# Dilution Calculations

Solution Dilution - Real Chemistry

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

You multiply the original concentration by the dilution factors for each dilution.

#### Explanation:

A serial dilution is any dilution in which the concentration decreases by the same factor in each successive step.

In serial dilutions, you multiply the dilution factors for each step.

The dilution factor or the dilution is the initial volume divided by the final volume.

$D F = {V}_{i} / {V}_{f}$

For example, if you add a 1 mL sample to 9 mL of diluent to get 10 mL of solution,

$D F = {V}_{i} / {V}_{f}$ = $\left(1 \text{mL")/(10"mL}\right) = \frac{1}{10}$. This is a 1:10 dilution.

Example 1

What is the dilution factor if you add 0.2 mL of a stock solution to 3.8 mL of diluent?

${V}_{f}$ = 0.2 mL + 3.8 mL = 4.0 mL

$D F = {V}_{i} / {V}_{f}$ = $\left(0.2 \text{mL")/(4.0"mL}\right) = \frac{1}{20}$. This is a 1:20 dilution.

Example 2

If you did the above dilution four times, what would be the final dilution factor?

Solution 2

Remember that serial dilutions are always made by taking a set quantity of the initial dilution and adding it successively to tubes with the same volume. So you multiply each successive dilution by the dilution factor.

You would transfer 0.2 mL from Tube 1 to 3.8 mL of diluent in Tube 2 and mix. Then transfer 0.2 mL from Tube 2 to 3.8 mL of diluent in Tube 3 and mix. Repeat the process until you have four tubes.

The dilution factor after four dilutions is

DF = 1/20 × 1/20 × 1/20 × 1/20 = 1/160000 = 1:160 000

If the concentration of the original stock solution was 100 µg/µL, the concentration in Tube 4 would be

100 µg/µL × $\frac{1}{160000}$ = 6.25 × 10⁻⁴ µg/µL

Hope this helps.

The more the solution of a reactant is diluted, the slower the reaction will occur.

#### Explanation:

The key thing you need to understand here is that chemical reactions depend on reactant particles bumping into each other (collision theory).

The more concentrated a reactant is, the more likely it will be to bump into other reactants and product chemical change.

Here is video of a quick lab performed to test this. In the video, varying concentrations of $N {a}_{2} {S}_{2} {O}_{3}$ are allowed to react with HCl. A timer is included in the video so you can see how the changing concentration changes the overall speed of the reaction!

video from: Noel Pauller

Hope this helps!

• This key question hasn't been answered yet.

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