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Topic: redox titration  (Read 9859 times)

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Offline nikita

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redox titration
« on: October 04, 2008, 05:20:11 PM »
A certain amount of hydrogen peroxide was dissolved in 100. mL of water and then titrated with 1.68 M  KMnO4 . How much H2O2was dissolved if the titration required 19.8 mL of the KMnO4 solution?

2KMnO4(aq)+H2O2(aq)+3H2SO4 (aq)  :rarrow: O2(g)+2MnSO4(aq)+K2SO4+4H2O

heres what i did and i am coming up with the wrong answer.


100mL is .100 L. so

.0166mol H202/.100 = .166M H2O2

Answer call for grams, so i
.166M H202 x 34.016g = 5.65g H2O2

i tried the answer in various ways, but it keeps telling me to check my mL to L conversion, but i know theyre all right.  What am i doing wrong?


Offline nikita

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Re: redox titration
« Reply #1 on: October 04, 2008, 05:33:14 PM »
am i just not thinking about this correctly?  am i supposed to put .565 g for some reason?  if the program is telling me my mL to L conversion is off, then my answer must be wrong by a /1000 or *1000

if i leave the mols at .0166 and multiply by the molar mass, then i get .565 g.  if that is correct, i still dont understand why.

Offline Borek

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Re: redox titration
« Reply #2 on: October 04, 2008, 06:28:48 PM »
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline nikita

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Re: redox titration
« Reply #3 on: October 04, 2008, 06:46:27 PM »
oooooh, in terms of balancing after the electrons.  ugh, once again i am an idiot.  its called redox titration and i didnt do any redox balancing here at all.  heh, thanks!

Offline nikita

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Re: redox titration
« Reply #4 on: October 04, 2008, 07:12:41 PM »
can i ask another assenine question?  if i have MnSO4, rather than setting the whole compound to 0, I take SO4 by itself and Mn by itself to get a charge of -2 for the SO4 and 2 for the Mn?  and the same for K2SO4?  i had been setting this all equal to 0, unless they were free floating polyatomic ions. 

maybe i can make myself more clear.

like in SO4, i would set the whole thing equal to -2 and get -2 for oxygen and 6 for Sulfur. 

but when i do MnSO4 am i supposed to set the whole thing equal to 0, and take SO4 all together as a -2 charge, thereby making Mn a +2 charge?  or should i post this in a new topic.  im posting it here because im still trying to balance that equation.

Offline Borek

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Re: redox titration
« Reply #5 on: October 05, 2008, 04:50:50 AM »
but when i do MnSO4 am i supposed to set the whole thing equal to 0, and take SO4 all together as a -2 charge, thereby making Mn a +2 charge?  or should i post this in a new topic.  im posting it here because im still trying to balance that equation.

Molecule is neutral, ions are not.

http://www.chembuddy.com/?left=balancing-stoichiometry&right=oxidation-numbers-method
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline nikita

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Re: redox titration
« Reply #6 on: October 05, 2008, 05:57:21 PM »
thank you for taking the time to answer my questions and giving me links to info!

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