ch2sol - CH201
SI:
CHAPTER
2
PRACTICE
TEST


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Unformatted text preview: CH201
SI:
CHAPTER
2
PRACTICE
TEST
 1.
Nitric
acid
(HNO3,
Mm
=
63.02
g/mol)
is
a
strong
acid
that
fully
dissociates
into
 H1+
and
NO31‐
in
water.

A
2.000%
by
volume
aqueous
solution
of
Nitric
Acid
(HNO3,
 Mm
=
63.02
g/mol)
has
a
density
of
1.012
g/mL.

What
is
the
molarity
of
NO31‐
ions?
 2.000%
by
vol.
HNO3
=
2.000
mL
HNO3
/
100.0
mL
Solution
 100.0
mL
Solution
*
1.012g*mL­1
=
101.2
g
Solution
 If
the
solution
is
2
%
by
volume
HNO3,
then
100.0
mL
solution
contains
98.00
mL
water
 98.00
mL
water
*
1.0
g*mL
water
(density
of
water)
=
98.00
g
water
 So,
100mL
of
solution
has
a
mass
of
101.2
g,
98.00
g
of
which
is
water.
 101.2g
solution
–
98.00
g
water
=
3.2
g
HNO3
/63.02
g*mol­1
HNO3
=
0.051
mol
HNO3
 0.051
mol
HNO3
*
(1
mol
NO31­/1
mol
HNO3)
=
0.051
mol
NO31­/
0.100
L
=
0.51
M
 
[NO31­]
=
0.51
M
 2.

If
30mL
of
the
solution
from
#1
requires
17.32
mL
of
a
NaOH
solution
to
 completely
neutralize,
what
is
the
molarity
of
OH1‐
in
the
NaOH
solution
(NaOH
 completely
dissociates
into
Na1+
and
OH1‐
ions
in
water).
 Neutralization
Reaction:


 H1+
+
OH1‐

H2O

 0.51
mol*L­1
NO31­
*
0.030
L
=
0.0153
mol
NO31­
*
(1
mol
H1+/1
mol
NO31­)
=

 0.0153
mol
H1+
*
(1
mol
OH1­/1
mol
H1+)
=
0.0153
mol
OH1­
/
0.01732
=
0.8834
M

 [OH1­]
=
0.8834
M
 3.
What
mass
of
Ca(OH)2
is
required
to
make
I.00
L
of
solution
with
the
same
 concentration
of
OH1‐
as
#2?

 0.8834
mol*L­1
OH1­
*
1.00
L
=
0.8834
mol
OH1­
*
(1
mol
Ca(OH)2/2
mol
OH1­)
=

 0.4416
mol
Ca(OH)2
*
74.08
g*mol­1
Ca(OH)2
=
32.71
g
Ca(OH)2
 32.71g
Ca(OH)2
 4.
What
is
the
mole
fraction
of
OH1‐
ions
in
the
solution
from
#3?
 Choose
1.00L
solution
to
make
the
calculations
easier.


 0.4416
mol
Ca(OH)2*
(1
mol
Ca2+/1
mol
Ca(OH)2)
=
0.4416
mol
Ca2+
 0.4416
mol
Ca(OH)2*(2
mol
OH­1/1
mol
Ca(OH)2)
=
0.8834
mol
OH1­
 1000.
mL
H20*
1
g*mL­1
H2O
=
1000.
g
HzO
/
18.00
g*mol­1
H2O
=
55.55
mol
H2O
 XOH1­
=
mol
OH1­/
total
mol
solution
=
mol
OH1­/
(mol
Hz0
+
mol
of
Ca2+
+
mol
H20)
 =
0.8834
/
(55.55
+
0.4416
+
0.8834)
=
0.0155
 
 XOH1­
=
0.0155
 ...
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This note was uploaded on 01/27/2011 for the course CHEM 201 taught by Professor Wilson during the Spring '10 term at N.C. Central.

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