Every morning, before the CTD cast (when we drop the Niskins into the ocean and collect our samples), my main job is to make the standard nitrite solutions. Unlike most solutions, which are made in deionized water, the nitrite solutions are made using seawater we collect from the previous day’s cast. Seawater contains tons of bacteria – these bacteria will consume nitrite and change the actual concentration of nitrite in the bottle, making it completely useless. This is why they must be made daily.
Everything I need to make my dilutions is seen above: an original solution, a pipette, and a volumetric flask. The hardest part of a dilution is the calculation. Remember, concentration is
In dilutions, the number of moles remains unchanged so we can rearrange the equation to solve for moles: mol = M × V.
Why are moles constant in dilutions?
Because moles stay constant, moloriginal = moldilution , or mol1 = mol2. Substituting in molarity and volume gives the dilution equation, M1V1 = M2V2. This is equation Andrew and I use to create our standards. We use 10µM and 1mM standards, then dilute them to create the concentrations we need for the experiment. For nitrite, we use 100nM, 200nM, 400nM, 800nM, and 1000nM standards. So, in addition to the dilution calculation, we also have a lot of unit conversions between nM, mM, and µM. You can never forget about them!!
Here is a sample problem for you:
I have a 10µM solution of NaNO2. I need to make 0.100L of a 400nM NaNO2 solution. What volume of the original 10µM solution do I need?
So, to make this dilution, you would pipette out 0.004L (4mL) of your 10µM solution and put that in a volumetric flask. You’d then fill the flask to 100mL to make the desired 400nM solution. These concentrations are very small – in class, you won’t be using µM and nM concentrations, but typically M concentrations.
Now here’s one for you to work out on your own:
You want to make 10.0L of a 4.50M HCl dilution from a 12.0M HCl solution. How much of the original solution will you need?
Try this problem – the answer, so you can check your work, is V1 = 3.75L
Now, both of these problems showed you how to solve for V1. You will be solving for any of the four variables, so make sure you do your homework!
Until next time,