H09_pH curves Indicators - krohn(tek335 H09 pH curves...

krohn (tek335) – H09: pH curves Indicators – Mccord – (52445) 1
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Multiple-choice questions may continue on
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before answering.
001 10.0 points
At the stoichiometric point in the titration of
0.130 M HCOOH(aq) with 0.130 M KOH(aq),
1. [HCO
-
2
] = 0.130 M.
2. [HCOOH] = 0.0650 M.
3. the pH is less than 7.
4. the pH is greater than 7. correct
5. the pH is 7.0.
Explanation:
002 10.0 points
What is the pH at the half-stoichiometric
point for the titration of 0.22 M HNO
2
(aq)
with 0.01 M KOH(aq)? For HNO
2
, K
a
=
4 . 3 × 10
- 4
.
1. 7.00
2. 3.37 correct
3. 2.16
4. 2.01
5. 2.31
Explanation:
003 10.0 points
For the titration of 50.0 mL of 0.020 M aque-
ous salicylic acid with 0.020 M KOH(aq), cal-
culate the pH after the addition of 55.0 mL of
KOH(aq). For salycylic acid, p K
a
= 2.97.
1. 11.26
2. 12.02
3. 10.98 correct
4. 12.30
5. 7.00
Explanation:
004 10.0 points
Consider the titration of 50.0 mL of 0.0200 M
HClO(aq) with 0.100 M NaOH(aq). What is
the formula of the main species in the solution
after the addition of 10.0 mL of base?
1. ClO
2
2. NaOH
3. ClOH
4. HClO
2
5. ClO
-
correct
Explanation:
005 10.0 points
You have a solution that is buffered at pH =
2.0 using H
3
PO
4
and H
2
PO
-
4
(p K
a1
= 2 . 12;
p K
a2
= 7 . 21; p K
a3
= 12 . 68). You decide to
titrate this buffer with a strong base. 15.0 mL
are needed to reach the first equivalence point.
What is the total volume of base that will
have been added when the second equivalence
point is reached?
1. > 30 mL correct
2. A second equivalence point in the titra-
tion will never be observed.
3. 30 mL
4. < 30 mL
Explanation:
006 10.0 points
50.0 mL of 0.0018 M aniline (a weak base) is
titrated with 0.0048 M HNO
3
. How many mL
of the acid are required to reach the equiva-
lence point?

krohn (tek335) – H09: pH curves Indicators – Mccord – (52445) 2
1. Need to know the K
b
of aniline.
2. 4.21 mL
3. 18.8 mL correct
4. 133 mL
5. Bad titration since HNO
3
is not a strong
acid.
Explanation:
V
aniline
= 50 mL [Aniline] = 0.0018 M
[HNO
3
] = 0.0048 M
Aniline is a monobasic base ( i.e. , it pro-
duces one OH
-
in solution). Thus you can
expect that aniline and HNO
3
will react in a
one-to-one fashion.
With this ratio, we can determine how
much HNO
3
will be required to react with
all of the aniline.
First, convert 50.0 mL aniline into L of aniline:
50.0 mL aniline
parenleftBig
1 L
1000 mL
parenrightBig
= 0.0500 L aniline
Then use the ratio to determine the volume
of HNO
3
needed:
(0.0500 L aniline)
parenleftBig
0 . 0018 mol aniline
1 L aniline
parenrightBig