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I was doing homework for my chemistry class, and I came across a question asking to complete and balance the following acid-equation:

$$\ce{Cu(OH)2 (s) + HClO4 (aq) -> ?}$$

At first I thought it was an acid-base reaction, but clearly $\ce{Cu(OH)2}$ is a precipitate. Thus, it could not be an acid-base reaction, and I was under the impression there was no reaction at all.

However, it turns out that the answer was to the question was (courtesy of the back of the textbook):

$$\ce{Cu(OH)2 (s) + 2 HClO4 (aq) -> Cu(ClO4)2 (aq) + 2 H2O (l)}$$

How is this possible? The answer to the question was exactly the same as an acid-base reaction, despite a precipitate instead of a base in the reactants.

Anyone know why this is the answer, or if it is even correct?

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    $\begingroup$ What makes you think it cannot be an acid-base reaction because one of the reactants is in solid state? $\ce{Ca(OH)2}$ is also poorly soluble in water, but I don't think you'd doubt it reacts with $\ce{HCl}$ reagrdless. $\endgroup$
    – andselisk ♦
    Commented Oct 6, 2017 at 1:04
  • $\begingroup$ Since its not soluble, how can it produce OH- in a solution? $\endgroup$
    – Frank
    Commented Oct 6, 2017 at 1:07
  • $\begingroup$ Why it should produce $\ce{OH-}$ in solution? The reaction takes place at the interface. $\endgroup$
    – andselisk ♦
    Commented Oct 6, 2017 at 1:10
  • $\begingroup$ If no $\ce{OH-}$ is produced, how is it a base? $\endgroup$
    – Frank
    Commented Oct 6, 2017 at 1:15
  • $\begingroup$ What is it then, if not a base? $\endgroup$
    – andselisk ♦
    Commented Oct 6, 2017 at 1:20

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An acid or base can react with a solid. For instance HCl will dissolve $\ce{CaCO3}$ with much fizzing of $\ce{CO2}$ gas.

You also miss the point that $\ce{Cu(OH)2}$ has a solubility product. So there is a small amount of $\ce{Cu^{2+}}$ and $\ce{OH^-}$ in solution.

$\ce{K_{sp} = [Cu^{2+}][OH^-]^2}$

The perchlorate anion is a powerful oxidizing agent, but since $\ce{Cu^{2+}}$ is fully oxidized no redox reaction occurs.

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  • $\begingroup$ Oh ok, thanks. That clears up my confusion a lot, but since $\ce{Cu(OH)2}$ is a weak base (I'm assuming it is), shouldn't the neutralization of it with perchloric acid be in the form $\ce{HA+B->A-+HB+}$, rather than forming water? $\endgroup$
    – Frank
    Commented Oct 6, 2017 at 2:10
  • $\begingroup$ @Frank you have essentially the right equation in your question $$\ce{Cu(OH)2 (s) + 2 HClO4 (aq) -> Cu(ClO4)2 (aq) + 2 H2O (l)}$$ The only quibble I'd have is that the $$\ce{Cu(ClO4)2 (aq)}$$ is solvated , so it would be $$\ce{Cu^{2+} + 2ClO4^-}$$ in solution. Also I think the copper ion would coordinated with water molecules rather than the perchlorate anion. $\endgroup$
    – MaxW
    Commented Oct 6, 2017 at 6:15

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