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.

Answer 1

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).