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We are currently trying to work out the double replacement solubility reaction for Nickel (III) Nitrate and Rubidium Sulfate.

$$\ce{Ni(NO3)3 + Rb2SO4 -> Ni2(SO4)3 + RbNO3}$$

Balancing gives:

$$\ce{2Ni(NO3)3 + 3Rb2SO4 -> Ni2(SO4)3 + 6RbNO3}$$

We believe all four compounds are salts and therefore split all four into their cations and anions:

$$\ce{2Ni^{3+}(aq) + 6NO3-(aq) + 6Rb+(aq) + 3SO4^{2-}(aq) -> 2Ni^{3*}(aq) + 3SO4^{2-}(aq) + 6Rb+(aq) + 6NO3-(aq)}$$

At this point, we want to eliminate the matching spectator ions. But they are the same on both sides of the equation and we're left with nothing and no precipitate.

We suspect there is something or things that we are misunderstanding. We don't even know if it's ok for these types of reactions to not have a precipitate and if it is ok, if that is actually the case here with these two compounds.

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    $\begingroup$ "We believe all four compounds are salts and therefore split all four into their cations and anions: 2Ni+6NO3+6Rb+3SO4⟶2Ni+3SO4+6Rb+6NO3" Charges are missing. In solution, you'll just have a mix of ions. If you want to isolate any species, check the solubility of all compounds (take in account that there might be several forms with different water content) and try fractionated crystallisation. $\endgroup$ – imalipusram Apr 24 '20 at 8:16
  • $\begingroup$ We did look up the charges, but I omitted them in the question because we had pretty low confidence in those as well. We did check the solubility of all and all four compounds "seemed" to be soluble according to the limited solubility chart we were given. $\endgroup$ – David Bandel Apr 24 '20 at 8:23
  • $\begingroup$ Regarding solubility, do you have access to the CRC Handbook of Chemistry and Physics? Another source to check out is PubChem (pubchem.ncbi.nlm.nih.gov). Regarding charges, as there is no redox involved, give it a try. Hint: Rb behaves like sodium (e.g. NaCl), sulphate is 2- $\endgroup$ – imalipusram Apr 24 '20 at 8:28
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    $\begingroup$ Most importantly, do $\ce{Ni(NO3)3}$ and $\ce{Ni3(SO4)3}$ exist? Most of nickel compounds are in +2 oxidation state. For instance, there is no solubility date for $\ce{Ni(NO3)3}$ and $\ce{Ni3(SO4)3}$ to be found. $\endgroup$ – Mathew Mahindaratne Apr 24 '20 at 12:23
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    $\begingroup$ If it's a true double displacement reaction, not everything will dissociate. Otherwise, it's just two dissociation reactions mashed together. $\endgroup$ – Zhe Apr 24 '20 at 12:49
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This problem is impossible, because $\ce{Nickel(III) nitrate}$ and $\ce{Nickel(III) sulfate}$ do not exist. You loose your time looking for their solubilities. The only known compounds where Nickel has an oxidation number $+3$ are Nickel oxide $\ce{Ni_2O_3}$ (which is insoluble) and the complex ion $\ce{[NiF_6]^{3-}}$. Simple $\ce{Ni^{3+}}$ ions do not exist in water. As a consequence, no corresponding nitrate or sulfate exist. And no double replacement may occur with these non-existing substances.

See N. N. Greenwood and A. Earnshaw, Chemistry of the Elements, Pergamon Press, Oxford, England, 1986, p. 1340

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  • $\begingroup$ Thanks for your help. It did turn out that that was what we were meant to find. And basically the fact that everything cancels out is an indicator of that. $\endgroup$ – David Bandel Apr 25 '20 at 9:58

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