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Standard addition involving HPLC

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data

    The concentration of retinol (vitamin A) in serum was determined by HPLC with UV absorbance detection. A 1.00 mL sample of serum was placed in a 10 mL test tube. To precipitate the proteins, 800.0 microL of ethanol was added, and the test tube was agitated. Next, a 2.00 mL solution of 90:10 n-hexane:chloroform was added to the test tube, which extracts the fat-soluble vitamin A into the organic solvent. The solution was agitated, and briefly centrifuged to separate the layers. The organic solvent was removed, evaporated to dryness, and the sample was reconstituted in 400.0 microL of methanol. A 1.00× 10^2 microL sample of this solution was injected onto the HPLC column, and a signal of 873.0 was obtained. In a fresh test tube, a second 1.00 mL sample of serum was spiked with 10 microL of 80.0 microg/mL vitamin A. The test tube was subject to the same sample preparation protocol, with 100.0 microL being injected on the HPLC column. This solution gave a signal of 1289.0.

    What is the concentration of vitamin A in the serum sample?


    2. Relevant equations

    The formula for standard addition is the concentration of analyte in unknown/(concentration of analyte+concentration of standard in mixture)= signal from unknown/signal in mixture.


    3. The attempt at a solution

    I believe the concentration of the analyte (vitamin A) is 1mL*x/0.4 mL (x in micrograms per mL), since it appears to me that because of the " evaporated to dryness" step would negate all the other liquids up until that point so all that matters is the initial number of grams of unknown from 1mL*x, being dissolved into the 400 microlitres of methanol=0.4 mL.

    Similarly the concentration of standard would be (0.01*80)/0.4 mL.

    This leads me to believe that if I stick that info into the fomrula and solve it for x, I should be left with the concentration of vitamin a in the serum, with units of micrograms/mL. However doing that, I've gotten an answer of 1.67 micrograms/mL, which was incorrect.

    So, what am I missing?
     
  2. jcsd
  3. Oct 8, 2009 #2
    Question closes in around 5 hours. Anyone?
     
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