Suppose you were asked you
to make one hundred milliliters of a 25 parts per billion (ppb) solution
from a five hundred parts per million (ppm) standard. Don't panic! If
you can do this off the top of your head you are the exception to the
rule. The object of this exercise is to get you to a point where you
can easily and simply perform serial dilutions. One of the problems
with performing dilutions is that there is no handy dandy, all-purpose
formula you can easily refer to. The strategy you use is pretty much
left up to you. It is often dependent upon the volume of your available
receptacles, pipettes, and the concentration of your initial and final
solutions.
CONCENTRATIONS
The most crucial fact to understand is that percent, ppm, and ppb are
all concentrations. Since we are dealing with solutions (liquids) in
this case, the term concentration, simply says that there is a certain
amount of solid "stuff" dissolved in a certain amount of liquid.
DEFINITIONS
These following definitions
will form the absolute foundation for serial diluting.
MEMORIZE THESE!!!!
a. A ppm is equal to 1 ug
per milliliter
b. A ppb is equal to 1 ng per milliter
c. 1000 ppb equals 1 ppm
d. 1000 ng equals 1 ug
PRACTICAL DEFINITION
OF PPB CONCENTRATION
What exactly
does it mean when we say that a solution has a concentration of 1 ppb?
Well, the term 1ppb should say to you that in this vessel of liquid,
for every milliliter of liquid I take out of it , there will be 1ng
of "stuff" dissolved in it.
OK. With the definitions
under our belts let's try that dilution we mentioned at the beginning
of the article.
From a 500 ppb standard lets
make 100 mls of a 25 ppb solution. First we need a practical method
to approach this problem. You must first ask yourself: How much "stuff"
do I need? To find the answer to this we just look at what we were asked
for. Now remember our practical definition: in a 1 ppb solution, for
every milliliter, there is 1ng of "stuff" in that ml . That
is to say, every time we take 1ml of the solution out of its container,
1 ng of "stuff" will be in that 1 ml . Therefore, for a 25
ppb solution we would have 25 ng of "stuff" in 1 ml of liquid..
But we don't need 1ml of a 25 ppb solution; we need 100 mls. So in total
we need (100 x 25ng) or 2500 ng of stuff. Now since we know how much
stuff we need where do we get it from? We get it from the bottle of
500ppm solution that has been provided to us. Using our practical definition
again we know that for every ml of solution we take out of this bottle
there will be 500ug of stuff in it. One ml of the 500 ppm solution has
500 ug or by the memorized definitions 1ug = 1000ng therefore 500ug
= 500,000ng. Great! That's more than enough. We can easily get our 2500
ng from this. But we only need 2500 ng. (This is where it gets a little
tricky and you have to decide what math you will use to figure out how
to get your 2500 ng from the 500,000ng. You can use fractions or percentages
or proportions. It's up to you. ) We only need that fraction of the
ml that will give you 2500ng or 2500ng/500,000ng of a ml. This is equal
to 1/200 of a ml of the 500 ppm solution. We then pipette out 1/200
of a ml of the 500ppm solution, place it in a 100 ml volumetric flask,
and fill to the mark. (In real world dilutions always try to make the
volume, of whatever is being diluted, no less than 1ml)
Congratulations. You have
completed your first dilution.
MAKING
A SOLUTION FROM A COMPOUND
Suppose you
were asked to perform the previous dilution but with a slight twist.
Suppose you were asked to make one hundred milliliters of a 25 parts
per billion (ppb) solution of lead from a five hundred parts per million
(ppm) lead standard but instead of being given the 500 ppm Pb standard
the instructor hands you a bottle of lead oxide, PbO. What do you do?
Well the first thing you need to know is how much lead is in this lead
oxide. First find the formula weight of lead oxide or PbO. Calculate
it by simply adding the molecular weight of all the elements in the
compound. The formula weight for lead is about 233 grams 207 grams of
which is lead and 16 grams of which is oxygen. So we then set up a proportion.
You say for every 233 grams of PbO there are 207 grams is Pb therefore
for every x grams of PbO there will be y grams of Pb in it . It should
look something like this:
So now you have
a way of figuring out how much lead you can get out of the lead oxide.
Now you have to figure out how much lead you need to make your 500 ppm
lead standard. Again, recall your definition. For every ml of standard
there will be 500 ug of lead dissolved in it. Lets say you decide to
make 100 mls of the standard so therefore you will need 50,000 ug ,or
50 mg, or .05 grams of lead. So we plug this into the proportion
We now simply
weigh out .0563 grams of the PbO into a 100 ml volumetric flask and
bring it up to the mark with water. Congratulations! You have made your
standard 500 ppm solution of lead from PbO and can now carry out the
previous dilution.
The contents of this page
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