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I performed an experiment where I did this:

Add one spatula of malachite to a test tube, add dilute hydrochloric acid to the malachite until the test tube is one third full. Note observations and identify gas produced.

The green solid turned to a blue solution and there was an effervescence of hydrogen. However, I now realise that copper doesn't react with acid and malachite is a copper carbonate hydroxide mineral... I researched online and found that

It dissolves in sulfuric acid to give Copper sulfate, Carbon dioxide and Water. http://wiki.answers.com/Q/What_happens_when_malachite_is_dissolved_in_sulphuric_acid?classic=true

So how did I get that malachite reacted with acid and did produce hydrogen with a squeaky pop test?

Thanks in advance

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P.S: Please could you include the formula to explain? Thanks again ;) –  Tim Timmy Mar 9 at 18:32

1 Answer 1

up vote 2 down vote accepted

No, pure malachite $\ce{Cu2CO3(OH)2}$ usually doesn't produce hydrogen because it is a copper carbonate. Copper is already in a higher oxidation state (+2) so it can't reduce hydronium to hydrogen. The right reaction is the following: $$\ce{Cu2CO3(OH)2 +2H2SO4 -> 2CuSO4 + CO2 ^ + 3H2O}$$

However if there are traces of metallic copper some hydrogen can be produced but not with sulphuric acid because:

$$\ce{Cu + 2H2SO4 -> CuSO4 + 2H2O + SO2 }$$

Nor with $HCl$ (copper doesn't react with $HCl$) but only with nitric acid or aqua regia you can produce hydrogen from the metallic form. So in fact the gas you see is carbon dioxide.

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@Tim Timmy edit approved, thanks a lot for the corrections! –  G M Mar 9 at 18:50
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@TimTimmy the up arrow $\ce{ ^ }$ indicates that the compound become a gas and can escape from the solution the down arrow $\ce{ v }$ indicates that the compound precipitate. Have a look at the bottom here for latex reference and here for a general introduction to the symbols used in chemistry. –  G M Mar 9 at 19:11
    
Just one last issue - in your last equation, copper reacts with sulphuric acid - why? I know that it is lower than hydrogen in the reactivity series, so how? Thank you very much for your time! –  Tim Timmy Mar 9 at 19:31
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@TimTimmy yes in fact if you use sulphuric acid you normally add some hydrogen peroxide as oxidant agent however hot concentrate sulphuric acid is a good oxidant agent so it can dissolve copper. However is not very effective (see here for reference). –  G M Mar 9 at 20:51
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Last thing hydrogen is not involved in this reaction so the effective oxidant agent in concentrate sulphuric acid I think is $SO_4^{-}$ that is reduce to $SO_2$, this is the point. –  G M Mar 9 at 21:14

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