How can one determine (at home) if the product shown below is Potassium Alum? I held a piece of it in a gas flame (assuming that the flame would become violet if potassium were present), but there was no change in the color of the flame. [Got product for use as astringent and antiseptic after-shave -- and people say (do not know why) that for such use, it has to be potassium-(not something else)-alum.]

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Update: As mentioned in the comments to the chosen answer (by @James Gaidis), the product is ammonia alum. Just to be double sure, I repeated the tests with McCormick's Alum powder whose ingredient is potassium alum. (1) The picture of the flame test is below.

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(2) With the McCormick Alum, the test involving sodium bicarbonate resulted in sound and definitely no ammonia gas. (3) The solution of Fatkari had a salty taste that became sour in a second or so; the K-alum was tasteless and became sour in a second or so.

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    $\begingroup$ Take some muddy water and see if it settles down faster. Alum is generally used for puriyfing water in this way. $\endgroup$ Commented Sep 6, 2020 at 12:24
  • $\begingroup$ Need to know if the 'X' in @gaidis XAl(SO4)2⋅12H2O, is potassium or something else. $\endgroup$ Commented Sep 6, 2020 at 13:33
  • $\begingroup$ See if it aids in the coagulation of blood loss in minor cuts like Potassium Alum is supposed to, if you fancy an experiment $\endgroup$ Commented Sep 6, 2020 at 21:58
  • $\begingroup$ @Autrey it that a definite test -- meaning, is it known that other Alums -- sodium, ammonia, chromium, and ferric -- do not aid in coagulation? $\endgroup$ Commented Sep 6, 2020 at 23:11

1 Answer 1


From Wikipedia:

An alum is a type of chemical compound, usually a hydrated double sulfate salt of aluminum with the general formula $\ce{XAl(SO4)2 · 12 H2O},$ where $\ce{X}$ is a monovalent cation such as potassium or ammonium.

There are also chromium alums and ferric alums, but the color in your sample does not seem to suggest either of them.

The alums melt at relatively low temperatures, because they already have a lot of water in the crystal structure, so ammonium alum and potassium alum can not be differentiated on the basis of melting points. Solubilities are fairly close, too:

$$ \begin{array}{llc} \hline \text{Compound} & \text{mp}/\pu{°C} & \text{Solubility}/\% \\ \hline \ce{(NH4)Al(SO4)2 · 12 H2O} & 93 & 15 \\ \ce{KAl(SO4)2 · 12 H2O} & 92.5 & 11 \\ \hline \end{array} $$

If there is no violet color to a flame, there may be no potassium, but the test can be sensitive and may be indeterminate. Eliminate the possibility that the alum is an ammonium salt by dissolving some in water and adding a strong alkali, then sniffing it gently. If you smell ammonia, you have an ammonia alum; it will still work as a flocculant for muddy water, but is different chemically. A strong alkali readily available at home, well, at the grocery store anyway, is sodium carbonate (washing soda).

  • $\begingroup$ In steel vessel, added water and put crystal in smaller steel cup inside vessel; heated water to under 95C while continuously moving smaller cup (with tongs). Melted -- then solidified before water got to 75C. MP for Na-alum is 61. For the ammonia-odor test, does pure baking soda (100% Na-bicarbonate) constitute strong alkali? Got product for use as astringent and antiseptic after-shave -- and people say (do not know why) that for such use, it has to be K-alum. $\endgroup$ Commented Sep 6, 2020 at 13:57
  • $\begingroup$ Put crystal in water and stirred vigorously for over a minute. Removed crystal. Water had no odor; taste was not as strong as I had expected. Dumped in sodium bicarbonate. Reaction made sound and did emit ammonia odor. $\endgroup$ Commented Sep 6, 2020 at 14:13
  • $\begingroup$ @gaidis please consider updating the post with the equation for the reaction between the X-alums and sodium bi-carbonate. $\endgroup$ Commented Sep 7, 2020 at 17:50

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