CLOates

OCCC APPM 1313

Chuck’s Quick and Dirty
Dilution Technique

CQDDT

__CQDDT Explained__. When
you’re asked to synthesize (build) a dilution stage that yields a 1/10 dilution
in a 99 mL water blank, you can figure the number of milliliters to pour INto the 99 mL water blank by the following formula:

_{} , (EQN 1)

where IN is the volume of liquid in mL to be poured INto the dilution stage.

To achieve a 1/10
dilution in a 99 mL water blank, evaluate the formula this way:

_{} (STEP
1a)

_{} (STEP
1b)

_{} (STEP
1c)

This
technique works as long as the numerator (top number) of the dilution factor is
one, as in 1/10. That’s all you *really* have to know, but read on.

__Verification of CQDDT Result__. We can show that this
value of IN really does achieve the desired dilution of 1/10, as follows:

_{} and
(EQN 2)

_{}, where (STEP 2a)

IN is defined as in EQN 1, TOTAL is the
amount of liquid in the vessel after volume IN is poured into the water blank, and THERE is the volume of liquid originally in
the water blank.

Substituting the value of 11 mL = IN, as
calculated above, and the value of 99 mL = THERE, we
have:

_{} (STEP
2b)

_{}. (STEP
2c)

Dividing
both numerator and denominator by 11 we have

_{} and (STEP 2d)

_{} as we expected. (STEP 2e)

This
technique works for any size water blank (9.0 mL, 9.9 mL, 99 mL, etc.) and for
any dilution factor with a one in the numerator ( 1/2 , 1/3, 1/4, 1/10, 1/50,
1/100, etc.). The only practical
restriction is that the calculated value of IN must be large enough to be
measured conveniently ( > 0.05 mL or so). That’s all you *have* to know. If you’re not
allergic to algebra, read on to see why this works.

__Derivation
of Chuck’s Quick and Dirty Dilution Technique__. The above technique is
based on the algebraic solution of the dilution factor equation,

Dilution_Factor = _{} _{} (EQN 3)

To solve EQN 3 for IN, let’s cross-multiply
to obtain

(STEP
3a)

Dilution_Factor_Numerator x ( THERE
+ IN ) =
IN x Dilution_Factor_Denominator .

But if the Dilution Factor Numerator is
one, this becomes

( THERE + IN ) = IN x Dilution_Factor_Denominator . (STEP 3b)

Subtracting IN from both sides of the
equation yields

THERE = (IN x Dilution_Factor_Denominator)
- IN (STEP 3c)

Factoring
out IN on the right side yields

THERE =
IN x (Dilution_Factor_Denominator
– 1) . (STEP 3d)

Dividing
both sides by the Dilution_Factor_Denominator - 1
yields

_{} (STEP 3e)

and, finally, swapping the left and right
sides of the equation yields

_{}, (STEP 3f)

which, if we substitute the definition of THERE,
becomes

_{} , (STEP 3g)

and that is essentially EQN 1, the formula we
were trying to develop.