sphperry
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Hi guys, I was wondering whether anyone could help me with this calculation. I've figured out the half equations and redox equation but I'm stuck.

1.713g of sodium ethanedioate were dissolved in water and made up to 250cm3. When 25cm3 aliquots were acidified and titrated against potassium manganate(VII) solution, 23.9cm3 of this were required to complete the reaction. Calculate the molarity of potassium manganate (VII) solution.

My attempt was the find the moles of Na2C2O4, then find the concentration, then multiply it by then, then use n=MV/1000 , use the 2:5 ratio and find conc by doing M=nx1000/23.9 and I got 0.214M... Not sure whether any of that is at all right, any help would be appreciated guys
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TSR Jessica
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Sorry you've not had any responses about this. Are you sure you’ve posted in the right place? Posting in the specific Study Help forum should help get responses.

I'm going to quote in Puddles the Monkey now so she can move your thread to the right place if it's needed. :yy:

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GDN
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(Original post by sphperry)
Hi guys, I was wondering whether anyone could help me with this calculation. I've figured out the half equations and redox equation but I'm stuck.

1.713g of sodium ethanedioate were dissolved in water and made up to 250cm3. When 25cm3 aliquots were acidified and titrated against potassium manganate(VII) solution, 23.9cm3 of this were required to complete the reaction. Calculate the molarity of potassium manganate (VII) solution.

My attempt was the find the moles of Na2C2O4, then find the concentration, then multiply it by then, then use n=MV/1000 , use the 2:5 ratio and find conc by doing M=nx1000/23.9 and I got 0.214M... Not sure whether any of that is at all right, any help would be appreciated guys
you are out by a factor of 10 - the answer is 0.0214M
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