Bodet (ngb299) – Homework 5 – sutclife – (51630)
1
This printout should have 20 questions.
Multiplechoice questions may continue on
the next column or page – Fnd all choices
be±ore answering.
Please note: ’p’ just equals ’negative log o±’
so we can have pH, pOH , pK
a
etc. This HW
combined with HW 3 4 will cover us up to
exam 2, so salts, neutralizations, bufers and
indicators will be on Exam 3 (that’s section
7.8, etc.)
HW due the evening be±ore the
exam.
001
10.0 points
How many grams o± HNO
3
are needed to give
apHo
±3
.
75 in an 11
.
5Ltank? Themo
lar
mass o± HNO
3
is 63
.
1g/mol.
1.
0.0112 g
2.
0.000178 g
3.
0.0000324 g
4.
1.36 g
5.
0.00204 g
6.
0.466 g
7.
5.49 g
8.
0.129 g
correct
9.
2.19 g
Explanation:
pH = 3
.
75
V
=11
.
5L
MM = 63
.
1g/mol
[H
+
]=10

pH
=10

3
.
75
=0
.
000177828
Since HNO
3
is a strong acid,
HNO
3
(aq)
→
H
+
(aq) + NO

3
(aq)
.
One HNO
3
→
one H
+
,so
[H
+
]=[HNO
3
]=0
.
000177828 M
and
(11
.
5L)
×
0
.
000177828 mol HNO
3
1L
.
00204502 mol HNO
3
.
Thus
(0
.
00204502 mol HNO
3
)
×
63
.
1gHNO
3
1mo
lHNO
3
.
129041 g HNO
3
002
10.0 points
The pH o± 0
.
1MHC
lO
2
(chlorous acid)
aqueous solution was measured to be 1
.
2.
What is the value o± p
K
a
±or chlorous acid?
1.
3.91
2.
2.57
3.
0.11
4.
1.20
5.
0.96
correct
6.
1.40
Explanation:
M
.
1M
pH=1
.
2
Analyzing the reaction with molarities,
HClO
2
+H
2
O
±
²
H
3
O
+
+ClO

2
0
.
1

00

x

xx
0
.
1

x

[H
3
O
+
]=[ClO

2

pH

1
.
2
.
0630957 mol
/
L
The
K
a
is
K
a
=
[H
3
O
+
][ClO

2
]
[HClO
2
]
=
(0
.
0630957)
2
0
.
1

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

log(0
.
107876) = 0
.
967077
.
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2
003
10.0 points
A7
.
2
×
10

3
MsolutionoFaceticacidis5.0%
dissociated. In a 7
.
2
×
10

4
Mso
lut
ion
,the
percent dissociation would be
1.
zero.
2.
<
5%.
3.
None oF these
4.
the same.
5.
>
5%.
correct
Explanation:
004 (part 1 of 3) 10.0 points
AgivenweakacidHZhasa
K
a
=3
.
1
×
10

6
.
What is the H
3
O
+
concentration oF a solution
oF HZ that has a concentration oF 0
.
07 mol/L?
Correct answer: 0
.
000465833 mol
/
L.
Explanation:
K
a
.
1
×
10

6
[HZ] = 0
.
07mol
/
L
The expression For
K
a
here is
K
a
=
[H
+
][Z

]
[HZ]
Substituting in values For the concentrations
and
K
a
,wehave
3
.
1
×
10

6
=
(
x
)(
x
)
(0
.
07

x
)
As HZ has a small
K
a
,w
ecana
s
sumetha
t
x
is very small compared to 0
.
07, so we can
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 Spring '07
 Holcombe
 pH, Pk, Acids

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