Question

To calibrate a gas chromatograph for blood alcohol assays, the technician must prepare a standard 0.08%(w/v) alcohol solution. Describe how to prepare 100mL of 0.08%(w/w) from a “140 proof” ethanol-water commercial standard. Hint: 140 proof = 70%(w/v). Comment on the total volume of the 0.08%(w/v) calibration standard: Is 100 mL a sufficient volume to allow accurate preparation of the calibration standard… or would 1,000 mL be better? Defend your answer.

Answer 1

Ans. Using                              C1V1 = C2V2

C1 = Concertation, and V1 = volume of initial solution ;           stock solution

C2 = Concertation, and V2 = volume of Final solution ;           diluted soln.

Putting the values in equation 1-

            C1V1 (70 % w/v solution) = C2V2 (0.08 % w/v solution)

            Or, 70 % x V1 = 0.08 % x 100.0 mL

            Or, V1 = (0.08 % x 100.0 mL) / 70 % = 0.1143 mL = 114.3 uL

Hence, volume of 70% w/v ethanol required = 0.1143 mL

Preparation: To a 100 mL standard volumetric flask, transfer 114.3 uL of 70% ethanol using a suitable micropipette. Add sufficient ultrapure water to the flask to make the volume upto the mark. Mix well. The resultant solution is 100 mL of 0.08 % w/v ethanol.

#. The amount of sample to be introduced into the gas chromatogram is generally few microliters. So, 100.0 mL volume of the standard solution is sufficient for the standardization procedure (only few mL will actually be used).

There is no need of preparing 1000.0 mL standard solution.