# Exact dilutions

Which exact dilution ratios can you make using only 2 dilutions?

*There are three parts to this question, which require increasing amounts of thought. However, all are based on the standard mathematics of dilutions.*

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You are provided with a large quantity of a solution of a chemical at a standard concentration and a large supply of water with which to dilute the solution. You have three 100ml beakers in which to mix dilutions and a 10ml syringe to transfer fluids from the large beakers to the small beakers or between the small beakers. The 10ml syringe can only measure exactly 10mls.

Part 1:

For certain applications you require non-standard dilutions. Try to use the three beakers and syringe to create a quantity of dilutions in each of these cases (in increasing order of complexity)

- 1/3
- 1/7
- 1/18
- 5/12
- 29/35
- 71/882

Part 2:

Once you have a feel for the way in which to create dilutions, try to create dilutions 1/2, 1/3, 1/4, 1/5, 1/6, 1/7 ... and so on up to 1/19, 1/20. Which can you make and which can you not make? Why? Do you think that an exact dilution 1/362880 will be possible?

Part 3:

I mix a dilution with a concentration $c_A$ in the first beaker and a smaller concentration $c_B$ in the second beaker. I pour some liquid (unmeasured amounts) from beakers A and B into beaker C to make a new dilution with concentration $c_C$. Use common sense to understand and algebra to prove that $$c_A> c_C> c_B$$

On a scrap of paper I found this method for making a 1/13 dilution using the above configuration of beakers:

- Fill beaker A with 100ml of water.
- Put 10ml of solution and 20ml of water into beaker B.
- Roughly fill beaker B with fluid from beaker A.
- Roughly fill beaker A with fluid from beaker B.
- Repeat steps 3. and 4. 10 times.

How accurately will this make the dilution? You might like to use a spreadsheet to work this out. Can you prove how repeating this mixing process will eventually provide the dilution to any required level of accuracy?

Can you extend this method to produce a way of mixing a 1/23 dilution using the above configuration of beakers?

Expressing numbers as a product of prime factors should prove useful in decomposing these dilutions...

in this problem there are two different mathematical operations which can be used: You can dilute a dilution, which corresponds to multiplying fractions; you can mix two dilutions, which corresponds to averaging two fractions.

Part 1:

1) To create a solution of $\frac{1}{3}$, pipette 10ml of solution then 2 x 10ml of water into the 100ml beaker.

2) To create a solution of $\frac{1}{7}$, pipette 10ml of solution followed by 6 x10ml of water into the 100ml beaker.

3) To create a solution of $\frac{1}{18}$ first create a solution of $\frac{1}{9}$ by pipetting 10ml of solution followed by 8 x 10ml of water into the 100ml beaker. Then, pippette 10ml of this solution into another beaker, followed by 10ml of water to give the required concentration.

4) To create a solution of $\frac{5}{12}$ first create a solution of $\frac{5}{6}$ by pipetting 5 x 10ml of solution followed by 10ml of water into the 100ml beaker. Then pipette 10ml of this solution into the second 100ml beaker, followed again by 10ml of water.

5) To create a solution of $\frac{29}{35}$ first note that $\frac{29}{35}=\frac{2}{5}+\frac{3}{7}$. This means that $\frac{29}{35}$ is the average of $\frac{4}{5}$ and $\frac{6}{7}$, so our strategy is to mix equal amounts of these solutions. First make a $\frac{4}{5}$ solution in beaker A by pipetting 4 x 10ml of solution followed by 10ml of water. Then make a $\frac{6}{7}$ solution in beaker B by pipetting 6 x 10ml of solution followed by 10ml of water. Finally, pipette 10ml of the solution in beaker A and 10ml of the solution in beaker B into beaker C, creating 20ml of a $\frac{29}{35}$ solution.

6) To create a solution of $\frac{71}{882}$ requires a few steps. First, create a $\frac{4}{7}$ solution by pipetting 4 x 10ml into the beaker, followed by 3 x 10ml water. Then pippette out 10ml of this into another beaker, followed by 6 x 10ml of water, to give a solution of $\frac{4}{49}$. Discard the original $\frac{4}{7}$ solution and create a $\frac{5}{7}$ solution by pipetting 5 x 10ml solution then 2 x 10ml water. Pipette 10ml of this out into the third beaker, then add 8 x 10 ml of water to give a $\frac{5}{63}$ solution. Now, mix equal amounts of this solution with the $\frac{4}{49}$ solution to give the required $\frac{71}{882}$.

NB there is more than one way to make each solution!

Part 2:

The concentrations of $\frac{1}{2}$ to $\frac{1}{10}$ are trivially created by adding 10ml of solution, followed by (n-1) x 10ml of solution, where n is the denominator of the concentration to be created.

For dilutions less than this, it is not possible to create those which have a denominator which is a PRIME number $p$. Indeed, it is not possible to multiply together two numbers (i.e dilute a dilution) to give this number, and averaging two fractions (i.e mixing two existing dilutions) with denominators which aren't divisible by $p$ gives a fraction whose denominator is not divisible by $p$.

To create the dilutions which are non-prime involves diluting an already made solution. For example, creating $\frac{1}{18}$ involves taking the $\frac{1}{9}$ solution and mixing it in a 1:1 ratio with water.

To see whether it is possible to create a dilution of $\frac{1}{362880}$ it is best to express 362880 as a produce of primes, and to see whether any of these primes are greater than 10. If so, then the solution cannot be made, but if not, then it can be made readily.

$362880 = 2^9 \times 5 \times 7 \times 9^2$. Thus it is possible to make!

Part 3:

Common sense tells us that by mixing a solution of concentration $c_A$ and $c_B$ we will obtain a solution with an intermediate concentration. It can be seen that we will not obtain a more concentrated solution than either, since there are insufficient moles being added for this to be possible! Also, to get a more dilute solution, water would need to be added.

This can be proved as follows:

Suppose I mix $a$ units of $A$ with $b$ units of $B$.

Using the general concentration formula I get $$\frac{ac_A+bc_B}{a+b} = c_C$$

I can rearrange this to give $$ c_A = \frac{(a+b)c_C-bc_B}{a} $$

Next note that $c_A> c_B$, so I have $$ \frac{(a+b)c_C-bc_B}{a}> c_B $$

This reduces to $c_C> c_B$ (using positivity of and b, of course)

Below is a spreadsheet for the creation of the $\frac{1}{13}$ dilution. Notice how tiny the error is after only 10 repeats of steps 3 and 4!!!

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You might be wondering how this actually works. Overall we have placed 10ml of solution among 120ml of water, which is the required concentration. We have then essentially mixed it all together as much as possible until the concentration is constant between both beakers, which will give an overall concentration of $\frac{1}{13}$.

To get a concentration of $\frac{1}{23}$ requires the use of the third beaker. Again, we must add 10ml of solution to 220ml of water. This can be facilitated by having beaker A and B full of water, and Beaker C having 10ml of solution and 20ml of water. B is used to fill C, and then A is used to fill B. After this C is used to fill A, and the cycle repeated. Below is a spreadsheet for 10 iterations:

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