Thompson, Margaret – Homework 7 – Due: Oct 29 2007, 11:00 pm – Inst: James Holcombe
1
This
printout
should
have
24
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
The curve for the titration of hypochlorous
acid (HOCl) with NaOH(aq) base is given
below.
0
2
4
6
8
10
12
14
0
20
40
60
80
100
120
140
Volume of base (mL)
pH
Estimate the p
K
a
of hypochlorous acid.
C
a
= 0
.
56,
C
b
= 0
.
5544, and the volume
of HOCl is 100 mL.
1.
49
.
5
2.
99
3.
7
.
5
correct
4.
10
.
2
5.
None of these
Explanation:
K
a
= 3
.
5
×
10

8
C
a
= 0
.
56
C
b
= 0
.
5544
V
HOCl
= 100 mL
0
2
4
6
8
10
12
14
0
20
40
60
80
100
120
140
Volume of base (mL)
pH
(99,10
.
2)
(49
.
5,7
.
5)
The equivalence point of this titration is
when the curve is at an inflection point (nearly
vertical);
i.e.
, at a volume of 99 mL
.
The pH at the equivalence point of this
titration is 10
.
2 pH
.
The p
K
a
can be found at onehalf the vol
ume of the equivalence point;
i.e.
, at 49
.
5 mL.
The p
K
a
is 7
.
5 pH from looking at the graph.
The formula is
p
K
a
=

log (
K
a
)
=

log
‡
3
.
5
×
10

8
·
=
7
.
45593 pH
.
Note
:
The p
K
a
is the pH when the mole
fraction is 0.5.
002
(part 1 of 1) 10 points
It required 25.0 mL of 0.333 M NaOH solution
to completely neutralize 15.0 mL of H
2
SO
4
so
lution. What was the molarity of the H
2
SO
4
?
1.
1.11 M
2.
0.555 M
3.
0.278 M
correct
4.
0.200 M
Explanation:
V
NaOH
= 25.0 mL
[NaOH] = 0.333 M
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Thompson, Margaret – Homework 7 – Due: Oct 29 2007, 11:00 pm – Inst: James Holcombe
2
V
H
2
SO
4
= 15.0 mL
The balanced equation for the reaction is
H
2
SO
4
+ 2 NaOH
→
Na
2
SO
4
+ 2 H
2
O
We determine the moles of NaOH used.
? mol NaOH = 0
.
025 L soln
×
0
.
333 mol NaOH
1 L soln
= 0
.
00832 mol NaOH
Using the mole ratio from the balanced
chemical equation we calculate the moles of
H
2
SO
4
needed to react with this amount of
NaOH:
? mol H
2
SO
4
= 0
.
00832 mol NaOH
×
1 mol H
2
SO
4
2 mol NaOH
= 0
.
00416 mol H
2
SO
4
This is the amount of H
2
SO
4
that must
have been in the 15.0 mL sample. Molarity is
moles solute per liter of solution:
? M H
2
SO
4
=
0
.
00416 mol H
2
SO
4
0
.
0150 L solution
= 0
.
277 M H
2
SO
4
003
(part 1 of 3) 10 points
Below is the pH curve of a weak acid (HA)
titrated with strong base.
Answer the fol
lowing question based on the interpretation
of this pH curve. Be as accurate as possible.
0
5 10 15 20 25 30 35 40 45 50 55 60
0
1
2
3
4
5
6
7
8
9
10
11
12
13
Titration Curve
Amount of
Base
added (mL)
pH
What is the pH at the equivalence point of
this titration? Your answer must be within
±
10%.
Correct answer: 8
.
2 pH.
Explanation:
The inflection points are shown below.
0
5 10 15 20 25 30 35 40 45 50 55 60
0
1
2
3
4
5
6
7
8
9
10
11
12
13
Titration Curve
Amount of
Base
added (mL)
Acid/Base
concentration (pH)
(18
.
5, 4
.
8)
(37, 8
.
2)
004
(part 2 of 3) 10 points
How much base much be added to make the
solution equalized?
Your answer must be
within
±
10%.
Correct answer: 37 mL.
Explanation:
005
(part 3 of 3) 10 points
What is the p
K
a
for this acid? Your answer
must be within
±
5%.
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 Spring '07
 Holcombe
 Chemistry, pH, ml

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