Identifying a green hydrated nickel salt by EDTA and qualitative tests
A 0.315 g sample of a green hydrated salt dissolves in water to give a pale green solution. With dilute NaOH it gives a green precipitate that does not dissolve in excess NaOH. With excess aqueous ammonia the precipitate dissolves to a deep blue solution. A white precipitate forms with acidified BaCl2 that is insoluble in dilute HCl. A 25.0 mL portion of a solution containing the whole 0.315 g sample requires 24.0 mL of 0.0500 M EDTA to reach the end point. The teacher asks the group to calculate the moles of metal ion, propose a formula from the candidates NiSO4·6H2O (M = 263), NiCl2·6H2O (M = 238) and Ni(NO3)2·6H2O (M = 291), and explain the ammonia test without writing a vague ‘complex is formed’.
1. Calculate the moles of Ni2+ in the sample and the molar mass of the hydrated salt. Use the qualitative tests to choose among the three candidate formulas.
Read the worked answer
EDTA reacts with Ni2+ in a 1:1 mole ratio:
[ n_{\mathrm{EDTA}}=(0.0240\ \mathrm{L})(0.0500\ \mathrm{M})=1.20\times10^{-3}\ \mathrm{mol} ]
Therefore (n_{\mathrm{Ni^{2+}}}=1.20\times10^{-3}) mol. The whole 0.315 g sample was present in the 25.0 mL portion, so:
[ M=\frac{0.315\ \mathrm{g}}{1.20\times10^{-3}\ \mathrm{mol}} =262.5\ \mathrm{g\ mol^{-1}} ]
This is very close to the molar mass of NiSO4·6H2O, 263 g mol−1. The white precipitate with acidified BaCl2 that is insoluble in dilute HCl is BaSO4, confirming sulphate. Chloride would give a white precipitate with AgNO3, and nitrate would not give BaSO4. Hence the salt is NiSO4·6H2O.
2. Write equations for the NaOH and ammonia tests and explain why excess NaOH does not dissolve the precipitate while excess ammonia does.
Read the worked answer
With dilute NaOH:
[ \mathrm{Ni^{2+}(aq)+2OH^-(aq)\rightarrow Ni(OH)_2(s)} ]
green precipitate.
With excess ammonia, the precipitate dissolves to give a blue ammine complex, commonly written as:
[ \mathrm{Ni(OH)_2(s)+6NH_3(aq)\rightarrow [Ni(NH_3)_6]^{2+}(aq)+2OH^-(aq)} ]
or, allowing for water ligands,
[ \mathrm{Ni(OH)_2(s)+4NH_3+2H_2O\rightarrow [Ni(NH_3)_4(H_2O)_2]^{2+}+2OH^-} ]
The key point is competition for Ni2+. Ni(OH)2 has a very small Ksp, so adding OH− alone precipitates it; Ni(OH)2 is not amphoteric in the way that Al(OH)3 or Zn(OH)2 is, so excess OH− does not form a soluble hydroxo complex. Ammonia is a Lewis base that forms a stable soluble ammine complex with a large formation constant. Excess ammonia lowers the free Ni2+ concentration enough to dissolve the precipitate. The deep blue colour is characteristic of the ammine complex.
3. The deep blue ammonia solution is diluted with a large volume of water and becomes pale green again. Explain this observation using equilibrium, and predict what happens if excess NaOH is then added.
Read the worked answer
The ammine complex is in equilibrium with its dissociation products:
[ \mathrm{[Ni(NH_3)_6]^{2+}\rightleftharpoons Ni^{2+}+6NH_3} ]
Dilution lowers the concentration of NH3 much more sharply than it lowers the concentration of the complex, because six NH3 molecules are released per complex ion. By Le Chatelier’s principle, the equilibrium shifts to the right, freeing Ni2+. The pale green colour of the hexaaquanickel(II) ion, (\mathrm{[Ni(H_2O)_6]^{2+}}), then reappears.
If excess NaOH is added to this diluted solution, the free Ni2+ will be precipitated again as green Ni(OH)2. The ammonia concentration after dilution is too low to keep the nickel soluble as the ammine complex, so the hydroxide precipitate forms. If enough concentrated ammonia were added afterwards, the precipitate could redissolve.