Transcript SKILLS Project Dilution Calculations Why should I care? • Solution preparation is one of the most common activities is chemistry lab.

SKILLS Project
Dilution Calculations
Why should I care?
• Solution preparation is one of the
most common activities is chemistry
lab. After all, most chemical reactions
occur in water.
• This unit will teach you:
– How to dilute more concentrated
solutions to produce a certain molarity.
How do we do it?
• Dilutions are achieved by adding
additional solvent, usually water, to
more concentrated solutions.
• As a result, a greater volume of
solution is produced with a lower
molarity.
The Equation
C 1 V 1 = C 2V 2
• Where C1 and C2 are the starting
and ending concentrations
(usually molarities) and….
• V1 and V2 are the starting and
ending volumes.
Example 1: Dilution of HCl
• How much 16.2 M HCl will be required to
produce 2.44L of 0.200 M HCl solution?
C1V1 = C2 V2
(16.2 M)(V1) = (0.200 M)(2.44 L)
First
thing’s
setitup
your
We’re
watering
some
amount
Then,
we’re
diluting
down
toequation!
a final
According
tofirst:
thedown
problem,
we’re
•• V
1 = 0.0301L
Remember,
you
can
use
this
(volume)
of the
HCl.
concentration
of1concentrated
0.200
MM.
andequation
volume
starting wih
a
C
of 16.2
whenever
you’re
changing
This
volume,
V1, is
what webetween
are
of
2.44L.
two
different
solving
for. molarities and volumes.
Example 2: Dilution of NaF
• How much 0.240 M NaF solution could be
made from 0.250L of a concentrated 1.00 M
solution of NaF?
C1V1 = C2 V2
(1.00 M) (0.250 L) = (0.240 M)(V2)
Once
again,
we’ll
use
thewith
dilution
Finally,
we’re
looking
for
the
of
We’re
making
athis
solution
a
The
concentration
of
our
starting
volume
of
same,
starting
In
general,
its
often
easier
tovolume
always
•• V
=
1.04L
2
equation.
the resulting
solution.
This
isvolume
the
concentration
ofM.
0.240
Mand
NaF.
solution
1.00
Remember
to “how
pair
solution,
0.250L.
place
theisisconcentration
of
much”
we’ll
be solving
for.
up
your
volumes
and concentrations
the
more
concentrated
solution at the
correctly.
front.
Example 3: Ammonia
• How much water has to be added to 0.300L
of 5.60 M NH3 to make the concentration
0.100 M?
C1V1 = C2 V2
(5.60 M) (0.300 L) = (0.100 M) (V2)
Start
out
withsolution
the We’ll
dilution
equation.
As
The
“ending”
solution
should
be to
Lastly,
we’re
looking
for
the
“ending”
Our
starting
has
a
volume
(V
•• V
=
16.8L.
need
1)
2
always,
this
will
form
the
for
diluted
to(Ca
of
concentration
)molarity
ofis
5.60
Mto0.100
NH
volume.
Note:
this
theframework
total
final
of
0.300down
L.
We’ll
adding
this
1be
3. M
add
16.5
L
of
water.
solving
problem.
NH
. this
volume.
volume
to produce
a lower molarity.
3
Example 4: Milliliters
• What is the initial concentration of a solution if
350 ml can be diluted to a final volume of
1.40L and concentration of 0.150 M?
C1V1 = C2 V2
(C1)(350 ml) = (0.150 M)(1400 ml)
final
C
, the
is is
stated
The
For
Vconcentration,
, we
states
need
that
tovolume
be
we’re
to
Once
again,
we’ll
with
dilution
This
time
our
in to
•• C
=problem
0.600
Mstart
2sure
2around,
1 our
be
0.150
M.
searching
match
units
forwith
thegeneral
our
initial
V1.concentration,
So,
equation
as
our
“framework.”
milliliters
instead
of
liters.
Wewe’ll
canuse
use
C1. as
1400
mllong
instead
of are
1.40consistent
L.
“ml”
as we
as
Remember:
1.0L
= 1000 ml.
use
ml in V2 as
well.
Example 5: Preparations
• If 50.0g HI is dissolved in 500. ml of water,
how much additional water must be added to
produce a final concentration of 0.0200 M HI?
C1V1 = C2 V2
(0.781 M)(500 ml) = (0.0200 M)(V2)
We’re
trying
to
produce
a
final
As
usual,
we
start
with
our
dilution
And
lastly,
we’re
trying
to
find
the final
We
can
also
assume
that
the
total
To
find
our
C
,
we’ll
have
to
solve
for
•• V
=
19525
ml.
An
1
2
concentration,
CM
0.0200
Mhow
HI.500
equation.
soUsing
we starting
can
volume
of
our
solution
is
molarity.
=determine
mol/L,
we
find
that
2, of
additional
19025
ml
(or
much
water solution
will needhas
to be
ml.
the
starting
a added.
19.025L)
be Madded.
concentrationwill
of 0.781
HI.
Practice Problems:
• What is the final concentration of a
solution made by adding 250 ml of
water to 10 ml of 12.0M H2SO4?
0.46 M H2SO4
• What is the initial concentration of an
HCl solution if 0.50L can be diluted
down to 3.00L with a concentration of
1.20 M?
7.2 M HCl