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Our FPLC works with two solutions:

A: 20 mM Tris.HCl, 0.2 M NaCl, pH 7.8

B: 20 mM Tris.HCl, 1.6 M NaCl, pH 7.8

I'm interested in sample volume from 102 mL to 117 mL. The FPLC will have a linear concentration gradient: At 24 mL it will be 75% A and at 144 mL it will be 5% A (the rest is filled up with B).

For visualization, I made a graph (y = -7/1200*x + 0.89): enter image description here

Now, I have a sample with the whole output between the two indicated points (102 mL and 117 mL, so I have 15 mL in my tube) and would like to calculate the NaCl concentration.

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Percentage of both A and B would change 70% from 24 to 144mL, with A from 75% to 5% and B from 25% up to 95%. The change rate would be 70% during the 120mL period or $0.70/120$ per mL. At any given point $x$ mL, between 24 and 144mL, concentration of $\ce{NaCl}$ in $\ce{mol/L}$ is: $$\ce{[NaCl]} = (0.75 - 0.7\times (x-24)/120)\times0.2 + (0.25 + 0.7\times (x-24)/120)\times1.6$$

For a 15mL solution between 102 and 117mL, the concentration should be the average of the two points when x = 102, and 117mL.

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  • $\begingroup$ I try to find out the concentration over a certain time (namely between 102 to 117 mL), so in the end, I have 15 mL in a tube. I want to find the concentration of this 15 mL, not a certain infinitesimally small point along the gradient. I hope it is clear. $\endgroup$ – ste Sep 28 '15 at 8:11
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    $\begingroup$ Sorry, didn't see it carefully. In that case, the concentration should be the average of the two points when $x$ = 102, and 117mL. Did I miss something else? $\endgroup$ – Dejian15 Sep 28 '15 at 13:47
  • $\begingroup$ Yes, that's it, the result of 1.24826 seems right. If you edit it in your post, I'll accept it for closure. $\endgroup$ – ste Sep 28 '15 at 14:02
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In addition to Dejian15's answer, which is probably the simplest, I also figured out a solution for a general curve (also non-linear ones). $$ \frac{\int^{117}_{102}(\frac{-7}{1200}x+0.89)*0.2+(1-(\frac{-7}{1200}x+0.89))*1.6~\mathrm{ d}x }{15} =1.24826$$

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