[rock candy]

Suppose you have 100ml of a solution from which you remove 2mL and place in a cuvette. You then add 3mL of buffer at pH7.0 and determine the %T to be 63. How much of the original solution will contain 4 absorbance units?

i know 2 mL of a solution will have 2 absorbance units ; 3mL of solution will have 3 absorbance units , etc .

i think im suppose to use this formula :

c = A / El ; its lambert and beers law

c: is the concentration in Molarity
l : is the length in cm and is always 1 cm
E: extinction coefficient which is equal to the absorbance (A) of a 1M solution of the substance


so basically i did this ...


c = ( 2- log 63) / (2-log63)(1)
c = 1.00M

[ARGOS++]

Dear rock candy;

Yes!,  -  You need to know the Beer-Lambert Law.

i know 2 mL of a solution will have 2 absorbance units ; 3mL of solution will have 3 absorbance units , etc . 

From where you know that?,  -  IMHO: It’s wrong!

Suppose you have 100ml of a solution from which you remove 2mL and place in a cuvette. You then add 3mL of buffer at pH7.0 and determine the %T to be 63. How much of the original solution will contain 4 absorbance units? 

Is that the original question or did you re-phrase it?

In this kind it is an absolutely silly question, because in a 1.0 cm Cuvette you can’t get a higher Absorbance (AU) than the original solution has!, and that’s quite less than 1.0 AU!!!
It makes more sense, if the question asks for the required path-length to measure 4.0 AU!

Hint for a possible start:   
     A.)    Do you know how to convert %T into AU?
     B.)    Then what is the AU of the original solution in the identical cuvette?
     C.)     ..

I hope to have anyway been of some help to you.
Good Luck!
                    ARGOS⁺⁺