CHEM 10123/10125, Exam 2
March 7, 2012 (50 minutes)
Name______________________
(please print)
Please box your answers, and remember that significant figures, phases (for chemical equations),
and units do count!
1. (13 points) SHOW ALL WORK. Ascorbic acid, H2A, has Ka1 = 8.0 x 10–5 and Ka2 =
1.6 x 10–12. Determine the following values for a 2.5 M solution of this acid. Briefly
show your work in the space below. (Reactions need not be shown here)
a. [HA–] ______1.4 x 10–2 _________ (x2) / (2.5) = 8.0 x10–5
b. pH _______1.85_________ = – log (1.4 x10–2 )
c. [A2–]_____1.6 x10–12 M __________
2. (16 points) Briefly define the following terms:
Lewis base — electron pair donor
Indicator — a substance whose color depends on the pH
Arrhenius base — a substance that produces OH– ions in aqueous solution
Buffer range-- the pH values for which a buffer system is the most effective. (usually
±1 pH unit of either side of pKa)
3. (8 points) For the amphoteric molecule HCO3–, write one equation showing it acting
as an acid, and another equation showing it acting as a base (label which is which).
Answer:
Acid: HCO3–(aq) ⇌ H+(aq) + CO32–(aq)
Base: HCO3–(aq) + H2O(l) ⇌ H2CO3(aq) + OH–(aq)
4. (5 points) Among the following, circle the weakest Bronsted-Lowry acid.
HClO HClO2 HClO3 HBrO HBrO2 HBrO3
Answer:
HBrO
5. (5 points) Among the following, circle the strongest Bronsted-Lowry base.
HS– HO– HSe– Br– F– Cl–
Answer:
HO–
1
6. (15 points) Arrange the following 0.010 M solutions in order of increasing pH:
NH3, HNO3, NaNO2, HC2H3O2, NaOH, NH4C2H3O2, NH4ClO4
Answer:
HNO3<HC2H3O2<NH4ClO4<NH4C2H3O2<NH3<NaNO2<NaOH
7. (13 points) For each reaction, identify the Lewis acid and Lewis base, then predict the
reaction product. Draw the complete and correct Lewis structure of the product for
reaction (a) only.
OH–(aq) + CO2(aq)  HCO3–(aq)
base
acid
3+
Cr (aq) + 6 H2O(l)  [Cr(H2O)6]3+(aq)
acid
base
Al(OH)3(aq) + OH–(aq)  Al(OH)4–(aq)
acid………….base
8. SHOW ALL WORK. A 20.00 mL sample of 0.150 M HF is titrated with 0.250 M
NaOH. The initial pH is 2.02. Calculate the pH when the neutralization is
a) (12 points) 25% complete [note we used Ka = 3.0 x 10–8 instead of HF Ka!]
Answer:
20.00 mL x (0.150 mol HF)/(1000 mL HF soln) = 0.00300 mol HF
When it is 25% complete, that means that (0.25)(0.00300 mol) = 0.000750 mol of NaOH
have been added, and 0.00300 – 0.000750 = 0.00225 mol of HF remain.
It is a buffer system, with Ka = 3.0 x 10–8.
3.0 x 10–8 = [H+][F–]/[HF] = [H+](0.000750)/(0.00225)
[H+] = 9.0 x 10–8 M
pH = 7.05
b) (5 points) 50% complete
Buffer system, still, and if it’s 50% complete then [HF] = [F–], so
Ka = [H+] = 3.0 x 10–8
pH = 7.52
c) (20 points) 100% complete
At this point, 0.00300 mol NaOH have been added and all of the acid has been
neutralized to form 0.00300 mol F-. It is a solution of a weak base.
The chemical equation is
F–(aq) + H2O(l) ⇌ HF(aq) + OH–(aq)
To get the total volume,
0.00300 mol OH– x (1000 mL NaOH soln)/(0.250 mol NaOH) = 12.0 mL NaOH soln
12.0 mL + 20.0 mL = 0.0320 L soln
(0.00300 mol F–)/(0.0320 L soln) = 0.09375 M F–
2
For the solution of a weak base we will need Kb = (1.0 x 10–14)/(3.0 x 10–8) = 3.333 x 10–7
F–
HF
OHI 0.09375 M
0
0
C -x
+x
+x
E 0.09375 - x M
x
x
3.333 x 10–7 = [HF][OH–]/[F–] ≈ (x2)/(0.09375) (assume x is small; clearly safe in this case)
x = 1.77 x 10–4 M OH-, so pOH = 3.75, and pH = 10.25.
9. (12 points) Fill in the blanks.
