How can I make a black precipitate without a sulfide?

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](https://en.wikipedia.org/wiki/Copper_monosulfide), but I don't have any source of the sulfide ion here. Is it actually possible to make a black precipitate?




Hint: Manganese dioxide is pretty black.


Look up [potassium permanganate](https://en.m.wikipedia.org/wiki/Potassium_permanganate) and note its reactions especially the reduction in neutral solutions.


 4

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](https://hal.science/hal-02503180/document))


 3

**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.


 2