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### Topic: acid base and titration stoich  (Read 4111 times)

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#### _cheers

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##### acid base and titration stoich
« on: February 23, 2007, 03:45:21 PM »
Calculate the number of molecules of water in oxalic acid hydrate, from the following: 5.00 g of the coumpound is made up to 250  ml and 25 ml of the solution requires 15.90 ml of .5 M NaOH to reach neutralization;

What I have so far is the moles of NaOH: .00795
Moles of oxalic acid in 250 ml= .3702 mol and in 25 ml = .009255 mole.

If I subtract the amount of moles of acid (25 ml) from the amound of NaOH, I end up with .001305 mole of acid that is used up as it looks like a one to one ratio? no?

H2C2O4 x H20 + NaOH ->  H+ + 2 CO2 + Na+ + 2H20

Am I on the right track? but then what?

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#### Borek

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##### Re: acid base and titration stoich
« Reply #1 on: February 23, 2007, 06:27:19 PM »
Oxalic acid is diprotic, whet you get after neutralization is sodium oxalate - (COONa)2.
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#### _cheers

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##### Re: acid base and titration stoich
« Reply #2 on: February 23, 2007, 07:06:17 PM »
ok, does that make it 2:1 (NaOH:acid)? and then would I just use those coefficients to find the moles acid used? then find the mass? So, I end up with 0.4293 g of oxalic acid hydrate? it doesnt seem right....am I right off track?
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#### xiankai

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##### Re: acid base and titration stoich
« Reply #3 on: February 23, 2007, 11:30:52 PM »
start from the beginning: how are you going to find the number of moles of water per mole of oxalic acid hydrate? (hint look at the formula, H2C2O4.xH2O)

Quote
What I have so far is the moles of NaOH: .00795
Moles of oxalic acid in 250 ml= .3702 mol and in 25 ml = .009255 mole.

Quote
ok, does that make it 2:1 (NaOH:acid)?

yes, so you better change your calculations.
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