I am trying to find a reagent that keeps either $\ce{MnO2}$ or $\ce{Cr(OH)3}$ solid and make other soluble so I can centrifuge or use filtration. I thought of $\ce{NaOH},$ but I could not find any information about how $\ce{MnO2}$ reacts with $\ce{NaOH}.$
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1$\begingroup$ Float separation? A solution of density ~ 4 g/ml would work. $\endgroup$– Todd MinehardtOct 22, 2021 at 15:53
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$\begingroup$ Gut feeling: try (at small scale) if $\ce{Cr(OH)3}$ enters again into solution by addition of base (think like $\ce{[Cr(OH)6]^{3-}}$) while $\ce{MnO2}$ remains untouched; two small scale tests for each of the two in absence of the other, one small scale test where the two are present at the same time (low temperature, of course). Use then a centrifuge to separate $\ce{MnO2}$ (don't forget washing to remove the lye), and carefully precipitate $\ce{Cr(OH)3}$ again. All with caution and protection for yourself (heavy metals, chromium salts; lye). And report back to chemistry.se. Good luck. $\endgroup$– ButtonwoodOct 22, 2021 at 17:11
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1 Answer
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Add at least twice its volume of $\pu{2 M}$ $\ce{NaOH}$ and stir. The $\ce{Cr(OH)3}$ precipitate will be dissolved if the solution is cold. It will make a solution of sodium chromite $\ce{Na[Cr(OH)4]},$ and $\ce{MnO2}$ will not react.
$$\ce{Cr(OH)3 + OH- <=> [Cr(OH)4]^-}$$
After filtration, $\ce{MnO2}$ is not easy to filtrate on paper. It can be centrifuged. The solution of chromite is then heated and diluted with a volume of water equal to four times the volume of the solution. The last equation is an equilibrium and will be reversed at high temperature, so that $\ce{Cr(OH)3}$ will precipitate again, and can be filtrated.
