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I'm very new to chem — taking an intro to chemistry course where we are currently focused on substance identification. I have two substances, both aqueous solutions of either $\ce{NaOH},$ $\ce{KSCN},$ or $\ce{NaCH3COO}.$ Both substances are colorless with a pH higher than 7.

When mixed with $\ce{NiCl2},$ substance A forms a precipitate, and substance B does not. I know that you would expect a precipitate to form when $\ce{Ni^2+}$ ions are exposed to $\ce{OH-}$ ions, and I know that you would not expect a precipitate to form when $\ce{Ni^2+}$ ions are exposed to $\ce{CH3COO-}$ ions, so if $\ce{KSCN}$ was not a potential substance, I would know that substance A was $\ce{NaOH}$ and substance B was $\ce{NaCH3COO}.$

However, I am confused as to how $\ce{KSCN}$ fits into this. It's my understanding that since $\ce{SCN-}$ is the anion of thiocyanic acid, which is a weak acid, alkali metal compounds with $\ce{SCN-}$ anions are weak bases.

I have done some research and concluded that $\ce{NiCl2 + KSCN}$ will not form a precipitate. Is substance A $\ce{KSCN}$ or $\ce{NaOH}?$ Is there even a way to distinguish the two with this data?

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I know that you would expect a precipitate to form when $\ce{Ni^2+}$ ions are exposed to $\ce{OH−}$ ions

Yes, you are right, they will indeed form a green precipitate, nickel(II) hydroxide:

$$\ce{NiCl2 + 2NaOH -> Ni(OH)2 + 2NaCl}$$

Nickel(II) hydroxide does not dissolve in excess $\ce{NaOH}$.

I know that you would not expect a precipitate to form when $\ce{Ni^2+}$ ions are exposed to $\ce{CH3COO−}$ ions

I haven't found any reaction like that to form nickel(II) acetate as it is generally formed from the reaction of nickel(II) carbonate and acetic acid but a little bit of searching gave me a site which shows the formation of nickel(II) acetate from various nickel(II) salts and acetic acid and various salts of acetic acid. The closest reactions I could find from that site is the reaction between acetic acid/lead acetate and nickel(II) chloride and reaction between sodium acetate and nickel(II) nitrate. Technically, the reaction can be written and balanced on pen-paper as it is a simple double displacement reaction but the reaction hasn't been performed practically. Also, I don't know the credibility of the site, so take that site as a reference in your discretion.

I have done some research and concluded that $\ce{NiCl2 + KSCN}$ will not form a precipitate

Theoretically, it can be formed from a double displacement reaction, but practically it is found to be unfeasible as the end products get contaminated with unreacted reactants and byproducts to the point where yield is somewhat negligible. From a paper1:

Compounds containing nickel(II) thiocyanate are prepared mostly by metathesis between nickel(II) chloride or nitrate with alkali metal thiocyanates. However, in such syntheses if precautions are not taken, the final product may be contaminated with starting materials. Methods known for preparing this compound include: treatment of a dilute thiocyanic acid solution with nickel hydroxide or carbonate, followed by evaporating the mixture at $\pu{150 ^\circ C}$; mixing and boiling aqueous solutions of ammonium thiocyanate and barium hydroxide, followed by addition of an aqueous nickel sulphate.

Conclusion

A: $\ce{NaOH}$

B: $\ce{CH3COONa}$, as it is somewhat a weak base, pH = 8.9

Reference

  1. Sarma, K.P., Poddar, R.K. A convenient method of preparing nickel(II) thiocyanate and its use in synthesis. Transition Met Chem 9, 135–138 (1984). DOI: 10.1007/BF00935928 (PDF)
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  • $\begingroup$ I'm still confused as to how we know substance B is the sodium acetate, as opposed to KSCN. KSCN is also a weak base, no? $\endgroup$ – Tornado547 Sep 2 at 18:32
  • $\begingroup$ @Tornado547 KSCN has a pH in the range of 5.3-8.7, so, it varies more in the range of neutral pH. $\endgroup$ – Nilay Ghosh Sep 3 at 3:33
  • $\begingroup$ And theoretically, all 3 compounds would form a precipitate with NiCl2. It's just that the latter two reactions has not been performed experimentally (at least I haven't seen them). $\endgroup$ – Nilay Ghosh Sep 3 at 4:09

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