To solve this problem, one needs to determine dilution factors as well as the internal standard response factor, F. The appropriate formula for internal standard calculations is:

(analyte signal) / (analyte concentration) = F x (internal standard signal) / (internal standard concentration)

F is determined by doing the above calculations based on a known concentration of analyte. It can then be used to determine the concentration of the analyte in an unknown. For this problem, iodoacetone is the analyte and p-dichlorobenzene is the unknown. The signals are peak areas for GC.

(400 / 1.0 x 10-7) = F x (800) / (2.0 x 10-7)

F = (400 x 2.0 x 10-7)/(800 x 1.0 x 10-7) = 1

For the unknown analysis, ignoring the technician error:

The concentration of p-dichlorobenzene in the unknown mixture is (2.0 x 10-5 M / 200) or 1.0 x 10-7 M, where 200 is the dilution factor in dissolving 0.100 mL in 20.00 mL.

(700 / unknown conc) = 1 x (500 / 1.0 x 10-7)

Unknown conc of iodoacetone in diluted sample = (700/500) x (1.0 x 10 -7 M) = 1.4 x 10 -7 M

The original sample was concentrated by a factor of 10 using the MIBK extraction and then diluted by a factor of 2 in adding the internal standard. Therefore, the concentration in the sample that was analyzed using the GC must be divided by 5 to determine the actual concentration in the river water. That concentration is 2.8 x 10-8 M iodoacetone.

Did the technician's blunder cause an error in the analysis? That depends on how good an internal standard the p-dichlorobenzene is for iodoacetone. If the two compounds evaporate at the same rate from solution, there will be no effect on the results because the ratio of internal standard to analyte will remain constant. On the other hand, if one of the compounds evaporates from the solution at a greater rate, the measurement will be biased by the difference in evaporation rates.

Would the results have been biased if the technician had spilled 5.0 mL of the sample after the addition of the internal standard (and equilibration) and then, to avoid being caught in an error, filled the volumetric flask up to 20.00 mL with pure MIBK?

Would the results have been biased if the technician had spilled part of the original MIBK sample extract, before the addition of the internal standard, and then, to avoid being caught in an error, filled the volumetric flask up to 20.00 mL with pure MIBK?

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