CLOates

18 Nov. 2006

OCCC APPM 1313

Chuckís Quick and Dirty Dilution Technique

CQDDT

CQDDT Explained.When youíre 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.†† Thatís 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).Thatís all you have to know.If youíre not allergic to algebra, read on to see why this works.

 

Derivation of Chuckís 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, letís 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 Ė 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.