# How would I calculate the Normality of a 0.00167 molar solution of KIO₃?

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.

• 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. – ManishEarth Jun 11 '12 at 19:00
• 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. – ManishEarth Jun 12 '12 at 0:58
• Hmm, are you sure that it is 0.00167 M ? I'm getting 0.0002 M. I may be wrong, though – ManishEarth Jun 13 '12 at 0:56
• 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. – DQdlM Jun 13 '12 at 2:36
• One source with details such as given by OP is here ocw.mit.edu/courses/earth-atmospheric-and-planetary-sciences/… – MaxW Apr 1 '19 at 15:42

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.