A protein solution (0.2 ml) of unknown concentration was diluted with 0.8 ml of water. To 0.5 ml of this diluted solution 4.5 ml of biuret reagent was added and the color allowed to develop. The absorbance of this mixture taken in a test tube of 1cm diameter at 540 nm was observed to be 0.20. 0.5 ml of BSA (4 mg/ml) solution plus 4.5 ml of biuret gave an absorbance of 0.20 when measured as above. What is the protein concentration (mg/ml) in the undiluted unknown solution?
1. 20
2. 40
3. 50
4. 80
Standard solution: 0.5 mL BSA (4 mg/mL) + 4.5 mL biuret; A = 0.2
unknown diluted solution: 0.5 mL (?? mg/mL) +4.5 mL biuret; A = 0.2
Therefore the unknown DILUTED solution has a concentration of 4 mg/mL. What is the concentration of the unknown BEFORE dilution. The 0.2 mL was diluted to 1.0 mL (see note below) so the initial concentration was 4 mg/mL x (1.0 mL/0.2 mL) = 4 mg/L x 5 = ? mg/mL
NOTE: In practice, when dilutions are made, usually a portion of the unknown, say 0.2 mL, is added to a test tube and it is diluted with water UP TO A FINAL VOLUME OF 1.0 ML. If it is done as the problem suggests then WE ARE ASSUMING that the final volume will be 1.0 mL but that is so ONLY IF the volumes are additive; i.e., 0.2 + 0.8 = 1.0 but technically that is not correct. From a practical point it may be true but there are examples galore where it is not true. For example, a 5 mL sample of ethanol added to a 5 mL sample of water will NOT be equal to a final volume of 10 mL. It will be closer to 9.5 mL. I think the problem should have been written as follows because any good analytical chemist would have done it that way and not the way the problem states it.
"A protein solution (0.200 ml), of unknown concentration, was diluted with distilled water to a final volume of 1.00 mL)." The moral to this story is that volumes should never be assumed to be additive, except of course, for the sake of simplicity.