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I need a 0.01 N solution of $\ce{KIO3}$ to use as a standard in a Winkler titration. The protocol that I am using specifies that this is 0.3567 g L-1, which I have calculated to be 0.00167 M based on 214 g mol-1 for $\ce{KIO3}$.

I like to understand the calculations in the protocols that I am using, so how would I calculate that a 0.00167 M solution of $\ce{KIO3}$ is 0.01 N?

EDIT (More detail on the reactions)

The $\ce{KIO3}$ solution is added to an $\ce{NaOH} \cdot \ce{NaI}$ solution under acidic conditions for: $\ce{IO3- + 5I- + 6H+ \rightarrow 3I2 + 3H2O}$

The $\ce{I2}$ is then titrated to standardize a thiosulfate solution.

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    $\begingroup$ Do you know what the iodine in $\ce{KIO3}$ is becoming? In all redox reactions, the final oxidation states are important. Looks like it can become a mix of iodide and iodine, which makes this problem slightly more complex. $\endgroup$
    – ManishEarth
    Commented Jun 11, 2012 at 19:00
  • $\begingroup$ In this case, the reaction was particularly important since it's a Comproportionation reaction, which is wierder than usual. I'll write up an answer if I get time. $\endgroup$
    – ManishEarth
    Commented Jun 12, 2012 at 0:58
  • $\begingroup$ Hmm, are you sure that it is 0.00167 M ? I'm getting 0.0002 M. I may be wrong, though $\endgroup$
    – ManishEarth
    Commented Jun 13, 2012 at 0:56
  • $\begingroup$ I calculated it as 0.3567 g $\ce{KIO3}$ / L * mol \ 214 g $\ce{KIO3}$. So 0.3567/214=0.00167. Thanks for your help with this. $\endgroup$
    – DQdlM
    Commented Jun 13, 2012 at 2:36
  • $\begingroup$ One source with details such as given by OP is here ocw.mit.edu/courses/earth-atmospheric-and-planetary-sciences/… $\endgroup$
    – MaxW
    Commented Apr 1, 2019 at 15:42

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Hmm.. I think that your data is wrong, I get that the normality:molarity ratio must be 5 in this case.

This ratio (I call it the $\eta$-factor) for a compound participating in a reaction is the net change of its oxidation state.

In this case, one iodine in $\ce{IO3-}$ (+5) becomes one iodine in $\ce{I2}$ (0). The net change in oxidation number is 5, so $\rm normality=5\times molarity$. Normality is like the reducing/oxidising "power" of a solution--or the concentration of electrons that it can furbish/absorb.

Your density, your $\mathrm{g~L^{-1}}$, or your value for normality is wrong here.

Note that one needs to be careful with this reaction. It is comproportionation, where the same element is oxidising itself (Basically, two forms of the element in different oxidation states are reacting with each other). We can have some trouble when dealing with the products of such a reaction, though dealing with the reactants is fairly simple, as shown above. I'm writing a detailed self-answered post on this later.

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