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

Question

You are required to prepare standards for a calibration curve. All of the glassware and pipettes your method requires have a tolerance (RSD) of 0.15%. You start with a working solution of 0.17M analyte (assume there is no error in the concentration of the working solution). Remove 10.0 mL of that solution with a pipette, add it to a 100.0 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 500.0 mL flask with uncertainty (don't forget units).

The correct answer is 0.00085 +/- 2.55 x 10^-6. I know how to correctly calculate the 0.00085 but cannot correctly calculate the error part of the question. Any help is greatly appreciated. Thank you!

Explanation / Answer

initial concentration = 0.17 M

Initial error according to RSD = 0.0015*0.17 = 0.000255M

After first dilution the concentration will be 0.17 * 10 / 100 = 0.017 M

concentration after second dilution = 0.017 *25/500 = 0.00085 M

Total error after second dilution = 0.00085*0.0015*2 = 0.000001275 * 2 = 2.55 * 10 ^-6 M

In the total error calculation the factor of 2 is coming by the fact that percent RSD is the quantity responsible for error determination. So , for two subsequent dilution the error gets added up and the final error will be 0.30% .

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