Asked by Abdul
If a 0.5-mL air bubble is present in the tip of a buret, what percent error in 10-mL, 20-mL, and 40-mL samples will be result if the air bubble is dislodged during the dispensing of the samples?
Answers
Answered by
DrBob222
(0.5/10)*100 = ??%
(0.5/20)*100 =
(0.5/40)*100 =
The point to the question is to show you that you need to titrate samples so that as much of the titrant that can be used is used. That isn't always possible. What can be done, and I have done it several times in my career, is to titrate an unknown sample, find that the replicates (I usually did at least 3) are low in the percent X and used only 10 or 15 mL from the buret. Two possibilities are (1) dilute the titrant so that it takes more (2)use more of the unknown so the titrant will be more. Again, the idea is to use 40 or 45 mL of titrant so that the reading error and the other errors of the same kind will be mimimized.
(0.5/20)*100 =
(0.5/40)*100 =
The point to the question is to show you that you need to titrate samples so that as much of the titrant that can be used is used. That isn't always possible. What can be done, and I have done it several times in my career, is to titrate an unknown sample, find that the replicates (I usually did at least 3) are low in the percent X and used only 10 or 15 mL from the buret. Two possibilities are (1) dilute the titrant so that it takes more (2)use more of the unknown so the titrant will be more. Again, the idea is to use 40 or 45 mL of titrant so that the reading error and the other errors of the same kind will be mimimized.
Answered by
Anonymous
5%, 2.5%, 1.25%
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