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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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diluting a solution


The objectives of thisvideo 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

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

50 mL


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

let us first examine the relationship between volume and concentration
Let us first examine the relationshipbetween volume and concentration.

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

Now, since:

n1 = n2


C1V1 = C2V2

here is a different situation
Hereis a different situation.

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,

100 mL.

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!

so we will only take some of our initial solution how much
So, wewillonlytakesome of our initial solution.How much?

This meansthatwe are actuallylooking for V1.

V1 = C2 x V2


V1 = 0.01M x 50 mL

0.1 M

V1 = 5 mL

we again need to find how much water to add volume to add v 2 v 1 50 ml 5 ml 45 ml
Weagainneed to find how much water to add: Volume to add = V2 – V150 mL – 5 mL = 45 mL

In conclusion, weuse 5 mL of our initial solution

and add 45 mL to make 50 mL of solution (final).