Asked by Raskin
Deriving Chem Beer's Law's Equation
By comparing the absorbance of each trial solution, Aeq, to the absorbance of the standard solution, Astd, [FeNCS2+]eq can be determined. Two Beer’s Law equations can be written, one for the trial (Aeq = ε[FeNCS2+]eql), one for the standard (Astd = ε[FeNCS2+]stdl).
Set these two equations equal to each other to derive equation (2).
Note : the "l" is a special symbol as well
The derived equation looks like this : [FeNCS2+]eq = Aeq / Astd *[FeNCS2+]std
I'm having trouble deriving it myself? Any tips or help would greatly be appreciated!
By comparing the absorbance of each trial solution, Aeq, to the absorbance of the standard solution, Astd, [FeNCS2+]eq can be determined. Two Beer’s Law equations can be written, one for the trial (Aeq = ε[FeNCS2+]eql), one for the standard (Astd = ε[FeNCS2+]stdl).
Set these two equations equal to each other to derive equation (2).
Note : the "l" is a special symbol as well
The derived equation looks like this : [FeNCS2+]eq = Aeq / Astd *[FeNCS2+]std
I'm having trouble deriving it myself? Any tips or help would greatly be appreciated!
Answers
Answered by
DrBob222
The hint tells you how to do it. Set the two equations equal to each other and solve for [FeNCS2+]eq.
Answered by
Anonymous
I don't understand how its Aeq / Astd * [FeNCS2+]std
What am I setting equal exactly?
ε[FeNCS2+]eql)/Aeq = ε[FeNCS2+]stdl/Astd ?
What am I setting equal exactly?
ε[FeNCS2+]eql)/Aeq = ε[FeNCS2+]stdl/Astd ?
Answered by
Anonymous
solve for the epsilon and then plug them in.
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