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?

  • 2
    $\begingroup$ 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
    Oct 6, 2017 at 1:04
  • $\begingroup$ Since its not soluble, how can it produce OH- in a solution? $\endgroup$
    – Frank
    Oct 6, 2017 at 1:07
  • $\begingroup$ Why it should produce $\ce{OH-}$ in solution? The reaction takes place at the interface. $\endgroup$
    – andselisk
    Oct 6, 2017 at 1:10
  • $\begingroup$ If no $\ce{OH-}$ is produced, how is it a base? $\endgroup$
    – Frank
    Oct 6, 2017 at 1:15
  • $\begingroup$ What is it then, if not a base? $\endgroup$
    – andselisk
    Oct 6, 2017 at 1:20

1 Answer 1


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.

  • $\begingroup$ 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
    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
    Oct 6, 2017 at 6:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.