Show calculation for the following K’s. Also, make a note if the experimental re
ID: 1033802 • Letter: S
Question
Show calculation for the following K’s. Also, make a note if the experimental result is consistent/inconsistent with your predicted K value and why. For your information, Nickel (II) hydroxide dissolution/precipitation equation (Eq.1) and Nickel sulfide, NiS dissolution equation in water by forming Ni-water complex (Eq.2) are shown below:
Ni(OH)2(s) + 6H2O =[Ni(H2O)6]2+ + 2OH- Ksp = 2.8 ×10-16 Eq.1
NiS(s) + 6H2O = [Ni(H2O)6]2+ + S2- KNiS=1 ×10-22 Eq. 2
[Ni(NH3)6]2+ + 2OH- = Ni(OH)2 + 6NH3 K=
[Ni(en)3]2+ + 2OH- = Ni(OH)2 + 3en K=
[Ni(dien)2]2+ + 2OH- = Ni(OH)2 + 2dien K=
NiS + 6NH3 = [Ni(NH3)6]2+ + S2- K=
Explanation / Answer
The equilibrium constant for Ni(OH)2 + 6NH3 = [Ni(NH3)6]2+ + 2OH- is 4.8 * 107
Therefore, the K value for the reverse equilibrium [Ni(NH3)6]2+ + 2OH- = Ni(OH)2 + 6NH3 is 1/(4.8 x 107) = 2.083*10-8
The equilibrium constant for Ni(OH)2 + 3en = [Ni(en)3]2+ + 2OH- is 2.0 * 1018
Therefore, the equilibrium constant for the reverse equilibrium [Ni(en)3]2+ + 2OH- = Ni(OH)2 + 3en is 1/(2.0 x 1018) = 5*10-19
The equilibrium constant for [Ni(dien)2]2+ + 2OH- = Ni(OH)2 + 2dien is much less than 5*10-19, i.e. K << 5*10-19
Reason: In all the above three cases, the equilibrium always shifts to the Ni--N (i.e. [Ni(NH3)6]2+, [Ni(NH3)6]2+ and [Ni(dien)2]2+ complexes). Hence, the equilibrium constant is very very low.
The equilibrium constant for NiS + 6NH3 = [Ni(NH3)6]2+ + S2- is 2.5 x 10-16
Reason: The reason for the low equilibrium constant is shown as follows.
NiS = Ni2+ + S2- Ksp = 1.4 * 10-24
Ni2+ + 6 NH3 = [Ni(NH3)6]2+ Kf = 1.8 * 108
NiS +6 NH3 --> [Ni(NH3)6]2+ + S2- K = Ksp * Kf = 1.4 x 10-24 * 1.8 * 108 = 2.52 * 10-16