Calculate the percentage error arising in an experiment if 1-mL, 5-mL, and 10-mL pipets are used for transfer and each pipet contains 5 drops of water adhering to the inside of the barrel. A single drop of water has an approximate volume of 0.05 mL.

Answer 1

5 drops H2O x (0.05 mL/drop) = 0.25 mL
Then (0.25/1.00)*100 = 25% error.

The others are done the same way. The purpose of this problem is to demonstrate that an absolute error (in this case one of 0.25 mL) is a larger percentage of the small pipet than of a larger pipet.
For 10 mL it is
(0.25/10)*100 = 2.5% error.

And of a 100 mL pipet is
(0.25/100)*100 = 0.25%

The lesson out of this is to use larger volumes to reduce volumetric errors.

For a buret the reading error is about 0.02 mL and you read twice (once at the beginning and once at the end) so that is a combined reading error f 0.04.
If you titrate with 10 mL that reading error is (0.04/10)*100 = 0.4% or 4 parts/thousand (ppt)
But out of a 50 mL titration that is just 0.8 ppt (0.08%)