[H+]
[OH–]
–5
4.5 x 10 M
2.2 x 10–10
5.2 x 10–12
1.9 x 10–3
pH
4.35
11.28
pOH
9.65
2.72
10. (6 points) Coal can be used to generate hydrogen gas ( a potential fuel) by the
endothermic reaction:
C(s) + H2O(g) ⇌ CO(g) + H2(g)
If this reaction mixture is at equilibrium, predict whether each disturbance will result in
the formation of additional hydrogen gas, the formation of less hydrogen gas, or have no
effect on the quantity of hydrogen gas.
a.) adding more C to the reaction mixture
no change
b.) adding more H2O to the reaction
mixture
c.) raising the temperature of the reaction
mixture
d.) increasing the volume of the reaction
mixture
e.) adding a catalyst to the reaction mixture
favors H2
f.) adding an inert gas to the reaction
mixture
no change
favors H2
favors H2
no change except speed
11. (20 points) A solution is formed by mixing 200.0 mL of 0.100 M KOH with 100.0
mL of a solution that is both 0.200 M in CH3COOH and 0.050 M in HI. Identify in the
final solution
a) all the solute species present
b) the two most abundant solute species, and
c) the two least abundant solute species.
200.0 mL of 0.100 M KOH = 0.0200 mol KOH
100 mL of 0.200 M HC2H3O2 = 0.0200 mol HC2H3O2
100 mL of 0.050 mol HI = 0.0050 mol HI
Take the solute ions one at a time:
The K+ is a spectator ion and will not react. There are 0.0200 mol K+.
3
The OH– starts at 0.0200 mol, but reacts with the 0.0050 mol H+ formed from the HI.
This leaves 0.015 mol OH–. The 0.015 mol OH– then reacts to neutralize the HC2H3O2,
converting 0.015 mol of HC2H3O2 to C2H3O2–.
Therefore, we now have 0.005 mol HC2H3O2 left, and the 0.015 mol C2H3O2– that was
formed.
The HI dissociated when dissolved into H+ and I–. The H+ was neutralized by the OH– as
described above, leaving 0.0050 mol I–.
a) K+, OH–, H+, HC2H3O2, C2H3O2–, I–
b) K+, C2H3O2c) H+, OH–
CHEM 10123/10125, Exam 2
March 7, 2012 (50 minutes)
Name______________________
(please print)
Please box your answers, and remember that significant figures, phases (for chemical equations),
and units do count!
1. (13 points) SHOW ALL WORK. Carbonic acid, H2A, has Ka1 = 4.4 x 10–7 and Ka2
= 4.7 x 10–11. Determine the following values for a 2.5 M solution of this acid. Briefly
show your work in the space below. (Reactions need not be shown here)
a. [HA–] ______1.0 x10–3 M _________ (x2) / (2.5) = 4.4 x 10–7
b. pH _______2.98_________ = – log (1.0 x10–3 )
c. [C6H6O62–]_____4.7 x10–11 M __________
2. (16 points) Briefly define the following terms:
Bronsted acid – proton donor
Equivalence point – In a titration, when number of moles of base is stoichiometrically
equal to the number of moles of acid
Buffer capacity — is the amount of acid or base you can add without causing a large
change in pH
Lewis acid – electron pair acceptor
3. (8 points) For the amphoteric molecule HS–, write one equation showing it acting as
an acid, and another equation showing it acting as a base (label which is which).
4
Answer:
Acid: HS–(aq) ⇌ H+(aq) + S2–(aq)
Base: HS–(aq) + H2O(l) ⇌ H2S(g) + OH–(aq)
4. (5 points) Among the following, circle the strongest Bronsted-Lowry base.
ClO3– ClO4– ClO2- BrO2– BrO4– BrO3–
Answer:
BrO2–
5. (5 points) Among the following, circle the weakest Bronsted-Lowry acid.
CH4 H2S SiH4 PH3 H2O
NH3
Answer:
CH4
6. (15 points) Arrange the following 0.010 M solutions in order of decreasing pH:
NH3, HNO3, NaNO2, HC2H3O2, NaOH, NH4C2H3O2, NH4ClO4
Answer:
NaOH>NaNO2>NH3>NH4C2H3O2>NH4ClO4>HC2H3O2>HNO3
7. (13 points) For each reaction, identify the Lewis acid and Lewis base, then predict the
reaction product. Draw the complete and correct Lewis structure of the product for
reaction (a) only.
