[xanthocyanopia]

In an instrumental analysis lab, we performed standard addition, however my results did not come out so I had to use someone else's data, and they did not do it correctly.

Instead of adding a minimal amount of a solution to the unknown to "spike" the sample, the amount added sometimes would comprise ~30% of the total solution (by volume).  I wanted to know whether or not this was mathematically correctable.

For example, 25 mL of Tonic Water was used in each case, and a different amount of 0.1 M NaCl was added.  This solution was then fed into the atomic absorption and the absorbance was read.  The data set goes as follows

Amount NaCl Added                  Absorbance
0                                                 024
2                                                 584
4                                                 656
6                                                 920

Now I know if done correctly I could just plot Absorbance vs. Amount Added to find the concentration of unknown, however (especially in the latter cases, some which I excluded) the unknown became diluted by the salt solution.  What would I plot though if I wanted to correct for this dilution?

I was messing around with the A=ebc (e for molar absorptivity) and using c = [25x + 0.1(mL NaCl used)]/(25+mL NaCl used),
but I don't know how I would plot this.  I don't even know if it is mathematicaly possible.  Any help would be appreciated.

Thanks,
Ray

[Albert]

Generally speaking, when you plot data, in this kind of problem, you should use the quadratic mean.
However, according to my experience, it's almost impossible to draw a line that fits ALL the points.

[Borek]

Instead of adding a minimal amount of a solution to the unknown to "spike" the sample, the amount added sometimes would comprise ~30% of the total solution (by volume).  I wanted to know whether or not this was mathematically correctable.

Correcting for dilution is an easy task. Amount of the substance doesn't change, only volume changes. C₁V₁ = C₂V₂. That's all.