Version 001 – Exam 2 – laude – (51635)
1
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001
6.0 points
Consider the titration of equal volumes of
0.1 M HCl and 0.1 M HC
2
H
3
O
2
with 0.1 M
NaOH. Which of the following would be the
same for both titrations?
1.
Two of the other answers are correct.
2.
the initial pH
3.
the volume of NaOH added to reach the
equivalence point
correct
4.
the pH at the half equivalence point
Explanation:
Both HCl and CH
2
C
3
O
2
H
2
are monoprotic
acids and so if the volume concentrations are
identical, the amounts of replaceable H
+
in
both solutions are identical.
002
6.0 points
Oxalic acid is a relatively strong diprotic acid
with
K
1
= 5
×
10

2
and
K
2
= 5
×
10

5
. What
is the [HC
2
O

4
] concentration in an oxalic acid
solution that is 0.10 M?
1.
5
×
10

5
2.
0.10 M
3.
0.025 M
4.
0.075 M
5.
0.05 M
correct
Explanation:
K
1
= 5
×
10

2
K
2
= 5.3
×
10

5
In this case, since we’re considering a solu
tion of H
2
C
2
O
4
and HC
2
O

4
, we should use
the expression for
K
1
here:
K
1
=
[H
+
] [HC
2
O
4

]
[H
2
C
2
O
4
]
Substituting in values for the concentrations
and
K
1
, we have
5
×
10

2
=
(
x
) (
x
)
(0
.
10

x
)
As H
2
C
2
O
4
has a relatively large
K
1
, we
CANNOT assume that
x
is very small com
pared to 0.10 and consequently we must use
the quadratic equation to solve for
x
:
5
×
10

2
(0
.
10

x
) =
x
2
0
.
005

5
×
10

2
x
=
x
2
x
2
+ 5
×
10

2
x

0
.
005 = 0
Using the quadratic equation, we discover
that
x
is 0.05. Bearing in mind that we called
[HC
2
O

4
]
x
in the equation, [HC
2
O

4
] is 0.05
M.
003
6.0 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.
10.98
correct
2.
10.02
3.
12.02
4.
12.30
5.
7.00
Explanation:
004
6.0 points
100 mL of 0.50 M BaCl
2
is added to 100 mL of
0.50 M HCl. Which of the following equations
is the correct charge balance for this system?
1.
2[Ba
2+
] + [H
+
] = [OH

] + [Cl

]
correct
2.
2[Ba
2+
] = [Cl

]
3.
[Ba
2+
] = 2[OH

]
4.
[Ba
2+
] + [H
+
] = 3[Cl

]
5.
[Ba
2+
] + [H
+
] = [OH

] + [Cl

]
6.
2[Ba
2+
] + [H
+
] = 2[OH

] + 2[Cl

]
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Version 001 – Exam 2 – laude – (51635)
2
7.
2[Ba
2+
] + [H
+
] = 2[OH

] + [Cl

]
Explanation:
005
6.0 points
Estimate the pH of 0.10 M Na
2
HPO
4
(aq)
given p
K
a1
= 2
.
12, p
K
a2
= 7
.
21, and p
K
a3
=
12
.
68 for phosphoric acid.
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 Spring '07
 Holcombe
 N2 H4

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