic)
not a SE
(ii)
2 HC2H3O2(aq) + Ca2+(aq) + 2 OH-(aq)  Ca2+(aq) + 2 C2H3O2-(aq) + 2 H2O(l)
(iii)
2 HC2H3O2(aq) + 2 OH-(aq)  2 C2H3O2-(aq) + 2 H2O(l)
(which reduces to a 1 : 1 : 1 :1
stoichiometery shown in box above)
Titration, Stoichiometry with Molarity
14. 4.90
A 30.0 mL sample of an unknown H3PO4 solution is titrated with a 0.100 M NaOH solution. The equivalence point is
reached when 26.38 mL of NaOH solution is added. What is the concentration of the unknown H3PO4 solution?
The neutralization reaction is
H3PO4 + 3 NaOH → 3 H2O + Na3PO4
Answer: 0.0293 M H3PO4(aq)
Strategy: 1) Use V (converted to L) and M (mol/L) of NaOH(aq) to calculate moles of NaOH.
2) Use the balanced equation / stoichiometry to calculate the moles of H3PO4 reacted.
3) Use moles and V (in L) of H3PO4(aq) to calculate [H3PO4] (in M).
Execution of Strategy:
mol H PO  26.38 mL NaOH(aq) x
2
4
0.100 mol NaOH 1 mol H 3 PO 4
1L
x
x
 0.00087 93.. mol H 3 PO 4
1000 mL
1 L NaOH(aq)
3 mol NaOH
2)
1)
0.0008793.. mol H3 PO 4
 0.02931.  0.0293 M H3 PO 4
1L
30.00 mL x
1000 mL
PS6-10
Answer Key, Problem Set 6
Oxidation Numbers, Redox Language
15. 4.94. Assign oxidation states ( “numbers”) to each atom in each element, ion, or compound.
(a) Cl2
(b)
Fe3+
0 (this is an element)
+3 (if only one “atom”, oxidation state = actual charge)
(c) CuCl2
Cl is -1, so Cu is +2 (ionic  Cl is a -1 ion. If Cu is x: x + 2(-1)  0  x  +2)
(d) CH4
H is +1, so C is -4; x + 4(+1)  0  x  -4
(e) Cr2O72- O is -2, so Cr is +6; 2x + 7(-2)  -2  2x  +12  x  +6
(f) HSO4-
H is +1, O is -2, so S is +6; 1(+1) + x + 4(-2)  -1  x  -1 -1 + 8  +6
16. 4.96. What is the oxidation state ( “number”) of Cl in each ion?
O is -2, so Cl in each case is:
(a) ClO-
+1; x + (-2)  -1  x  +2 -1  +1
(b) ClO2-
+3; x + 2(-2)  -1  x  +3
ClO3-
+5; x + 3(-2)  -1  x  +5
(d) ClO4-
+7; x + 4(-2)  -1  x  +7
(c)
17. WO8.
For each of the following kinds of chemical species, state whether the oxidation number increases or decreases
when a redox reaction takes place:
(a) An oxidizing agent. Oxidation number DECREASES, because an oxidizing agent is “one that
oxidizes (somebody)” which means it “takes their electrons” and thus become more negative
in the process (because electrons are negative!). (Or you might remember that the oxidizing
agent is the one that gets reduced.) So oxidation number becomes more negative (it
decreases).
(b) A species undergoing oxidation. Oxidation number INCREASES, because “undergoing oxidation”
means “having your electrons taken from you” (or “getting oxidized”) which means you will
become more positive (oxidation number increases).
(c) A species that gives away electrons. Oxidation number INCREASES, because electrons are
negative! If a species gives away something negative, then it must become more positive.
So oxidation number becomes more positive (it increases).
(d) A species undergoing reduction. Oxidation number DECREASES because “undergoing reduction”
means “having electrons given to you” (or “getting reduced”) which means you will become
more negative (oxidation number decreases).
(e) A species that accepts (gains) electrons. Oxidation number DECREASES, because electrons are
negative! If a species gains something that's negative, it must become more negative. So
oxidation number becomes more negative (it decreases).
(f) A reducing agent. Oxidation number INCREASES, because a reducing agent is “one that reduces
(somebody)” which means it “gives electrons” and thus become more positive in the
process. (Or you may recall that the reducing agent is the one that itself gets oxidized.) So
oxidation number becomes more positive (it increases).
Identifying Redox Reactions, Redox Language
18. 4.98 Answer not in key. Problem is analogous to the next problem (4.91). See that entry for strategy.
PS6-11
Answer Key, Problem Set 6
19. 4.97
Determine whether each reaction is a redox reaction. For each redox reaction, identify the oxidizing agent and the
reducing agent.
NOTE: Only answers to (a) and (c) are shown below in this key at this time.
Answers: (a) is redox; O2 is the oxidizing agent; Li is the reducing agent
(c) is not a redox reaction. All oxidation numbers remain the same during chemical change.
