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I've bought $\ce{NaOH}$ or $\ce{KOH}$ (I don't remember which one!) for my electrolyzer and I've tested it. Now I want to improve its performance and I was searching for a better electrolyte when I realized I've forgotten what my current electrolyte is! It is white and has crystal shape, but which one is it, $\ce{KOH}$ or $\ce{NaOH}$? If they are both white how can I know?

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  • $\begingroup$ I edited your title to make the question a little more specific. I also suggest that you remove the last three sentences and make them a separate question - it will be easier to get an answer that way. $\endgroup$ – thomij Aug 5 '14 at 17:09
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    $\begingroup$ I'm sorry to say, you won't improve your mileage by using your alternator to electrolyse water then burning the hydrogen produced. Given the efficiency of the engine/alternator/electrolyser, you'll be lucky to get 1 litre of electrolytically produced hydrogen for every 5 litres of hydrogen (or the gasoline equivalent thereof) burned in your engine. $\endgroup$ – Level River St Aug 5 '14 at 21:17
  • $\begingroup$ Mind if I ask (out of curiosity), what did it end up being? ;) $\endgroup$ – Aaron Esau Aug 17 '17 at 3:46
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A flame test should tell you pretty quickly. Make a small wire loop (preferably steel or nichrome), heat in a bunsen burner or propane torch flame until red hot and all the contaminants are burned off, then dip it in your solution and put it in the flame. If sodium is present, you should get a bright yellow flame, and if it is potassium, the flame will be a dim purple (lilac) color. But a small amount of sodium present can mask that color.

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    $\begingroup$ Back when we used a piece of cobalt glass to filter out the sodium line. $\endgroup$ – Abel Friedman Oct 31 '14 at 19:30
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An alternative, non-destructive, physical method:
Potassium hydroxide (KOH) is radioactive due to the presence of naturally occurring 40K. With sufficient counting time, even a simple radiation detector can be used to distinguish the radiation of a nice big sample of potassium hydroxide from the natural background radiation in the room.

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