2011/07/27

One more...

----------
I have a question on #5b on the practice test. You posted the answer as

H3AsO4(aq) + 2 KOH(aq)  2 H2O(l) + K3HAsO4(aq)
(0.02500L H3AsO4(aq)) (0.127 M H3AsO4(aq)) (2mol KOH / 1mol H3AsO4) ( 1/0.03868L KOH(aq)) = 0.164M KOH(aq)

I am confused at where the .127 M H3AsO4 came from.
----------
The 0.127M came from the first part of the question. In 5b, you're using the arsenic acid solution you determined the concentration of in 5a to titrate a new KOH(aq) solution that has an unknown concentration.


No comments:

Post a Comment