**NOTE: See Problem 20 (WO9) for more challenging problems of this type.**
Strategy:
1) Assign oxidation numbers to all atoms (as stated in the problem!!).
2) Look to see which substance or species has an atom whose oxidation number changed.
3) Recognize that if the oxidation number increased (i.e., got more positive) in going from reactant to
(any one) product, the reactant substance or species lost electrons and was therefore oxidized,
making it the reducing agent.
4) Recognize that if the oxidation number decreased (i.e., got more negative), the reactant
substance or species gained electrons and was therefore reduced, making it the oxidizing agent.
Execution of Strategy:
(a) 4 Li(s) + O2(g) → 2 Li2O(s)
0
0
+1 -2
Li’s oxidation number went up (from 0 to +1)  it lost electrons and underwent oxidation, and so
Li is thus the reducing agent (it gave electrons to O2).
O’s oxidation number went down (from 0 to -2)  it gained electrons and underwent reduction,
and O2 is thus the oxidizing agent (it took electrons from Li)
(c) Pb(NO3)2(aq) + Na2SO4(aq)
+2 +5 -2
+1 +6 -2
→ PbSO4(s)
+ 2 NaNO3(aq)
+2 +6 -2
+1 +5 -2
Pb(NO3)2 → Pb2+ + 2 NO3-, so Pb is +2, and since O is assigned -2, N must be +5 (x + 3(-2)  -1)
Na2SO4 → 2 Na+ + SO42-, so Na is +1, and since O is assigned -2, S must be +6 (x + 4(-2)  -2)
PbSO4 → Pb2+ and SO42-, so Pb is +2 and S is +6 (prior line!)
NaNO3 → Na+ and NO3-, so Na is +1 and N is +5 (three lines back)
Pb remains +2; N remains +5, O remains -2, and S remains +6  not redox
(Actually it’s clear that this is an exchange reaction. All that happens is “ion swapping”. Nobody changes
their (fictitious) “charge”. Exchange reactions are not redox reactions.
20. WO9. (i) Identify the oxidizing agent and reducing agent in each reaction depicted by a chemical equation below (How?
By
assigning oxidation #’s to all atoms in all reactants and products. SHOW THIS WORK).
(ii) State which species gets oxidized, and which species gets reduced when reaction occurs in each case.
(iii) State which species takes (i.e., gains) electrons, and which species gives (i.e., loses) electrons in each case.
Strategy: See the list in prior problem.
(a) Na(s) + H2O(l)  NaOH(aq) + H2(g)
0
Answers:
+1-2
+1 –2 +1
0
(i) Oxidizing agent: H2O; Reducing agent: Na
(ii) Na got oxidized. H2O got reduced.
(iii) H2O takes electrons. Na gives electrons.
Rationale for assigning oxidation numbers:
 Atoms in elements are given an oxidation number of zero, so the Na atom in Na(s) and the H
atoms in H2 are given 0’s.
PS6-12
Answer Key, Problem Set 6
 O is (typically) given –2 in compounds or polyatomic ions. H is given +1 in compounds,
typically.
 You can get +1 for Na in NaOH two ways: 1) if you separate the (ionic) compound into Na+
ions and OH-, the oxidation number of Na is obviously +1 (the actual charge is +1 and there is
only one atom in the species!). 2) If you look at the formula unit, the sum of the oxidation
numbers of the O's and H's is –1 [1(-2) + 1(+1) = -1], so the oxidation number of Na must be
+1 (since the compound is overall neutral; the sum of all atoms' oxidation numbers in a
species must add up to the overall charge on that species).
Explanations: The oxidation number of Na went from 0 to +1 as reaction occurs, so each Na atom
(gets more positive and thus) loses an electron, which means each has an electron taken from
it, which means Na gets oxidized when the reaction occurs. Each Na atom gave an electron
(to the H atoms in H2O), and so Na is the reducing agent.
The oxidation number of H goes from +1 to 0 as reaction occurs, so each H atom gains an
electron, which means each has an electron given to it, which means H gets reduced. Each
H takes an electron (from Na), and so "H" is effectively the oxidizing agent. However, as your
book notes, "H" isn't an actual species in this reaction, so we consider H2O (the species that
contains H atoms) to be the species that gains an electron, gets reduced, and is the oxidizing
agent.
(b) 2 I2(s) + IO3(aq) + 10 Cl(aq) + 6 H+(aq)  5 ICl2(aq) + 3 H2O(l)
0
+5-2
Answers:
–1
+1
+1 –1
+1 –2
(i) Oxidizing agent: IO3; Reducing agent: I2
(ii) I2 got oxidized. IO3got reduced.
(iii) IO3takes electrons. I2 gives electrons.
Rationale for assigning oxidation numbers:
 Atoms in elements are given an oxidation number of zero, so the I atoms in I2 get 0.
 O is (typically) given –2 in compounds or polyatomic ions. H is given +1 in compounds,
typically.
 In H+, the oxidation number is obviously +1 for the same reason that Na+ is +1 in (a) (actual
charge and monatomic). Likewise, in Cl-, the actual charge is –1 and there is only one atom,
Cl, so its oxidation number is obviously -1.
