Question
please make an easy way to solve this problem??!!
Step 3 Transfer4.00 mDof supe to a fresh test tube and add 1.00 mL of solution containing ferrozine and buffer Measure the absorbance after 10 min. tep 4 TO establish each point on the calibration curve in Figure 17-9. use 1.00mL of standard containing2-9 ?8Feyn place of serum The blank absorbance was 0.038 at 562 nm in a 1,000.cm cell. A serum sample had an bsorbance o 0.129 After the blank was subtracted from each standard absorbance oints in Figure 17-9 were obtained. The least-squares line through the standard the p points is Absorbance 0.0670 x H Fe in initial sample) +0,001s According to Beer's law, the intercept should be 0, not 0,.001s. We will use the small. observed intercept for our analysis. Find the concentration of iron in the serum Solution Rearranging the least-squares equation of the calibration line and inserting the corrected labsorbance (observed blank 0.129-0.038 0.091 of unknown. we find absorbance 0.0015 0.091 0.001s ug Fe in unknown To find the Equation 4- 0.067o 0.067o The concentration of Fe in the serum is [Fe] moles of Fe/liters of serum 1.336 X 10 6 g Fe00x 55.845 g Fe/mol Fe 10M /( 1.00 × 10-3 L) =2.39-x10-M- orbance is0.200 and the hlank absorhance ik (0.049 est Yourself If the observed what is the concentration offe (ng/mL) in the serum? (Answer: 2.23 ?g/mL)
Explanation / Answer
The observed absorbance of the sample of serum is 0.200 and the absorbance of the blank is 0.049. The corrected absorbance of the serum sample = (0.200 – 0.049) = 0.151.
Use the regression equation
Absorbance = 0.0670*(µg iron in initial sample) + 0.0015
=====> µg iron in the initial sample = (0.151 – 0.001)/(0.067) = 2.23.
Note that we took 1.00 mL of the serum sample and diluted to a final volume of 5.00 mL. The micrograms of iron in the sample of serum will remain fixed, irrespective of the dilution. Hence, the concentration of iron in the serum sample = (micrograms of iron in the initial sample)/(volume of serum sample) = (2.23 µg)/(1.00 mL) = 2.23 µg/mL (ans).
