any1 who know back titration?need help!!

xiao.you

i m rushing for this assignment.but i totally not uds about this topic.and i been assigned to find a video regarding back titration.any 1 r familiar v this help me arrr!!
big thx

xiao.you

Hi, I found out explanation on back titration from internet. I think the explanation is clear to ...
lalazaza 发表于 2012-5-23 09:23 AM

    thx for ur explanation =DD ...but i need a video to illustrate the back titration..but i found only 1..which is not an experiment.....

lalazaza

i m rushing for this assignment.but i totally not uds about this topic.and i been assigned to find a ...
xiao.you 发表于 2012-5-22 08:30 PM

Hi, I found out explanation on back titration from internet. I think the explanation is clear to me, hope that it helps.
Please read the following explanation with example given.

Sometimes it is not possible to use standard titration methods. For example the reaction between determined substance and titrant can be too slow, or there can be a problem with end point determination.

In such situations we can often use a technique called back titration. In back titration we use two reagents - one, that reacts with the original sample (lets call it A), and second (lets call it B), that reacts with the first reagent. How do we proceed? We add precisely measured amount of reagent A to sample and once the reaction ends we titrate excess reagent A left with reagent B. Knowing initial amount of reagent A and amount that was left after the reaction (from titration) we can easily calculate how much reagent A was used for the first reaction.

Example:

1.435 g sample of dry CaCO3 and CaCl2 mixture was dissolved in 25.00 mL of 0.9892 M HCl solution. What was CaCl2 percentage in original sample, if 21.48 mL of 0.09312 M NaOH was used to titrate excess HCl?

During titration 21.48×0.09312=2.000 mmole HCl was neutralized. Initially there was 25.00×0.9892=24.73 mmole of HCl used, so during CaCO3 dissolution 24.73-2.000=22.73 mmole of acid reacted. As calcium carbonate reacts with hydrochloric acid 1:2 (2 moles of acid per 1 mole of carbonate), original sample contained 22.73/2=11.37 mmole of CaCO3, or 1.137 g (assuming molar mass of CaCO3 is 100.0 g). So original sample contained 1.137/1.435×100%=79.27% CaCO3 and 100.0-72.27%=20.73% CaCl2.