1
CH 301
Chapter 11a
Reactions in Aqueous Solutions II:
Calculations
111 Calculations Involving Molarity Review:
1.
If you know the
percent mass
and
density of a solution
, you can
determine the molarity
of a
solution:
E.g.
A 82.5 % by mass solution of hydrobromic acid has a density of 1.63 g / mL.
What is the molarity of the
hydrobromic acid solution?
3
82.5
1.63
)%
100.
82.5
1.63
1
1
16.6
0
100.
80.91
soln
KOH
gHB
r
rso
ln
mo
le
sHB
r
aM
a
s
s
D
M
g Hbr soln
mL HBr soln
L HBr soln
g HBr
g HBr soln
mol HBr
mL HBr soln
M
g HBr soln
mL HBr soln
g HBr
L H
M
Br soln
==
=
2.
If you want to
make a dilution
of the more concentrated HBr solution to a less concentrated HBr
solution, you can use the following relationship:
11
2 2
1
2
1
:
.
.
MV
Where
the original concentrated HBr solution
some volume of the original concentrated HBr solution that
you will add to water to make a less concentrated HBr solution
the diluted concentration of HBr solut
M
V
ntha
M
io
=
=
=
=
()
2
12
(
).
.
t you want the result of
adding some volume of the more concentrated solution to some
volume of water
the TOTAL VOLUME of the diluted solution V
V
V
=+
You CANNOT use this dilution relationship to solve stoichiometric
solution problems (that is, neutralization questions)!!!!!
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E.g.
You want to neutralize a 135 mL sample of 0.855 M calcium hydroxide.
However, the stock solution
(16.6 M HBr) is way too concentrated to use for this reaction.
Instead, you would like to use a much less
concentrated solution of HBr .
The only source of HBr you have is the stock solution, therefore, you must
perform a dilution of the more concentrated, 16.6 M HBr, solution to make the less concentrated HBr solution.
Procedure: You are going to take 25.0 mL of the 16.6M HBr solution and add it to 150 mL of water.
Now you
must determine the new concentration of HBr solution:
M
1
V
1
=
M
2
V
2
M
1
=
16.6 M (the original concentration)
(16.6 M) (.025 L) =
M
2
(0.175 L)
V
1
= 25.0 mL (1.0 L / 1000 mL)
= 0.025 L (the volume of the
2.37 M =
M
2
original 16.6 HBr solution that you use to add to the water.)
M
2
=
????
V
2
=
175 mL (1.0 L / 1000 mL)
= 0.175 mL (the volume of
original HBr solution plus the volume of the water
added—remember the two volumes are added together)
The final procedure is to neutralize the calcium hydroxide solution.
You want to know how many mL of the
2.37 M HBr solution is required to neutralize the entire sample of 0.855 M calcium hydroxide.
1.
Procedure:
You must write a balanced equation for the neutralization reaction:
22
2
2(
)
(
)
)
(
)
HBr aq
Ca OH
H O l
CaBr aq
+→
+
•
The balanced equation provides the stoichiometry of the reaction: that is; that 2 moles of HBr are
needed to neutralize 1 mole of Ca(OH)
2
.
2.
Determine the numbers of moles of calcium hydroxide there are in the entire volume of the 0.855 M
calcium hydroxide solution.
2
2
2
2
0.855
(
)
'
0.135
(
)
'
0.115
(
)
'
1.0
(
)
'
moles Ca OH
sol n
L Ca OH
sol n
moles of Ca OH
in the sol n
LCaOH so
ln
=
3.
Determine the number of moles of the HBr that are needed to neutralize 0.115 moles of Ca(OH)
2
.
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 Spring '07
 Fakhreddine/Lyon
 Molarity, Neutralization, Reaction, ml, Sodium hydroxide

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