DILUTIONS
Lecturer: Engr. Umer Hassan
Dilutions
Dilution is the process of reducing the
concentration of a solute in a solution by
adding more solvent. This results in a
larger volume of the solution, but with
the same amount of solute distributed
throughout. The total amount of solute
remains constant before and after
dilution.
Dilutions and Molarity
• Use this formula to make a more dilute
solution from a concentrated solution
Molarity1 x Volume1 =Molarity2 x
(Concentrated) Volume2
(before) = (Dilute)
(after)
M 1V 1 = M 2V 2
Example 1
Suppose you have a 2.0 M sulfuric acid
(H2SO4) solution and you want to make
500 mL of a 0.5 M sulfuric acid solution.
Example 1
Suppose you have a 2.0 M sulfuric acid (H2SO4)
solution and you want to make 500 mL of a 0.5 M
sulfuric acid solution.
M1=2.0M (Initial concentration)
V1=? (Initial volume, what we want to find)
M2=0.5M (Final concentration)
V2=500 mL=0.5 L (Final volume)
M1 V1 = M2 V2
2 x V1 = 0.5 x 0.5
V1 = 0.5 x 0.5 = 0.125 L
2
Example 2
How many liters of 2.5 M HCl are
required to make 1.5 L of 1.0 M HCl?
Example 2
How many liters of 2.5 M HCl are
required to make 1.5 L of 1.0 M HCl?
M 1V 1 = M 2V 2
M1 =
2.5 M V1 =
?
M2 =
1.0 M V2V=
(2.5M) = (1.0M) (1.5 L)
1
1.5 L V = (1.0M) (1.5 L)
1
= 0.60L
(2.5M)
Example 2 part b
M1 = 2.5M V1 = 0.60L M2 = 1.0 M V2= 1.5 L
How much water should you add to the volume
of 2.5M HCl you calculated above to make the
solution? (draw this in your notes)
1st add .60L of HCl
to measuring device.
Example 2 part b
M1 = 2.5M V1 = 0.60L M2 = 1.0 M V2= 1.5 L
How much water should you add to the volume
of 2.5M HCl you calculated above to make the
solution? (draw this in your notes)
Then add enough water to get to
1.5L of solution
V2 – V1 = Amount of water
1.5L – 0.60L = 0.90L water
Example 2 part b
M1 = 2.5M V1 = 0.60L M2 = 1.0 M V2= 1.5 L
How much water should you add to the volume
of 2.5M HCl you calculated above to make the
solution? (draw this in your notes)
Final solution is 1.5L of 1.0M HCl
Example 3
250.0 mL of a 0.500 M HCl solution needs
to be made from concentrated HCl. What
volume of the concentrated solution is
needed if its molarity is 12.0 M?
Example 3
250.0 mL of a 0.500 M HCl solution needs
to be made from concentrated HCl. What
volume of the concentrated solution is
needed if its molarity is 12.0 M?
M 1V 1 = M 2V 2
(12M) V1 = (0.5M) (0.250L)
V1 = (0.5M) (0.250L) = 0.0104L
(12M)
Example 3 part b
How much water would you add to
make the final solution?
M1 =
12.0M V1 =
0.104L M2
= 0.500M
V2 = 0.250L
Example 3 part b
How much water would you add to
make the final solution?
M1 =
12.0M V1 =
0.104L M2
= 0.500M
V2 = 0.250L
0.250L – 0.104L = 0.146L
