I wanted to see if I could get a chromium salt from stainless steel, so I put the stainless steel in HCL, turning the liquid green. Then I mixed the result with ammonia, and a green solid precipitated out. After drying it out, it was brown. Does anyone know what the formulas were for each of the reactions described, and what the resulting brown stuff was?
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Plenty of different stainless steel do exist. They all behave differently. But all these stainless steels contain a large proportion of Chromium $\ce{Cr}$ apart from Iron $\ce{Fe}$. Most stainless steels contain also Nickel $\ce{Ni}$. Read the long and detailed article "Stainless steel" in Wikipedia.
One of the most common stainless steels is the famous $18-8$, made of $18$% Chromium + $8$ % Nickel, and of course $74$% Iron. Dipped in concentrated hydrochloric acid, it produced some Hydrogen gas, and the solution contains the metallic ions $\ce{Fe^{2+}, Cr^{2+}, and Ni^{2+}}$ according to the following equations :$$\ce{Fe + 2 H+ -> Fe^{2+} + H2}$$ $$\ce{Cr + 2 H+ -> Cr^{2+} + H2}$$$$\ce{Ni + 2 H+ -> Ni^{2+} + H2}$$ If ammonia is added to such a solution, the excess of acid is neutralized first. Then the metallic ions are transformed into insoluble metallic hydroxides like $\ce{Fe(OH)2}$ (green), plus some $\ce{Cr(OH)2}$ (grayish) and $\ce{Ni(OH)2}$ (green) $$\ce{Fe^{2+} + NH3 + H2O -> Fe(OH)2 + NH4^+}$$ $$\ce{Cr^{2+} + NH3 + H2O -> Cr(OH)2 + NH4^+}$$ $$\ce{Ni^{2+} + NH3 + H2O -> Ni(OH)2 + NH4^+}$$ The obtained mixture looks greenish, because the most important hydroxide in the mixture is iron(II) hydroxide.
After drying out such a mixture, it turns brown, because the iron(II) and chromium (II) hydroxides quickly react with oxygen from air and produce iron(III) and chromium(III) hydroxides, according to the reactions : $$\ce{4 Fe(OH)2 + O2 + 2 H2O -> 4 Fe(OH)3}$$ $$\ce{4 Cr(OH)2 + O2 + 2 H2O -> 4 Cr(OH)3}$$ As $\ce{Fe(OH)3}$ is dark brown, and $\ce{Cr(OH)3}$ is green, the mixture takes the color of the most abondant product, namely $\ce{Fe(OH)3}$. The green nickel(II) hydroxide is not modified by air. It does exist in the mixture. But as it is not very abundant, it does not modify the brown color of the final mixture.
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$\begingroup$ I think Cr+2 is oxidized in acid solution to Cr+3; check the potentials. First step is removal of the oxide layer followed by attack on the metal. final product is possibly mostly chrome salt that could be isolated by its amphoteric character in strong base under a N2 atmosphere to minimize reaction with O2 to form chromate. $\endgroup$– jimchmstMay 26 at 22:22
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$\begingroup$ @jmchmst. Thank you for your message. Of course the oxide layer is first removed, but the weight of the salt so produced is negligible. I have not spoken of the other elements which are also present in the steel, like Carbon, or other metals which are present in traces. Remember that the problem was about the color of the precipitates, green in the beginning that becomes brown at the end. $\endgroup$– MauriceMay 27 at 10:26