You are required to prepare standard standards for a calibration curve. All the

ID: 104007 • Letter: Y

Question

You are required to prepare standard standards for a calibration curve. All the glassware and pipettes your method requires have a tolerance (RSD) of 0.05%. You start working solution of 0.2 M analyte. Remove 10.0 mL of that solution with a pipette, add it to a 100 mL volumetric flask and dilute it to the mark. Remove 25.0 mL of the newly prepared (diluted) standard, add to a 500.0 mL volumetric flask, and dilute it to the mark. Report the concentration of standard in the 500 mL flask with uncertainty.

Explanation / Answer

All the glasswares used in the experiment have a RSD of 0.05%. Report the absolute uncertainity and the relative uncertainty as below.

Glassware

Volume Delivered (mL)

Absolute Error (RSD*Volume Delivered) (mL)

Relative Error

10 mL pipette

10.00

(0.005)/(10.00) = 0.0005

100 mL volumetric flask

100.00

(0.05)/(100.00) = 0.0005

25 mL pipette

25.00

(0.0125)/(25.00) = 0.0005

500 mL volumetric flask

500.00

(0.25)/(500.00) = 0.0005

We take 10.00 mL of 0.2 M analyte and dilute to 100.00 mL in a volumetric flask; the dilution factor is (100.00 mL)/(10.00 mL) = 10.00. The concentration of the analyte is the 100 mL volumetric flask is (0.2 M)/(10.00) = 0.02 M.

25.00 mL of the prepared analyte is taken and diluted further to 500.00 mL; the dilution factor is (500.00 mL)/(25.00 mL) = 20.00. The concentration of the analyte in the final solution is (0.02 M)/(20.00) = 0.001 M.

Since we have a RSD of 0.0005 overall, the absolute error in the measured concentration is (0.0005)*(0.001 M) = 5.0*10-7 M.

The concentration of the final solution is reported as 1.0*10-3 5.0*10-7 M (ans).