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According to University of Oregon — Chemistry Interactive Demonstrations and Educational Resources — Copper-Ammonia Complex, the equilibrium constant for the formation of copper(II) hydroxide is relatively larger compared to the formation of tetraamminecopper(II).

This means that $\ce{Cu(OH)2}$ dissolve in ammonia solution poorly. Yet in fact, it does react really well. Where does this contradiction come from?

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  • $\begingroup$ Have you done the calculation involving the provided constants? // Comparison of constants must consider also different stoichiometries and present concentrations.// CH SE questions are supposed to be self-contained, involving all essential info, with the links serving just as additional reference original text or for further reading. Questing alone should be fully valid even if links stopped working. $\endgroup$
    – Poutnik
    Commented Jun 12 at 6:26
  • $\begingroup$ See also Schweitzer reagent and production of "copper silk". In the process, cellulose is dissolved in the ammonia solution of copper(II) hydroxide(dark violet). $\endgroup$
    – Poutnik
    Commented Jun 12 at 6:37
  • $\begingroup$ The equilibrium constant for the formation of copper(II) hydroxide can hardly be compared to the formation of tetraamminecopper(II), because these two constants don't have the same unit. $\endgroup$
    – Maurice
    Commented Jun 12 at 8:36
  • $\begingroup$ What do you mean by different unit? $\endgroup$
    – Shira
    Commented Jun 12 at 9:53
  • $\begingroup$ The equilibrium constant for the formation of copper(II) hydroxide is $\ce{[Cu^{2+}][OH-]^2}$, so it is expressed in $mol^3 L^{-3}$. The equilibrium constant for the formation of tetraamminecopper(II) is $\ce{\frac{[Cu(NH3)4^{2+}]}{[Cu^{2+}][NH3]^4}}$. It is expressed in $mol^{-4}L^{4}$ $\endgroup$
    – Maurice
    Commented Jun 12 at 19:43

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