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I'm supposed to make a black precipitate using only the following chemicals: copper sulfate(II), lead nitrate, acetic acid, barium chloride, sodium hydroxide, copper(II) carbonate, methanol, potassium permanganate, hydrochloric acid, hydrogen peroxide, magnesium sulfate, and potassium dichromate.

I know that solid sulfides (used in qualitative analysis) are typically black in colour, such as copper(II) sulfide, but I don't have any source of the sulfide ion here. Is it actually possible to make a black precipitate?

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  • $\begingroup$ The most common black pigment is charcoal. The chemical actions are interesting, but one might ask how color film was able to depict black without using a black pigment? All that is needed is a mix that absorbs all the visible frequencies. Please use care these are a bunch of nasty materials to play with. An interesting experiment: do paper chromatography on black ballpoint ink. There is much more to chemistry than just mixing chemicals. $\endgroup$
    – jimchmst
    Jul 24, 2023 at 21:25

3 Answers 3

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Hint: Manganese dioxide is pretty black.

Look up potassium permanganate and note its reactions especially the reduction in neutral solutions.

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It is possible to obtain a black precipitate of copper(II) oxide from an aqueous solution of $\ce{Cu(II)}$. Adding a solution of $\ce{NaOH}$ to an aqueous solution of $\ce{CuSO4}$ will initially give a blue precipitate of copper(II) hydroxide. This redissolves with an excess of $\ce{OH-}$ to form a deep blue solution of tetrahydroxocuprate(II). In strongly alkaline medium at room temperature, $\ce{CuO}$ can precipitate from this complex by means of a condensation reaction.

$\ce{Cu^2+ + 2OH- -> Cu(OH)2(s)}$

$\ce{Cu(OH)2(s) + 2OH- -> [Cu(OH)4]^2-}$

$\ce{[Cu(OH)4]^2- <=> CuO(s) + 2OH- + H2O}$

Alternatively, the precipitated $\ce{Cu(OH)2}$ can be converted into the oxide by boiling the mixture.

$\ce{Cu(OH)2(s) ->[\Delta] CuO(s) + H2O}$

Reference:

Yannick Cudennec, André Lecerf. The transformation of Cu(OH)2 into CuO, revisited. Solid State Sciences, 2003, 5 (11-12), pp.1471-1474. DOI: ff10.1016/j.solidstatesciences.2003.09.009ff. (Open Access Link)

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    $\begingroup$ In boiling water, the blue precipitate $\ce{Cu(OH)2}$ is quickly transformed into black $\ce{CuO}$ according to : $$\ce{Cu(OH)2 -> CuO + H2O}$$ $\endgroup$
    – Maurice
    Feb 13, 2023 at 19:54
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Disclaimer

This answer is somewhat an addition to the existing ones. Here, I will try to present a practically feasible method for obtaining lead(IV) oxide $\ce{PbO2}$ from the available reagents. It is a dark brown precipitate, almost black, as required in the question.

Required reagents

The following reagents from the list are needed in order to get the desired black precipitate of lead dioxide: lead(II) nitrate $\ce{Pb(NO3)2}$, hydrogen peroxide $\ce{H2O2}$ (the higher the concentration, the better) and sodium hydroxide $\ce{NaOH}$. All solids have to be dissolved in water while sodium hydroxide solution should not be too strong.

Procedure

At first, it is necessary to mix the solutions of sodium hydroxide and hydrogen peroxide, obtaining an alkaline $\ce{H2O2}$ solution. Then, when the alkaline solution of hydrogen peroxide is added to a $\ce{Pb(NO3)2}$ solution, a precipitation of dark brown solid occurs according to the following reaction equation: $$\ce{Pb(NO3)2 (aq) + H2O2 (aq) -> PbO2 (s) + 2HNO3 (aq)}$$

Heating can help to speed up the reaction, but too much heating can dissolve the precipitate: $$\ce{PbO2 (s) + 2OH- (aq) + 2H2O (l) \xrightarrow{\textrm{heating}} [Pb(OH)6]^2- (aq)}$$

Reference

  • Ronald L. Rich. Inorganic Reactions in Water (1st ed.): "Peroxide", p. 356.
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    $\begingroup$ A very good answer, however I doubt OP should use lead salts :D $\endgroup$
    – Mäßige
    Jul 24, 2023 at 9:21

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