DILUTING A SOLUTION. The objectives of this video are twofold : Using the dilution formula Establishing the steps to follow in order to realize the dilution. How to prepare a solution of 100 mL :. Pour around half of your solvent , in this case 50 mL.
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The objectives of thisvideo are twofold:
Using the dilution formula
Establishing the steps to follow in order
to realize the dilution
2. Put all of your solute into the half solution, making sure that no solute is lost.
3. Stir (dissolve) the solute completely
4. Complete the solution by topping it off with solvent to reach the 100 mL line. Remember, that it is 100 mL of solution not 100 mL of solvent.
5. Stir again
We have a 100 mL solution of Na2SO4, a salt solution, dissolvedin water. The concentration is 0.1 M (moles/ liter).Let’ssaywewant to diluteall of this solution to 0.01 M.
How much water must weadd?
The formula is: C1V1 = C2V2
where C1 = initial concentration
V1 = initial volume , or volume to use
C2 = final concentration
V2 = final volume, or volume to make
Here are someeasynumbers:
If C1 = 10
V1 = 4
C2 = 5
and C1 x V1 must alwaysequal C2 x V2,
Then10 x 4 = 5 x
10 x 4 / 5 =
Weseethat as C decreases ( 2 times less), 10x4=5x8
V must converselyincrease ( 2 times more) 40=40
In our case,C1 = 0.1 M ( 0.1 molar or 0.1 moles/liter)V1 = 100 mLC2 = 0.01 MV2 = ?There are 3 variables for whichwe have information.Wealso know that, as C decreases, V must increase.
Applying the formula, weget:
C1 X V1 = V2
0.1M x 100 mL V2 0.01M
1000 mL = V2
The question thatremainsis: How much water do weneed to add?In the lab, dilutingalwaysinvolvesaddingmore solvant.We have found V2, the volume to make; not thevolume to add.
This requires an additionnalstep:
V2 – V1 = Volume to add
1000 mL – 100 mL = 900 mL
Weneed to make 1000 mL of solution by adding 900
mL of solvant, water.
Dilution lowers the concentration of a substance by adding a solvent.
Let’s look at our previous example: 100 mL solution
a) Volume to use: V1: 100 mL First concentration: C1: 0.1 M
b) Final volume: V2: 1000 mL Second concentration: C2: 0.01 M
Adding 900 mL of solvent
C1 = n1
n1 = C1 V1
C2 = n2
n2 = C2 V2
Now with this formula, we can predict concentration and dilutions of solutions. Laws of concentration: C1V1 = C2V2
n1 = n2
C1V1 = C2V2
What if westillwant to dilutefrom 0.1M to
0.01M but wewant to make a small volume
of diluted solution, say 50 mL.
Westill have the initial solution atourdisposal,
C1 = 0.1 M
V1 = 100mL
C2 = 0.01 M
V2 = 50mL
As C decreasestenfold,
V must increasetenfold.
If weadd water to end
with 50, wecan’t use
all of our initial 100mL!
This meansthatwe are actuallylooking for V1.
V1 = C2 x V2
V1 = 0.01M x 50 mL
V1 = 5 mL
In conclusion, weuse 5 mL of our initial solution
and add 45 mL to make 50 mL of solution (final).