Two means to an N
Page 1: Intuitive method
First case
Consider: you have a solution that is 5 times the concentration you desire ("5x")
You will need to reduce it by what fraction to bring it to the right concentration? In other
words, what value, when multiplied by the current concentration, yields the desired outcome?
This value represents the fraction of the FINAL solution that should be contributed by the
stock/starter (5x) solution
That being the case, if the final volume desired is 150 ml, what is the desired fraction of the
final volume should be made from the concentrated starter? (example: 1/5 of 20 ml = 4 ml)
Second case
You have a solution that is 200 millimolar (mM; a measure of concentration expressed in
units of molecules/volume)
You desire a final concentration of 10 mM.
How 'too-concentrated' is the starter solution? In otherwords, it is has many TIMES more
molecules/volume than desired?
OK, now we're back on familiar territory. What FRACTIONAL value is needed to bring the
concentrated stock to the desired concentration?
(yeah, it's always the reciprocal of the 'over-concentration' of the starter solution)
OK, with that information in hand, if a final volume of 500 mls of solution is desired, how
much of it should be contributed from the 200mM starting solution?
Page 2: Deriving a general equation
Prob #1: You desire a solution of 100 mls containing 50 molecules/ml final concentration.
How many molecules is that?
What operation did you perform? (express you answer in terms of how you manipulated the
CONCENTRATION and VOLUME values)
So you have a solution that has 100 molecules/ml; how many mls. do you need to take to
achieve the number of molecules above?
Prob #2: You have a solution (#1) of 250 mls. containing 70 molecules/ml.
You desire 1000 mls of a solution at 14 molecules/ml. In total, then, How many molecules are
sought?
So how many mililiters of solution #1 contains that desired number of molecules?
NOTE that in both problems on this page, we went through an intermediate measurement-the number of molecules. In other words, in each case, the solution path was
(1) Desired # of molecules = (concentration of final solution) * (volume of final solution)
combined with
(2) (needed volume of stock solution) = (Desired # of molecules) / (Concentration of stock
solution)
Multiplying both sides of #2 by (concentration of stock solution) yields
(2a) (volume of stock solution) * (concentration of stock stock solution) = Desired # of
molecules
Setting (2a) and (1) equal due to the presence of 'Desired # of molecules' in each, we derive
(3) (concentration of final solution) * (volume of final solution) = (volume of stock solution) *
(concentration of stock stock solution)
Equation 3 has been formalized as C V = C V
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