ig996 – Homework 7 – Lyon – (53565)
1
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This homework is due Thursday, March 6
by 12:00pm.
001
10.0 points
A lake in central Texas contains 1 billion liters
of water.
The pH is normally 6.0, but it is
tested at 5.0 and the fish are in danger. How
much CaCO
3
should be dumped in to bring
the pH back to 6.0? Assume
CaCO
3
+ H
+
→
Ca
2+
+ HCO
3
−
The
molecular weight
of CaCO
3
is
100.1
g/mol.
1.
900 kg
correct
2.
157 kg
3.
90 kg
4.
900,000 kg
5.
157 g
Explanation:
V
= 10
9
L
pH
normal
= 6
.
0
pH
current
= 5.0
CaCO
3
= 100.1 g/mol
[H
3
O
+
]
needed
= [H
3
O
+
]
current

[H
3
O
+
]
normal
= (10
−
5
.
0

10
−
6
.
0
) M
= 9
.
0
×
10
−
6
M
CaCO
3
+ H
3
O
+
→
Ca
2+
+ HCO
−
3
+ H
2
O
n
CaCO
3
=
n
H
3
O
+
= (10
9
L)(9
.
0
×
10
−
6
mol
/
L)
= 9
.
0
×
10
3
mol
m
CaCO
3
= (9
.
0
×
10
3
mol)(100
.
1 g
/
mol)
= 9
.
0
×
10
5
g
= 900 kg
002
10.0 points
The pH of 0
.
1 M HClO
2
(chlorous acid)
aqueous solution was measured to be 1
.
2.
What is the value of p
K
a
for chlorous acid?
1.
2.57
2.
0.11
3.
0.96
correct
4.
1.20
5.
3.91
6.
1.40
Explanation:
M
= 0
.
1 M
pH = 1
.
2
Analyzing the reaction with molarities,
HClO
2
+ H
2
O
⇀
↽
H
3
O
+
+ ClO
−
2
0
.
1

0
0

x

x
x
0
.
1

x

x
x
[H
3
O
+
] = [ClO
−
2
] = 10
−
pH
= 10
−
1
.
2
= 0
.
0630957 mol
/
L
The
K
a
is
K
a
=
[H
3
O
+
][ClO
−
2
]
[HClO
2
]
=
(0
.
0630957)
2
0
.
1

0
.
0630957
= 0
.
107876
and the p
K
a
is
p
K
a
=

log(0
.
107876) = 0
.
967077
.
003
10.0 points
A solution is prepared by dissolving NaCl in
water. The pH of this solution is closest to
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ig996 – Homework 7 – Lyon – (53565)
2
1.
11.
2.
7.
correct
3.
5.
4.
9.
5.
3.
Explanation:
NaCl is a neutral salt. It does not alter pH.
004
10.0 points
A 0.28 M solution of a weak acid is 3.5%
ionized. What is the pH of the solution?
1.
0.55
2.
3.17
3.
2.01
correct
4.
5.25
5.
1.46
Explanation:
M
= 0
.
28 M
P
= 3
.
5%
3
.
5% of the 0
.
28 M is ionized (contributes
to pH), so
[H
+
] = (0
.
28 M)
×
3
.
5
100
= 0
.
0098 M
pH =

log[H
+
] =

log(0
.
0098) = 2
.
00877
005
10.0 points
What is the H
3
O
+
concentration of a 0.15 M
solution of NH
4
Cl in H
2
O at 25
◦
C? (
K
b
for
NH
3
= 1
.
8
×
10
−
5
)
1.
9
.
1
×
10
−
6
M
correct
2.
1
.
6
×
10
−
3
M
3.
1
.
1
×
10
−
9
M
4.
1
.
0
×
10
−
14
M
5.
4
.
9
×
10
−
5
M
6.
8
.
2
×
10
−
1
M
Explanation:
C
a
= 0
.
15
K
w
= 10
−
14
K
b
= 1
.
8
×
10
−
5
NH
+
4
⇀
↽
NH
3
+ H
+
K
a
=
K
w
K
b
=
10
−
14
1
.
8
×
10
−
5
= 5
.
56
×
10
−
10
Since NH
+
4
is a weak acid, use the weak acid
equation:
[H
+
] =
radicalbig
K
a
C
a
=
radicalBig
(5
.
56
×
10
−
10
)(0
.
15)
= 9
.
13
×
10
−
6
M
006
10.0 points
K
a
for HClO is 3
.
5
×
10
−
8
.
What is the
pH of a 0.3 M solution of HClO? You may
approximate the full quadratic for problem
because
K
c
is so small. The reaction is
HClO + H
2
O
→
H
3
O
+
+ ClO
−
K
w
= 1
×
10
−
14
.
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 Spring '08
 DavidVandenbout
 pH, kA

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