2 Diluting things
Very little math knowledge is needed to get by in the lab. However, the math that is needed is used frequently: the calculation to dilute a substance from a higher concentration to a lower concentration. This work is really driven only by a single equation:
\[ C_1\times V_1 = C_2 \times V_2 \]
2.1 Diluting Liquids
Suppose you have a drug that you know has a concentration of \(1\mu M\), and you want to perform an experiment with it. The questions you have to ask yourself first are:
- What concentration do I want the drug to be?
- How much volume of diluted drug to I need?
Suppose you ask yourself this and decide you need \(10mL\) of \(10nM\) drug.
Now we apply our equation - but first, let’s describe what the variables mean.
- \(C_1\): The concentration you start with. For us, this is \(1\mu M\)
- \(V_1\): The volume of the stock you will dilute with medium to get the final volume. We don’t know this.
- \(C_2\): The concentration you want. This is \(10nM\).
- \(V_2\): The volume you want. This is \(10mL\)
If you rearrange the equation, you can get:
\[ V_1 = \frac{C_2\times V_2}{C_1} \]
Which we substitute in our known values for:
\[ V_1 = \frac{10nM\times 10mL}{10\mu M} \]
Now we need to convert units - I find it’s (generally) easier to convert to a smaller unit than vice-versa - do what ever works best for you, just don’t make a mistake!
\[ V_1 = \frac{10nM\times 10mL}{10,000nM} = \frac{1mL}{100} = 0.01mL = 10\mu L \]
Ok, our answer is \(10\mu L\). But what does this mean?
This is the amount of stock (\(10\mu M\)) drug we are going to dilute to \(10mL\). Note that I did not say we would add to \(10mL\). While this would work (and this is in fact what we do in this case), the wording matters if, say, we wanted to dilute from \(10\mu M\) to \(5 \mu M\) with a final volume of \(5mL\):
\[ V_1 = \frac{5\mu M\times 10mL}{10\mu M} = 5mL \]
If we took our \(5mL\) of stock and added to \(10mL\), we would have a final volume of \(15mL\). Instead of having a final concentration of \(5\mu M\), we would have a final concentration of \(\sim 3.3 \mu M\). And woe to thee who is using this method to do a serial dilution, where the error will multiply!
So, back to our problem. We are going to dilute \(10\mu L\) of our stock to \(10mL\). We theoretically should dilute in \(9.99mL\) of whatever we’re diluting with. However, the difference is negligable. In fact, we can use our equation in another way to see just how little it matters:
\[ C_2 = \frac{C_1\times V_1}{V_2} = \frac{10\mu M \times 10\mu L}{10.01mL} = \frac{10\mu M \times 10\mu L}{10010\mu L} = 9.99001nM \]
To put that in perspective, you would get the same error if you pipetted \(9.99\mu L\) vs \(10\mu L\) of stock into your solution - a much more likely error!
So when SHOULD you care?
Generally, I typically care when the volume is \(\ge 1\%\) of my final volume.
2.1.1 Serial Dilutions
Suppose instead of in our previous example where we started with a \(10 \mu L\) stock, we instead started with a \(10mM\) stock. Then, when we do our calculation, we get:
\[ V_1 = \frac{10nM\times 10mL}{10,000,000nM} = \frac{1mL}{100,000} = 0.00001mL = 0.01\mu L \]
There’s simply no way that you’d be able to pipette \(<1\mu L\) - so we need to create an intermediate dilution. By looking at our final volume, we realize that this intermediate dilution must be at least \(100\times\) more dilute than the current stock. (because \(100\times 0.01\mu L = 1\mu L\), and \(1\mu L\) is the lowest we can pipette). To make this intermediate dilution, we again need to ask ourselves the two questions above:
- What concentration do I want the drug to be? (100x more dilute than stock - so \(100\mu M\), for example)
- How much volume of diluted drug to I need? (Enough to be able to make accurately and enough to use for our next (final) dilution - so \(\ge 100\mu L\) so that \(V_1 \ge 1\mu L\))
Unless there isn’t a lot of drug to go around, I usually just make \(V_2 = 1mL\). So to make a \(1:100\) dilution, I would add \(10\mu L\) \(10mM\) drug to \(990\mu L\) medium (if medium is the correct diluent). Then, vortex the sample - you can now start your calculation with a \(100\mu M\) stock instead of a \(10mM\) stock