Question

The working range of the total phenol test is 0-5.00 mg/L. You have at your disposal multiple 500 ± 0.2 mL and 100 ± 0.15 mL volumetric flasks for calibration standard preparation, a 2L Winchester of 99.95% phenol, unlimited distilled/deionized water, and standard 5 ± 0.015, 10 ± 0.025, 25 ± 0.05 mL pipettes:

question : Propose a methodology to produce a stock solution of ~ 5.00 mg/L (ρ(phenol) = 1.07 g/cm³) and the dilution steps taken to produce 6x calibration standards (~ 0, 1, 2, 3, 4, 5 mg/L pure phenol in solution). Determine the concentration of your standards with errors (Graduate Capability - Quantitative literacy)

Answer 1

Method to produce stock solution

m = 5/1.07 = 5.35 mg

We have to make this solution in deionized water. So to prepare stock solution of 5 mg/l. Weigh 5.35 mg on the analytical balance and then mix it in 1000ml of water. This solution has concentration of 5 ppm.

To make further dilutions

1 Sol. = 0 mg/l so you have to take it simply deionized water and use as sample.

2 Sol.= 1 mg/l Add 20 ml of phenol in 100 ml of water

3 Sol. = 2 mg/l Add 40ml of phenol in 100ml of water

4 Sol = 3 mg/ml Add 60 ml of phenol in 100 ml of water

5 Sol = 4mg/ml Add 80 ml of phenol in 100 ml of water

6 Sol = 5 mg/ml 100 ml of phenol

Error Calculation

In 0 mg/l = We take 100 ml of deionized water in 100 ml volumetric flask which have error 100 ± 0.15 mL

In 1 mg/l = We have to take 20 ml of Phenol and we have 10 mml of pipette so we take it 2 timezs by 10 ml of pipette which have error range of 10 ± 0.025 so error added two times and we again take in 100ml of flask which have error ± 0.15 mL.

So total error can be possible is 0.025 + 0.025 = 0.05 ml from pipette and it is added in 100ml of volumetric flask and then error of flask added.

2 mg/l = We have to take 40 ml of Phenol and we have 25 ml of pipette which have error 0.05 mL and then 10 ml of pipette so we take it by one time which have error range of 10 ± 0.025 and then remaining 5 ml by 5 ml pipette which have error ± 0.015 mL so all the error added and we again take in 100ml of flask which have error ± 0.15 mL.

So total error can be possible is 0.05 + 0.025 + 0.015 = 0.09 ml in 100 ml of flask and then error of flask added.

3 mg/l = We have to take 60 ml of Phenol and we have 25 ml of pipette which have error 0.05 mL take it by two times and then 10 ml of pipette so we take it by one time which have error range of 10 ± 0.025 so all the error added and we again take in 100ml of flask which have error ± 0.15 mL.

So total error can be possible is 2 X 0.05 + 0.025= 0.125 ml in 100 ml of flask and then error of flask added.

4 mg/l = We have to take 80 ml of Phenol and we have 25 ml of pipette which have error 0.05 mL take it by three times and then 5 ml of pipette so we take it by one time which have error range of 10 ± 0.015 so all the error added and we again take in 100ml of flask which have error ± 0.15 mL.

So total error can be possible is 3 X 0.05 + 0.015= 0.165 ml in 100 ml of flask and then error of flask added.

In 5 mg/l = We take 100 ml of phenol in 100 ml volumetric flask which have error 100 ± 0.15 mL