The nickel(II) ion forms a variety of colored complex ions, including green Ni(H2O)6 2+, violet Ni(N — Coordination Chemistry and Aqueous Equilibria Chemistry Question
Problem Context
The nickel(II) ion forms a variety of colored complex ions, including green Ni(H2O)6 2+, violet Ni(NH3)6 2+ (Kf = 2.0 × 108), and yellow Ni(CN)4 2– (Kf = 2.0 × 1031).
Explain the trend in colors among the three complex ions.
Model Answer
The complexes absorb the complementary colors of those shown, so green Ni(H2O)6 2+ absorbs red light, violet Ni(NH3)6 2+ absorbs yellow light, and yellow Ni(CN)4 2– absorbs violet light. Thus the complexes are listed in increasing energy of the absorbed light. These d8 complexes absorb light when an electron is promoted from a nonbonding d orbital to an antibonding d orbital. The strength of the bonding increases in the order of H2O < NH3 < CN– (increasing Lewis basicity, increasing ligand field strength), corresponding to a greater difference in energy between nonbonding and antibonding orbitals.
How many unpaired electrons does each of the three complex ions have?
Model Answer
Octahedral d8 complexes such as Ni(H2O)6 2+ or Ni(NH3)6 2+ have two unpaired electrons. Square planar d8 complexes such as Ni(CN)4 2– have no unpaired electrons.
To 1.00 L of a 0.01 M solution of Ni(NO3)2 is slowly added ammonia. Initially, a green precipitate of Ni(OH)2 (Ksp = 2.0 × 10–15) is formed. What is the pH of the solution at the point where the precipitate just begins to form?
Model Answer
At the point of precipitation, [Ni2+] = 0.010 M and
[Ni2+][OH–]2 = [0.010][OH–]2 = 2.0 × 10–15
[OH–] = 4.47 × 10–7
pH = 14 + log10[OH–] = 7.65
By the time 0.600 mol of NH3 have been added to the solution in c., the solution has become homogeneous and violet in color. Calculate the concentration of Ni2+(aq) in this solution.
Model Answer
Almost all the nickel will be complexed in this solution, so [NH3] = 0.600 – 6(0.010) = 0.54 M. (The solution is highly basic, so the amount of ammonia in the form of NH4+ is negligible.) From the formation equilibrium,
[Ni(NH3)6 2+] / ([Ni2+][NH3]6) = [0.010] / ([Ni2+][0.54]6) = 2.0 × 108
[Ni2+] = 2.0 × 10–9 M
Nickel sulfide is quite insoluble (Ksp = 4.0 × 10–20). 0.010 mol of NiS is suspended in 1.00 L of a solution buffered at pH = 10.00. Ammonia is then bubbled through this solution until the nickel sulfide just dissolves. Show that no solid Ni(OH)2 will be present at this point. For H2S, pKa1 = 7.05 and pKa2 = 19.0.
Model Answer
Since the nickel sulfide has just dissolved, all the sulfur is in solution either as H2S, HS–, or S2–. At pH = 10, almost all the sulfur will be in the form of HS–, so [HS–] = 0.0100 M. One can calculate [S2–] using the Henderson-Hasselbalch equation:
pH = pKa + log10([S2–]/[HS–])
10.00 = 19.0 + log10([S2–]/[0.0100])
[S2–] = 1.0 × 10–11 M
From the Ksp of NiS, which is valid because the NiS has just barely dissolved:
[Ni2+][S2–] = 4.0 × 10–20
[Ni2+][1.0 × 10–11] = 4.0 × 10–20
[Ni2+] = 4 × 10–9 M
At pH = 10.00, [OH–] = 1.0 × 10–4 M. One can calculate Qsp for Ni(OH)2:
Qsp for Ni(OH)2 = [Ni2+][OH–]2 = [4 × 10–9][1.0 × 10–4]2 = 4.0 × 10–17
Since Qsp < Ksp (= 2.0 × 10–15), no Ni(OH)2 will precipitate under these conditions.
Calculate the number of moles of NH3 added to the solution in part e. (The pKa of NH4+ is 9.25.)
Model Answer
From part e., [Ni2+] = 4 × 10–9 M. Since there is no solid, this means that almost all the nickel must be present in the form of Ni(NH3)6 2+, so [Ni(NH3)6 2+] = 0.010 M. From the complexation equilibrium:
[Ni(NH3)6 2+] / ([Ni2+][NH3]6) = [0.010] / ([4 × 10–9][NH3]6) = 2.0 × 108
[NH3] = 0.48 M
This is the amount of free ammonia present, but to determine the number of moles of ammonia added, one needs to account for the other forms of ammonia present. This includes NH3 bound to Ni2+, which amounts to 6 × (0.010 mol) = 0.060 mol. It also includes NH4+, which is related to [NH3] via the acid-base equilibrium:
pH = pKa + log10([NH3]/[NH4+])
10.00 = 9.25 + log10([0.48]/[NH4+])
[NH4+] = 0.085 M
Since the volume of solution is 1.00 L, the number of moles of ammonia that need to be added is 0.48 mol + 0.060 mol + 0.085 mol = 0.63 mol ammonia.