Page 5 Question 4. Riboflavin (vitamin B2) exhibits strong fluorescence at an em
ID: 590273 • Letter: P
Question
Page 5 Question 4. Riboflavin (vitamin B2) exhibits strong fluorescence at an emission wavelength of 530 nm when illuminated at 470 nm. A vitamin tablet is crushed and dissolved in 500.0 mL of 0.25 F acetate buffer (pH 6.5). A 10.0 mL aliquot of this solution is transferred to a 25.0 mL flask and diluted to the mark with buffer. The fluorescence of this solution was measured as 49.2. A second 10.0 mL aliquot of the vitamin solution is then mixed with 5.00 mL of a standard solution containing 6.00 ppm of riboflavin and this is diluted to 25.0 mL with buffer. The fluorescence reading for this solution was 85.6. A blank consisting of only acetate buffer reads 7.8. (a) [6 pts] Assuming linearity, how many milligrams of riboflavin are in the vitamin tablet?Explanation / Answer
The blank solution shows a fluorescence absorbance of 7.8. We need to subtract the absorbance for the blank from the recorded absorbances. Therefore, we have,
Absorbance for first solution = 49.2 – 7.8 = 41.4
Absorbance for second solution = 85.6 – 7.8 = 77.8
Let the vitamin tablet contain x mg Riboflavin; the concentration of Riboflavin in the original buffer solution is (x mg)/(500 mL) = x/500 mg/mL.
10.0 mL of this solution was taken and transferred to 25.0 mL to prepare the solution for fluorescence measurement.
Use the dilution law to find out the concentration of Riboflavin the solution used for the first measurement.
(10.0 mL)*(x/500 mg/mL) = (25.0 mL)*C
====> C = 8.0*10-4*x mg/mL.
Use Beer’s law to write the florescence absorbance as a function of concentration, i.e,
41.4 = k*(8.0*10-4*x mg/mL) …….(1)
10.0 mL of the original buffer solution will contain (10.0 mL)*(x/500 mg/mL) = 0.02*x mg Riboflavin. To this, we add 5.0 mL of a standard solution contain 6.00 ppm Riboflavin.
We know that 1.00 ppm = 1 mg/L = (1 mg)/[(1 L)*(1000 mL/1 L)] = (1 mg/1000 mL) = 0.001 mg/mL. Since, we take 5.00 mL of the standard, the mass of Riboflavin injected is (5.00 mL)*(0.001 mg/mL) = 0.005 mg.
The total mass of Riboflavin in the second solution is (0.02*x + 0.005) mg; this mass of Riboflavin is made up to a volume of 25.0 mL; therefore, the concentration of Riboflavin in the second solution used for the measurement is (0.02*x + 0.005) mg/(25.0 mL) = (0.02*x + 0.005)/25.0 mg/mL.
Again use Beer’s law and obtain
77.8 = k*(0.02*x + 0.005)/25.0 mg/mL ……(2)
Divide (2) by (1) and get
77.8/41.4 = [k*(0.02*x + 0.005)/25.0 mg/mL]/ k*(8.0*10-4*x mg/mL)
= (0.02*x + 0.005)/25.0/(8.0*10-4*x)
=====> 1.8792 = (0.02*x + 0.005)/0.02x
=====> 0.037584 = 0.02*x + 0.005
=====> 0.017584*x = 0.005
=====> x = 0.2843
The vitamin tablet contains 0.2843 mg Riboflavin (ans).
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.