 Letting x represent the unknown oxidation number in each case below:
IO3 : If each O is –2, then x + 3(–2) must equal –1 (the overall charge). Thus x = +5
ICl2:If each Cl is –1, then x + 2(–1) = –1 (overall charge). Thus x = +1
Explanations: The oxidation number of the I atoms in I2 goes from 0 to +1 as reaction occurs, so I
atoms in I2 (get more positive and thus) lose electrons, which means I2 has electrons taken
from it, which means I2 gets oxidized when reaction occurs. I2 gives electrons (to the I
atoms in IO3) and so I2 is the reducing agent.
The oxidation number of the I atom in IO3goes from +5 to +1 as reaction occurs, so the I atom
in IO3 gains electrons, which means IO3 has electrons given to it, which means IO3 is
reduced. IO3takes electrons (from I2), and so it is the oxidizing agent.
Cumulative Stoichiometry/Molarity/L.R./%Yield/EF Problems
21. WO10.
Carminic acid, a naturally occurring red pigment extracted from the cochineal insect, contains only carbon,
hydrogen, and oxygen. It was commonly used as a dye in the first half of the nineteenth century. It is 53.66% C and
4.09% H by mass. A titration required 18.02 mL of 0.0406 M NaOH to neutralize 0.3602 g carminic acid. Assuming
that there is only one acidic hydrogen per molecule (i.e., it has a generic formula “HA”), what is the molecular
formula of carminic acid? (This problem is from Zumdahl, 7th edition)
Answer: C22H20O13
PS6-13
Answer Key, Problem Set 6
Solution:
1) Find the molar mass using the titration data (NOTE: Rxn stoichiometry is 1 : 1 since carminic
acid has only one acidic hydrogen: HA + NaOH → NaA + H2O; HA = carminic acid)
(a)
18.02 mL
0.0406 mol NaOH 1 mol carminic acid
x
x
 0.0007316 mol carminic acid
1000 mL/L
L solution
1 mol NaOH
(b) Molar Mass of carminic acid =
# g acid
0.3602 g

 492.3..  492 g/mol
# moles acid 0.0007316..mol
2) Use % mass data, along with molar mass from (1) above, to get molecular formula (this is what you
did in problem #21. 3.130 on PS4, except there they gave you MM!):
NOTE: When I first did this problem on my own, I used 100 g of sample since % masses were given first (see below for that
solution, if interested). HOWEVER, it then occurred to me that since the molar mass is obtainable from the titration
data, it makes, by far, the most sense to pick a sample with one mole of compound (i.e., a 492 g sample) so that you
do not need to do the “ratio reduction” step and “MM/EM” step (which is quite awkward here since n turns out to be
13!?):
Mass of C in sample = 53.66% of 492 g = 0.5366 x 492 g = 263.0 g C
Mass of H in sample = 4.09% of 492 g = 0.0409 x 492 g = 20.12 g H
Mass of O in sample = Mass of sample – mass of C – mass of H = 492 – 263.0 – 20.12 = 208.9 g O
Since we know that these amounts are found in one mole of the compound, conversion of the above
mass amounts into moles of each element will give us the subscripts in the molecular formula. In
other words, the values obtained should already be whole numbers of moles of atoms:
263.0 g C
12.01 g/mol C
20.12 g H
1.01 g/mol H
 21.89 mol C  22 mol C
 19.92 mol H  20 mol H
208.9 g O
16.00 g/mol O
C22H20O13
 13.06 mol O
------------------------------------------------------------------------------------------------[Alternative approach (but more work at the end, and the fractional EM thing is likely not “intuitive”
for many of you)]: Assume an exactly 100 g sample of the acid.
It will contain:
53.66 g C
12.01 g/mol C
4.09 g H
1.01 g/mol H
53.66 g C, 4.09 g H, and (100 – 53.66 – 4.09) = 42.25 g O
 4.4679 mol C
 4.0495 mol H
42.25 g O
16.00 g/mol O
 2.6406 mol O
C4.4679H4.0495O2.6406
Divide all subscripts by the smallest to get
 C 4.4679 H 4.0495 O 2.6406  C1.692H1.533O1
2.6406
2.6406
2.6406
That ratio is not an easy ratio to turn into a whole number ratio! HOWEVER, you don't need to!
You have the molar mass! So calculate an "empirical mass" with your (fractional) subscripts and
treat as you did in problem 18. 3.96 (on PS4):
(pseudo) empirical mass = 1.692(12.01) + 1.533(1.01) + 1(16.00) = 37.869 g
How much "bigger" is one mole than this?
492 g
 12.99  13 times
37.869 g
PS6-14
Answer Key, Problem Set 6

C
H1.533O1 13  C13 x 1.692H13 x 1.533O13 x 1  C21.996H19.9O13  C22H20O13
1.692
22. MP The answer to this problem is not in this key at this time.
PS6-15