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Offline sven21

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Mole ratio
« on: May 03, 2009, 07:12:59 AM »
What is mole ratio between Bi and EDTA in Bismuth subnitrate

Offline Arkcon

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Re: Mole ratio
« Reply #1 on: May 03, 2009, 07:35:36 AM »
Well, lets start working on your question stepwise, so you learn more, and are ready for the next question of this type.  Can you write your reactants as chemical formulas?  EDTA is a little complicated, so you can skip that one.  But the formula of bismuth subnitrate, what is it?  And how does your particular problem relate it to EDTA?  I'm asking for a balanced chemical reaction here, again, you can just abbreviate EDTA for now.

Anyhow, a question for you, from me.

Downtown Manhattan, how do I get there, from where I am?  What are the exact times for the subways?

Not a very good question, I've asked there, is it?  I haven't even bothered to tell you where I've started from.  And I want the most specific information.  That is, in some ways, what your question looks like. So try to think it through, and ask a complete question.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline sven21

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Re: Mole ratio
« Reply #2 on: May 03, 2009, 01:21:31 PM »
well i think the formula is BiNO3(OH)BiO(OH) . I know the formula of edta but what matters is that it's two valent (2-)

so we have 2 Bi 3+ and 1 edta 2- .  So that would mean that we need 3 edta for 2 Bi.

But when i calculate with that ratio i'm getting wrong result in my calculation problem in which i need to calculate assay of bismuth in bismuth subnitrate

m (bismuth sunitrate) = 0.2455g
amount of 0.01 M edta - 17.5ml
f(0.01M EDTA) 0.9740
Ar(Bi) =209

And the result is 71.1% of Bismuth in sample of bismuth subnitrate

Offline Borek

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Re: Mole ratio
« Reply #3 on: May 03, 2009, 01:32:45 PM »
EDTA always react 1:1 with metals.

http://www.titrations.info/EDTA-titration
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline sven21

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Re: Mole ratio
« Reply #4 on: May 05, 2009, 07:32:45 AM »
thx :)

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