IBDP Chemistry
S1.4: Stoichiometry
Dilutions
When a solution is diluted by the addition of more water, the number of moles of solute –
the number of particles - does not change, they are just spread through a greater
volume.
DILUTION: water is added and
the solute particles are more
spread out, but the number of
particles remain constant.
Number of moles in initial solution = number of moles in the
diluted solution
i.e. η1 = η2
since η1 = C1 x V1 and η2 = C2 x V2
then C1 xV1 = C2 x V2
This relationship can be used to calculate the quantities needed to dilute stock solutions
to the desired concentrations. Any unit of volume can be used in this equation, as long
as it is the same on both sides.
Answer the following questions:
1. 25 cm3 of water is added to 125 cm3 of a 0.15 mol dm-3 NaOH solution. What is
the molar concentration of the diluted solution?
2. Water is added to a 100 cm3, 0.15 mol dm-3 NaOH solution until the final volume
is 150 cm3. What is the molar concentration of the diluted solution?
3. How much 0.05 mol dm-3 HCl solution can be made by diluting 250 cm3 of 10
mol dm-3 HCl?
4. How much water would need to be added to 500 cm3 of a 2.4 mol dm-3 KCl
solution to make a 1.0 mol dm-3 solution?
ANSWERS
1.
C1V1 = C2V2
(0.15 )(125 cm3) = x (150 cm3)
x = 0.125 mol dm-3
2.
C1V1 = C2V2
(0.15 )(100 cm3) = x (150 cm3 )
x = 0.100 mol dm-3
3.
C1V1 = C2V2
(10 C)(250 cm3) = (0.05 ) x
x = 50,000 cm3
4.
C1V1 = C2V2
(2.4 )(500 cm3) = (1.0 ) x
x = 1200 cm3
1200 cm3 will be the final volume of the solution. However, since there’s already
500 cm3 of solution present, you only need to add 700 cm3 of water to get 1200
cm3 as your final volume. The answer: 700 cm3.
