Suppose that to 50 cm3 of 0.2 N HCl was added 50 cm3 of 0.1 N NaOH. What volume of 0.5 N KOH should be added to neutralize the HCl completely?
Answers
CLEARING UP UNITS
First off, the unit of normality (
N
) means that the concentration is described based on the number of ions that dissociate, so for example,
H
2
SO
4
is
2 N
in
H
+
but
1 N
in
SO
2
−
4
.
For
HCl
,
NaOH
, and
KOH
, it is the same as saying molarity (
M
) in this case, because there is only one
H
+
and one
OH
−
going into solution for each
HCl
,
NaOH
, or
KOH
dissociating (with respect to their corresponding ions).
So, let's just say we have
0.2 M
HCl
,
0.1 M
NaOH
, and
0.5 M
KOH
. It's the same thing in this case, anyway, and more familiar to us.
After clearing up the units, this problem is basically a segmented equimolar neutralization of acid. All we have to do is:
Find out how much
H
+
was in solution before we did anything.
Find out how much
H
+
is now in solution after adding
50 cm
3
NaOH
.
Find out how much
KOH
needs to be added to neutralize the remaining
H
+
in solution.
STARTING CONCENTRATION OF PROTONS
The volume
V
HCl
of
0.2 M
HCl
is:
V
HCl
=
50 cm
3
=
50 mL
=
0.050 L
So the
mol
s of
HCl
that dissociated in water to give
H
+
is:
n
HCl
=
0.2 M
×
0.050 L
=
0.01 mols
of
HCl
to begin with.
CURRENT CONCENTRATION OF PROTONS
We know that
50 cm
3
of
NaOH
was added, so by adding
0.050 L
of
NaOH
, we have added
n
NaOH
=
0.1 M
×
0.050 L
=
0.005 mols
of
NaOH
,
meaning that half of the
HCl
has been neutralized; we went from
n
HCl
,
starting
=
0.01 mols
to
n
HCl
,
current
=
0.01
−
0.005 mols
=
0.005 mols HCl
.
HOW MUCH KOH IS NEEDED
Therefore, when we add
0.5 M
KOH
, we just need to neutralize
0.005 mols HCl
, irrespective of the solution's current volume---the
mol
s of something do NOT change with volume .
Since the
OH
−
produced by dissociating
KOH
is equimolar with the
H
+
produced by dissociating
HCl
, that means we need
0.005 mols
KOH
. Therefore, we need this volume:
V
KOH
=
0.005
mols KOH
×
1 L
0.5
mols KOH
=
0.01 L KOH
I looked up the answer online and found this. Since the actual answer was
10 cm
3
...
10 cm
3
=
10 mL
=
0.01 L
∴
our answer is correct.
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