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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.
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.
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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.
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