SO2(aq) + OH–(aq)  HSO3–(aq)
acid…….base
B(OH)3(aq) + OH–(aq)  B(OH)4–(aq)
acid……………base
Al3+(aq) + 6 H2O(l)  [Al(H2O)6]3+(aq)
acid……………base
8. SHOW ALL WORK. A 30.00 mL sample of 0.165 M propanoic acid (HC3H5O2) is
titrated with 0.300 M KOH. The initial pH is 2.83. Calculate the pH when the
neutralization is
a) (12 points) 25% complete
Answer:
30.00 mL x (0.165 mol HC3H5O2)/(1000 mL HC3H5O2 soln) = 0.00495 mol HC3H5O2
When it is 25% complete, that means that (0.25)(0.00495 mol) = 0.0012375 mol of KOH
have been added, and 0.00495 – 0.0012375 = 0.0037125 mol of HC3H5O2 remain.
It is a buffer system, with Ka = 1.3 x 10–5.
1.3 x 10–5 = [H+][C3H5O2–]/[HC3H5O2] = [H+](0.0012375)/(0.0037125)
[H+] = 0.000039 M
pH = 4.41
5
b) (5 points) 50% complete
Buffer system, still, and if it’s 50% complete then [HC3H5O2] = [C3H5O2–], so
Ka = [H+] = 1.3 x 10–5
pH = 4.89
c) (20 points) 100% complete
At this point, 0.00495 mol KOH have been added and all of the acid has been neutralized
to form 0.00495 mol C3H5O2-. It is a solution of a weak base.
The chemical equation is
C3H5O2–(aq) + H2O(l) ⇌ HC3H5O2 (aq) + OH–(aq)
To get the total volume,
0.00495 mol OH– x (1000 mL KOH soln)/(0.300 mol KOH) = 16.5 mL KOH soln
16.5 mL + 30.0 mL = 0.0465 L soln
(0.00495 mol C3H5O2–)/(0.0465 L soln) = 0.106 M C3H5O2–
For the solution of a weak base we need Kb = (1.0 x 10–14)/(1.3 x 10–5) = 7.6923 x 10–10
“C3H5O2–”
“HC3H5O2”
OHI 0.106 M
0
0
C -x
+x
+x
E 0.106 - x M
x
x
7.6923 x 10–10 = [HC3H5O2][OH–]/[C3H5O2–] ≈ (x2)/(0.106) (assume x is small; clearly safe
in this case)
x = 9.03 x 10–6 M OH-, so pOH = 5.04, and pH = 8.96
9. (12 points) Fill in the blanks.
[H+]
[OH–]
6.7 x 10–13
0.015 M
1.3 x 10–12
7.6 x 10–3
pH
1.82
11.88
pOH
12.18
2.12
10. (6 points) Coal can be used to generate hydrogen gas ( a potential fuel) by the
endothermic reaction:
C(s) + H2O(g) ⇌ CO(g) + H2(g)
If this reaction mixture is at equilibrium, predict whether each disturbance will result in
the formation of additional hydrogen gas, the formation of less hydrogen gas, or have no
effect on the quantity of hydrogen gas.
a.) raising the temperature of the reaction
favors H2
mixture
b.) removing some H2O from the reaction
less H2
mixture
c.) removing some of the C from the
no change
reaction mixture
d.) adding an inert gas to the reaction
no change
mixture
e.) adding a catalyst to the reaction mixture no change except rate
6
f.) decreasing the volume of the reaction
mixture
less H2
11. (20 points) A solution is formed by mixing 200.0 mL of 0.100 M KOH with 100.0
mL of a solution that is both 0.200 M in CH3COOH and 0.050 M in HI. Identify in the
final solution
a) all the solute species present
b) the two most abundant solute species, and
c) the two least abundant solute species.
Answer:
200.0 mL of 0.100 M KOH = 0.0200 mol KOH
100 mL of 0.200 M HC2H3O2 = 0.0200 mol HC2H3O2
100 mL of 0.050 mol HI = 0.0050 mol HI
Take the solute ions one at a time:
The K+ is a spectator ion and will not react. There are 0.0200 mol K+.
The OH– starts at 0.0200 mol, but reacts with the 0.0050 mol H+ formed from the HI.
This leaves 0.015 mol OH–. The 0.015 mol OH– then reacts to neutralize the HC2H3O2,
converting 0.015 mol of HC2H3O2 to C2H3O2–.
Therefore, we now have 0.005 mol HC2H3O2 left, and the 0.015 mol C2H3O2– that was
formed.
The HI dissociated when dissolved into H+ and I–. The H+ was neutralized by the OH– as
described above, leaving 0.0050 mol I–.
Finally, the autoionization of water happens in all aqueous solutions, though to a small
extent only. Therefore both H+ and OH– are present in very small concentrations. (you
don’t need to calculate them, but if you want to, you can use the buffer equation to get
the [H+], followed by Kw to get [OH–].)
This means the answers are:
a) K+, OH–, H+, HC2H3O2, C2H3O2–, I–
b) K+, C2H3O2c) H+, OH–
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