In the froth flotation process, the concentration of a sulphide ore takes place using $\ce{NaCN/KCN}$ as a depressant. It forms a soluble complex with the impurity sulphide.

Taking the particular case of lead extraction from Galena($\ce{PbS}$), $\ce{NaCN}$ reacts with the impurity $\ce{ZnS}$ present in it forming a soluble complex.
$$\ce{ZnS + 4NaCN → [Zn(CN)4]^{2-} + 4Na+ + S^{2-}}$$

We also know $\ce{Pb^2+}$ ions react with $\ce{NaCN/KCN}$ to form a white precipitate of $\ce{Pb(CN)2}$.

My question therefore is why when both $\ce{ZnS}$ and $\ce{PbS}$ are present in the solution only $\ce{ZnS}$ reacts whereas $\ce{PbS}$ doesn't? Or is it simply that it reacts but from an insoluble compound?

  • $\begingroup$ You have indicated that the lead sulfide does react and forms an insoluble precipitate, as opposed to zinc which forms a charged soluble complex. $$\ce{PbS(s) + 2CN- -> Pb(CN)2(s) + S^{2-}}$$ $\endgroup$
    – MaxW
    Apr 29, 2020 at 7:00
  • $\begingroup$ @MaxW Ikr, but that's what generally happens. That's basically a test for Pb2+ ions. But when you read through a metallurgy book you find 'No reaction' written. $\endgroup$
    – GouravM
    Apr 29, 2020 at 7:04
  • $\begingroup$ HUH?!? you wrote: "We also know $\ce{Pb^{2+}}$ ions react with NaCN/KCN to form a white precipitate of $\ce{Pb(CN)2}$." So do you KNOW that the reaction happens, or are asking IF the reaction happens? $\endgroup$
    – MaxW
    Apr 29, 2020 at 7:19
  • $\begingroup$ @MaxW Come on the test for Pb2+ takes place in an aqueous solution. But in the extraction of Pb the same reaction is taking place in a solvent made of water, oily components, collectors, activators, depressants etc. In which I'm asking about the particular reaction between the depressant and the principal ore. Anyone who has relevance to the topic will probably understand! I didn't mean you don't have, but please open your book it must be written. $\endgroup$
    – GouravM
    Apr 29, 2020 at 7:34

1 Answer 1


Ok, after having poked around a bit I doubt that $\ce{Pb(CN)2}$ forms.

I'm guessing that you're collecting the froth to get a concentrate of the galena $\ce{PbS}$ out of some finely crushed ore.

An aspect of the operation seems to be chemically reducing the concentration of $\ce{ZnS}$ by dissolving via the following reaction:

$$\ce{ZnS(s) + 4NaCN(l) -> Zn(CN)4^{2-}(l) + 2Na2S }$$

Now there are two ways that $\ce{ZnS}$ could be mixed in with the $\ce{PbS}$.

(1) The individual $\ce{Zn^{2+}}$ cations replace individual $\ce{Pb^{2+}}$ cations in the $\ce{PbS}$. In this case the random distribution of the $\ce{Zn^{2+}}$ cations wouldn't allow much extraction of $\ce{ZnS}$ as $\ce{Zn(CN)4^{2-}}$. The $\ce{Zn^{2+}}$ cations just are not going to diffuse through the solid galena particles to any appreciable extent.

(2) There could be also be that the crushing creates a mixture of relatively pure $\ce{PbS}$ and $\ce{ZnS}$ particles which are mixed together. In this case an appreciable amount of the Zn impurity could be removed.

Now for what seems to be the question. $\ce{PbS}$ is very insoluble with a $K_\mathrm{sp} = 9.04\times10^{−29}$. I couldn't find a solubility product for $\ce{Pb(CN)2}$ but it is doubtful that the solubility would be enough to cause the lead sulfide to dissolve and then form lead cyanide.

  • $\begingroup$ So it's the sulphide part that prevents such a reaction. The test is probably shown by soluble salts of Pb. But ZnS is too an insoluble salt in water (to show such reaction). That's exactly why I emphasized on the role of the sovent! $\endgroup$
    – GouravM
    Apr 29, 2020 at 9:14

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