Some musings on dilution
Brian Lamp
Last modified: October 29, 2007
Dilution by Mass versus Dilution by Volume
If you were to think about the process of diluting a sample from a higher concentration to a
lower concentration, most chemistry students (and likely most chemists) would envision doing
so volumetrically. We picture using the relationship CconcVconc = CdilVdil, to figure out the volume
(Vconc) of concentrated solution (Cconc) we would need to measure to prepare a fixed volume
(Vdil) of solution at a new concentration (Cdil). We could then go into the lab and, using things
like pipets and volumetric flasks, prepare the solution.
Many times this volumetric approach is appropriate and quite reasonable, but there are
situations where another approach works well. For example, if a material is not compatible with
glass pipets and flasks, then using them in the volumetric approach just won’t work. In cases
like this, we need an alternative approach. Often the easiest approach is to do our dilutions
based on mass, instead of volume. Remember, whenever we do a dilution, we are just scaling
the concentration of the more concentrated solution down by some dilution factor defined by the
ratio of the concentrations of the dilute and concentrated solutions. In the volumetric dilution
above, that factor is equal to the ratio of Vconc to Vdil, as shown below.
dil
conc
conc
dil
V
V
C
C
dilution factor = =
Measuring volumes is an easy way to define the dilution factor, but making mass measurements
allow us to do that as well. As a matter of fact, we can easily derive a relationship based on
dilution by mass: Cconcmconc = Cdilmdil, where we’ve used masses (mconc and mdil) instead of
volumes and concentrations with units of mass (or moles) solute per mass of solution.
Here’s an example: Let’s say that we have a 1000.0 ppm solution of lead and we need to make
35 ppm solution. In order to prepare this solution, our concentrated solution needs to be diluted
by a factor of Cdil/Cconc or 35/1000. How might we accomplish this both volumetrically and by
mass?
How do we attain this dilution factor volumetrically? On a volume basis, the 35/1000 factor
means that for every 35 mL of 1000 ppm solution, we would need to dilute the solution to a total
volume of 1000 mL (1L!). Practically, how might we accomplish this? One approach would be
to plan to prepare 1000 mL of our dilute solution. To do this would require delivering (precisely)
35 mL of our stock solution. The best way to do this would be with a pipet. After scouring the
lab, we realize that 35 mL pipets are hard to come by and we need to try another approach. We
could use multiple pipets, but each manipulation comes with an increase in uncertainty. Plus, if
we only need a few mL of solution, making 1 L may be wasteful. We know that 10 mL pipets
are readily available, so why not use one of those instead? If we do that, our 10 mL aliquot
would need to be diluted to: 10 mL(1000/35) = 285.71 mL. The best vessel to dilute this
solution in would be a volumetric flask, but good luck finding that 285.71 mL volumetric! It looks
like a volumetric approach isn’t convenient here. Will a mass approach be any better?
To do the dilution by mass, we first measure a mass of our concentrated solution and add
enough solvent until we have reached the appropriate dilution factor. Lets say we plan on
starting with 5.00 grams of our 1000 ppm solution, what mass of solution must we dilute this to
in order to get to 35 ppm? Using the dilution factor, 5.00g(1000ppm/35ppm) = 142.85 g. So, if
we go into the lab and place a clean, dry container on a balance, tare the balance, and add
5.00 g of our 1000 ppm solution, we need only add enough water so that the total mass is
142.85 grams in order to prepare a 35 ppm solution! Even if we weighed out 4.95 g of our
solution instead of 5.00 g, we could still prepare the diluted solution accurately. We just dilute to
4.95g(1000ppm/35ppm) = 141.43 g instead. As long as we know the initial and final masses,
we can readily calculate the final concentration. Mechanically, this is a simpler process than
pipetting and filling volumetrics, and leads to the same result!
By now you might probably be asking: Why not dilute by mass always? In principle, we could.
But, there are some practical limitations. If precision is critical, we need to take a step back and
determine which approach gives us acceptable precision. This is a case-by-case decision.
Since for many dilution factors, appropriate pipets and volumetric flasks are available at a lower
price than a good balance, so practicality plays a role. Also, we typically try to minimize
exposure of electronic equipment to liquids. So, if doing a dilution volumetrically is convenient,
we typically do so. But for situations where reagents are limited or where dilution factors
preclude the volumetric approach, dilution by mass is the way to go!
Serial Dilutions
Here’s another situation. Let’s say we want to prepare a 10 ppb solution from our 1000 ppm
lead solution above. How might we go about this? Our earlier discussion suggests that we
need a dilution factor of 10/1,000,000 = 1/100,000 (remember 1000 ppm = 1,000,000 ppb) to
make this happen. Doing this in a single step is pretty challenging and typically results in poor
accuracy in the concentration of the diluted solution. If we were going to do this dilution by
volume and prepare even 1L (1000 mL) of solution, we would need to begin with:
1000 mL(1/105) = 0.010 mL of our 1000 ppm stock solution. Even with micropipets, measuring
10 μL (0.010 mL) is difficult to do precisely. The dilution by mass approach also has some
problems here. If we were to start with 1.00 gram of our stock solution, we would need to dilute
it to 100,000 g to obtain our desired concentration. That’s about 100 L! Not very practical.
The way around this challenge is to perform the dilution in a series of steps, each one leading to
an intermediate concentration until we finally reach the target concentration. This process is
known as serial dilution. A serial dilution approach to our problem might go like this:
1. measure 1.00 grams of our 1000 ppm solution and dilute to 100.00 grams to prepare a
10.0 ppm solution. (dilution factor = 1/100)
2. measure 1.00 grams of the 10 ppm solution and dilute to 100.00 grams to make a
100. ppb solution. (dilution factor = 1/100 , overall dilution factor = 1/10,000)
3. measure 1.00 grams of the 100 ppb solution and dilute to 10.00 grams to prepare a
10.0 ppb solution. (dilution factor = 1/10, overall dilution factor = 1/100,000)
We could develop a volumetric approach as well, but the process is really the same. Ultimately,
we end up with the same dilution factor, but in a more practical manner. In order to maximize
precision and minimize error, we try to keep the number of intermediate concentrations to the
minimum, within the constraints of what is practical.