Ratios and Dilutions
Problem
Scientists often require solutions which are diluted to a particular concentration. In this problem, you can explore the mathematics of simple dilutions.
Imagine you have a beaker containing a solution with a concentration of 100 000 cells per millilitre of liquid. You can transfer some of this solution into a second beaker, in multiples of 10ml, and add water in multiples of 10ml to dilute the solution.
If you diluted 100ml of the original solution with 100ml of water, what would be the concentration, in cells/ml, of your new solution?
Investigate other dilutions that can be made.
You could use this interactivity to check your ideas. The interactivity lets you perform a series of up to four dilutions. To perform a single dilution, transfer 0ml of water and 100ml of solution for the last three dilutions.
Here are some questions to consider:
- Can you make solutions which are half the strength of the original?
One third of the strength? One quarter? One fifth? ... - What about fractions with a numerator greater than 1?
- Are there any concentrations you can make in more than one way?
- What can you say about the concentrations you can't make?
A series of dilutions can be performed, where a solution is diluted, and then the resulting solution is also diluted.
Find two dilutions which give a final concentration of:
- $50000$ cells/ml
- $33333.\dot3$ cells/ml
- $75000$ cells/ml
- $49000$ cells/ml
- $24000$ cells/ml
- $45000$ cells/ml
- $26666.\dot6$ cells/ml
How many different ways can you find to make a final concentration of $25000$ cells/ml?
You could use this interactivity to check your ideas. The interactivity lets you perform a series of up to four dilutions. To perform a series of two dilutions, transfer 0ml of water and 100ml of solution for the last two dilutions.
Find some concentrations which are impossible to create using two dilutions.
How can you convince yourself that they are not possible?
List the necessary criteria for deciding whether a concentration is possible or not.
You may wish to try the problems Investigating the Dilution Series and Exact Dilutions, which expand on the ideas in this
problem.
Getting Started
Think about how you can use these proportions to express the concentration of the new solution each time.
Student Solutions
Caogan looked for patterns in the mixtures first of all, using a trial and error to find the first few answers, and then proceed to generalise into a formula:
To make it simpler to record my results, first I decided to let $U$ be the volume of undiluted liquid added in, let $W$ be the volume of water added, and let $D$ be the concentraion of the new substance.
First, I did a few experiments
| U (ml) | W (ml) | D (cells/ml) | D (as a fraction of original) |
| 100 | 100 | 50,000 | $\frac{1}{2}$ |
| 50 | 100 | 33,333.$\dot{3}$ | $\frac{1}{3}$ |
| 100 | 50 | 66,666.$\dot{6}$ | $\frac{2}{3}$ |
| 10 | 30 | 25,000 | $\frac{1}{4}$ |
| 20 | 20 | 50,000 | $\frac{1}{2}$ |
| 30 | 10 | 75,000 | $\frac{3}{4}$ |
| 10 | 40 | 20,000 | $\frac{1}{5}$ |
| 20 | 30 | 40,000 | $\frac{2}{5}$ |
| 30 | 20 | 60,000 | $\frac{3}{5}$ |
| 40 | 10 | 80,000 |
$\frac{4}{5}$
|
With this formula we can see why there are more than one way of making most concentrations. Anything where there are equivilances to the fraction, then there are more than one way of making it. For example
$\frac{1}{2} = \frac{10}{10+10} = \frac{50}{50+50} = \frac{20}{20+20}$
$\frac{1}{4} = \frac{10}{10+30} = \frac{20}{20+60} = \frac{50}{50+150}$
$\frac{2}{3} = \frac{20}{20+10} = \frac{30}{30+15} = \frac{40}{40+20}$
So the concentraions we can't make are any where they cannot be expressed as a fraction. I have slightly cheated above as I have allowed $15 ml$ of water to be used, and in the example we can only use $10, 20, ... , 90, 100 ml$. So the concentrations we can't make are anything such as $\frac{7}{7+5}$ which is $58,333.\dot{3}$
Caogan did well, using trial and error to find patterns, and then linking them using a formula. Can you think of any other places where this method can be useful?
Jen and Tess built on Caogan's formula, extending it to the two dilution problem:
In Caogan's formula, you have the proportions of $U$ and $W$, which is the fraction of the original strength that the new substance is, then multiplied by the concentration of the original.
So with two dilutions you must have the proportion of the strength of the first dilution, times the fraction of the strength of the original. So if we notate using:
$U$ - volume of undiluted solution
$W$ - volume of water added in first dilution
$D$ - volume of diluted solution
$V$ - volume of water added in second dilution
$C$ - concentraion of solution after two dilutions
Then we get
$C = \frac{U}{U+W} \times \frac{D}{D+V} \times 100,000$
With Jen and Tess's formula, can any one now calculate the given dilutions to make?
Archie continues the problem:
The number of ways to make a final concentration of 25,000 cells/ml will be the same as the number of ways of making the fraction $\frac{1}{4}$ from a product of two fractions.
For example
| U (ml) | W (ml) | D (ml) | V (ml) | $\frac{U}{U+W} \times \frac{D}{D+V}$ | C (fraction of 100,000) |
| 50 | 50 | 50 | 50 | $\frac{50}{50+50} \times \frac{50}{50+50}$ | $\frac{1}{2}\times \frac{1}{2}$ |
| 40 | 60 | 50 | 30 | $\frac{40}{40+60} \times \frac{50}{50+30}$ | $\frac{2}{5} \times \frac{5}{8}$ |
I decided to see if I could come up with a formula for finding the concentration of a substance after arbritarily many dilutions. Using the pattern above, we have that the concentration after dilution will always be $\frac{\text{volume of diluted solution}}{\text{volume of diluted solution + volume of water added}} \times (\text{concentration of solution})$
So if we let $U_i$ be the volume of the solution used at step $i$, let $W_i$ be the volume of water used at step $i$, and let $C_i$ be the concentraion after i dilutions then the concentration after $n$ steps will be
$C_n=\prod_{i=1}^n (\frac{U_i}{U_i + W_i}) \times 100,000$
Excellent work Archie! Developing a formula for one, two and then many cases is extremely useful. Can anyone now use Archie's formula to answer the last few questions on what concentrations are possible?
Teachers' Resources
Why do this problem?
Possible approach
Then ask the class to imagine mixing 100ml of this solution with 100ml of pure water. Students could record any working out and their answer on individual whiteboards. Now show students the interactivity, and explain how it can be used to perform a single dilution, then use it to check their answer. Take time to discuss how they got to the correct answer.
Demonstrate that the interactivity can measure multiples of 10ml of liquid, up to 100ml - the scientific context of this could be using a dropper that measures 10ml at a time.
Ask students to come up with questions they would like to explore using the interactivity - some suggested questions appear in the problem. Then allow them some time to investigate, using the interactivity to check the predictions that they make.
Once students are competent at working with solutions created using one dilution, use the interactivity to perform a series of two dilutions. Perhaps start by giving them a couple of concentrations to work out, using individual whiteboards as before, and using the interactivity to check. At the end of the problem there are some suggested concentrations they could be asked to make.
Pairs of students could take it in turns to create a concentration using two dilutions, and then challenge their partner to work out the dilutions they used.
Finally, the problem challenges students to investigate impossible dilutions.
Key questions
Possible extension
Possible support
Mixing Lemonade investigates the strengths of different solutions informally and may provide a useful starting point.