I just got a final test and one of the question is:

$\ce {Ni^2+}$ + 6$\ce{NH_3}$ $\ce {<=>}$ $\ce{[Ni(NH3)6]^2+}$
If we add some aqueous $\ce{KOH}$ after the reaction reach equilibrium, where will the reaction shifts?

  • 1
    $\begingroup$ What are your initial thoughts? $\endgroup$ – jonsca Nov 28 '13 at 6:28
  • 4
    $\begingroup$ Think about complex stability, NH3 dissociation equilibrium and Ni hydroxide precipitation. $\endgroup$ – RFG Nov 28 '13 at 6:35
  • $\begingroup$ @jonsca i think NH3 is a base and KOH is a base, so it looks like add NH3 on reaction, so the reaction will shifts to right $\endgroup$ – lambda23 Nov 28 '13 at 6:42
  • $\begingroup$ I think @RauruFerro has the right idea. $\endgroup$ – jonsca Nov 28 '13 at 6:48

The equilibrium will shift to the left, as $\ce{Ni^2+}$ is removed from the equilibrium by the precipitation of $\ce{Ni(OH)2}$:

$$\ce{Ni^2+ + 2 OH- \rightleftharpoons Ni(OH)2\downarrow}$$

The precipitate forms when $[\ce{Ni^2+}]$ and $[\ce{OH-}]$ are sufficiently high to exceed the solubility product of $\ce{Ni(OH)2}$ ($K_{sp}=5.48 \times 10^{-16}$ at 25 °C, source). The concentration of $\ce{Ni^2+}$ before addition of $\ce{KOH}$ depends on the stability constants of the ammin complexes (there are several species in equilibrium with each other, as the ligands are exchanged stepwise, and not all at once). Nickel hydroxide will dissolve again when $\ce{NH3}$ is added, as this shifts the equilibrium of ammin complex formation to the right.


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