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I need to find the concentration of $\ce{KMnO4}$ as it is an unknown.

It is a titration experiment between $\ce{KMnO4}$ and sodium oxalate solution.

I assumed I had $25.00 \ \ce{ml}$ of $0.01003 \ \ce{M}$ of $\ce{Na2C2O4}$ in a beaker and I titrated it with $\ce{KMnO4}$.

In the burrete there was around $29.00 \ \ce{ml}$ of $\ce{KMnO4}$, therefore that is my initial volume. The final volume in the burette was $20.38\ \ce{ml}$, which gives me a net volume of $8.620\ \ce{ml}$ of $\ce{KMnO4}$ that was used (titrant). The final volume of the titrate was $33.62\ \ce{ml}$.

The equation is this:

$\ce{2MnO4- + 6H^+ + 5H2C2O4 -> 2Mn^2+ + 8H2O +10CO2}$

How do I calculate the concentration of $\ce{KMnO4}$?

I found the moles of the titrate:

Moles = concentration x volume Moles = $0.01003\ \ce{M} \times 33.62\ \ce{ml}$

Which gives me $0.33372086\ \ce{M}$

What is the molar ratio of sodium oxalate to $\ce{KMnO4}$ for me to find the concentration, because then when I find the moles of that I can divide it by volume to find the concentration. However, I am stuck in between as I do not know how to find the moles to $\ce{KMnO4}$.

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    $\begingroup$ Do all the calculations using the moles (absolute amount of a given substance). Only at the very beginning and very end take the amount of solution into account to convert from and to concentrations. And don't forget to take the correct quantities, i.e. 25 and 8.62 ml, no other number should appear. $\endgroup$ – ssavec Oct 18 '13 at 14:55
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    $\begingroup$ @ssavec - since this is a homework question, your comment has the makings of a fine answer $\endgroup$ – Ben Norris Oct 18 '13 at 22:53
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As already mentioned in the comments to your question, you can simplify your calculations and avoid careless mistakes by using moles of substances from the beginning and converting to concentrations at the very end. Furthermore, be careful to avoid messing up units. Multiplying a concentration in $\ce{M}$ ($\ce{mol/l}$) with a volume in $\ce{ml}$ will give you a mole value that is off by a factor of $10^3$. When you write down intermediate results of your calculations, make sure that the written values are correct, e.g. $0.33372086$ versus $0.3372086$, otherwise this will likely be a source of consequential errors. And as a last hint: the molar ratio of oxalate to permanganate can be derived from the reaction equation.

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