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I got some chromium(III) oxide ($\ce{Cr2O3}$) sold as pigment, called "Chrome Green". I've been having a hard time trying to react it with $\ce{HCl}$ to make $\ce{CrCl3}$. I thought it was just hard to break the $\ce{Cr-O}$ bond.

Now I'm just afraid that the pigment that I bought was actually cut with something else, like a silicate-type of pigment. How do I test for chrome(III) oxide, if I can't get it to become a salt in the first place?

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  • $\begingroup$ I was thinking of roasting it in excess amounts with either $NaCO_{3}$ or $NaOH$ to get sodium chromate or chromite (the chromite I can just further oxidize). Dissolve it all in water and then dry it. If I see any white specs I can assume them to be sodium silicate. But this only tests for silicates mixed in. I want to know if the store I bought it from mixed it with enough actual $Cr_{2}O_{3}$. Perhaps, a test specific to Chromium ions would be better $\endgroup$ – Dehbop Feb 7 '17 at 6:13
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The test for $\ce{Cr(III)}$ can be performed in the solid state using an oxidative melt. You will need

  • chromium(III) oxide (obviously)
  • sodium carbonate
  • potassium nitrate
  • a Bunsen or Teclu burner
  • a magnesia trough or a crucible
  • suitable pliers to hold the trough

Mix the green pigment with up to the threefold amount of sodium carbonate and potassium nitrate, place the mixture on the magnesia trough and melt it.

A yellow melt indicates the formation of sodium chromate according to

$\ce{Cr2O3 + 3KNO3 + 2Na2CO3 -> 2Na2CrO4 + 3KNO2 + 2CO2 ^}$

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Chromium(III) oxide contains $\ce{Cr^3+}$ which can be detected through salt analysis. Only specific test for $\ce{Cr^3+}$ is mentioned.

  1. Source 1*

$$\ce{Cr^3+(aq) + 3OH^{–}(aq) -> Cr(OH)3(s)}$$

Chromium(III) hydroxide is formed when $\ce{Cr^3+}$ is treated with ammonia or sodium hydroxide, which is soluble in excess alkaline solution (amphoteric, dissolves in acids too) but not soluble in excess ammonia . With sodium hydroxide, a dark green soluble hexahydroxo–complex ion is formed.

$$\ce{Cr(OH)3(s) + 3NaOH(aq) -> [Cr(OH)6]^{3–}(aq)}$$

  1. Source 2**

Chromium can be taken through a series of colored tests which leaves no doubt as to its identity. Chromium(III) forms a steel green hydroxide which dissolves in excess strong base to give a deeply green colored solution of the hydroxy complex. Treating this complex with 3% hydrogen peroxide gives the yellow solution of the chromate ion, which upon acidification with dilute nitric acid gives the orange color of dichromate. Treatment of the cold solution of dichromate with 3% hydrogen peroxide gives the intense blue color of a peroxide of chromium. (The actual composition of this peroxide is not known, but it is believed to have the empirical formula $\ce{CrO5}$). This peroxide readily decomposes to the pale violet color of the original hydrated chromium(III) ion. In low concentrations of dichromate the blue color is fleeting, and attention must be focused on the test tube during the addition of the hydrogen peroxide to avoid missing the color change.

$$\ce{Cr(OH)^{4-} (green) + H2O2 + OH^{-}→ CrO4^{2-}(yellow)}$$ $$\ce{CrO4^{2-} + H+→ Cr2O7^{2-} (orange)}$$ $$\ce{Cr2O7^{2-} + H2O2 + HNO3→ CrO5 (blue)→ [Cr(H2O)6]^3+ (violet)}$$

The following color changes are all indicative of $\ce{Cr^3+}$. Add an excess of 6 M $\ce{NaOH}$ to about one mL of test solution. To this green solution add 10 drops of 3% $\ce{H2O2}$. Heat the test tube in the water bath until the excess $\ce{H2O2}$ is destroyed as indicated by the cessation of bubbles. Acidify the yellow solution with 3 M $\ce{HNO3}$. Cool the resulting orange solution in an ice bath. To the cooled solution add a drop or two of 3% $\ce{H2O2}$ and observe the immediate fleeting blue color.

Reference

*http://www.docbrown.info/page13/ChemicalTests/ChemicalTestsc.htm

**http://www.wiredchemist.com/chemistry/instructional/laboratory-tutorials/qualitative-analysis

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  • $\begingroup$ Is the acidification with $HNO_{3}$ specific to the test? Is it because you need a strong oxidizing acid? Or would common $HCl$ suffice? $\endgroup$ – Dehbop Feb 8 '17 at 6:18

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