I am doing a chemistry report related to redox titrations (using $\ce{KMnO4}$ as the titrant), and one section of the report asks me to explain what would occur if I were to use sodium permanganate ($\ce{NaMnO4}$) instead of potassium permanganate ($\ce{KMnO4}$), and more importantly, what I would have to do differently in my experiment if I were to use this new chemical.

What I know as of now.

  • Sodium permanganate is not widely used because it is more expensive and is much more difficult to create than potassium permanganate.
  • Molar mass is different when performing stoichiometric equations, which leads to different results when calculating certain values.

I looked online and found that the chemical is "hygroscopic;" however, I don't know if this changes anything in the experiment.

Is there anything else I can "change" in my experiment if I were to use $\ce{NaMnO4}$ instead of $\ce{KMnO4}$?

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    $\begingroup$ Well... I think your answer is already pretty good. For the point on molar mass, what you could maybe say is that you will need to use a different mass of NaMnO4 in order to get the same concentration of NaMnO4. It's a titration, so it really shouldn't matter what mass you use as long as you record it accurately, but maybe that's what your teacher is looking for? $\endgroup$ – orthocresol Oct 13 '15 at 22:32
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    $\begingroup$ I think for the most part, it being hygroscopic just means you need to keep it properly (i.e. make sure to cap the thing after using it). But someone else might know better $\endgroup$ – orthocresol Oct 13 '15 at 22:35

You are right. Preparing sodium permanganate is more expensive. The second reason is, sodium permanganate is hygroscopic and occurs as a monohydrate. This means that sodium permanganate is more soluble than potassium permanganate. (Solubility of $\ce{KMnO_4}$ is $0.065$ g/ml and that of $\ce{NaMnO_4}$ is $0.9$ g/ml)

So if you require a larger concentration of $\ce{MnO_4^-}$ in the solution, you can go for $\ce{NaMnO_4}$.

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    $\begingroup$ Also sodium salt solutions are less stable. $\endgroup$ – Mithoron Oct 15 '15 at 19:59

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