Question

The solubility of solid nickel hydroxide, Ni(OH)₂, is governed by its Ksp= 6x10^-16. a) Write the equation for this dissolution reaction, and the equilibrium expression. b) Nickel ions undergo three complexations with hydroxide in basic solutions. Assume OH concentration to be 0.0010 M and pH to be 11, calculate the concentrations of Ni²⁺, NiOH⁺, Ni(OH)₂, and Ni(OH)₃^-.

logK1= 4.1

logK2= 3.9

logK3= 3

Answer 1

② The solubility product of Nicola is Kap= 6*10-16 ay The dissolution equation for solid Në (OH), is given as Ni (OH) (3) - Nihlag) + 204 Cag) The equilibrium expression for the above dissolution equation is. Requilibrium? [N: 24 Oral Ni Con ] We know the for solid compound active mass is one Reg= [Nel ray] [OH [an] =kge. akse b. given 1OH-] = 0.00109 logna 24.1 log K₂ = 3.9 log Kz 2 3 first we find out the concentration of Nile now that Ni (OH), Is Nie (ar) + 2014 (ar) Ropa Ni 28] [ot] a 6x10-16 - [NE 24 ] [ 2x 0. 0010 ) ? 6x10-16 & 6x10-16 m or. [Ni 2+] = = 0.00000 10. 0020)2 [1:24 = 1.5*10-10M. concentration of Nire ion is 15x10-70 m.

concentration of Ni COH)* :- The completation reaction is - Nihag & OH Cag) Ne (OM) * eas) The equilibrium expression is m. Ra [Ni (o H) t] [ni 2P] [OH]. or [Ni(ou)") = x (pez [ok]. Here loger, 241, 142 184!2 1.34109 Mi (OH)+] 2 1.3x104x1.541010m x 0.0010 21.95*10-9 M. concentration of NE COH)2 The complexation reaction is Nifot Ni 2+ 20H (ar) i i equilibrium equation is NiCOH), (29) R22 [Mi COH [Ni2] @H]

- [NE (0*.] a kex [Nier] [ou] Here.ly, a 3-9; R = 103.9 2 7-9x108 . [Ni (OH)] = 7.9 x 1081-5x10-10 (2+0.001) m 24.74 X10-12 M. concentration of Ni COM)s :- The complexation reaction is Ni 29 ass + 304 (aq) = Ni (OH)₂ (as) i equilibrium expression - Rga tancer Cour] or [(OH)) a Kg [1:2] [or] Here log Rga 3; Kg = 103 Here [N: (04) 3] 2 10 8 10-15 m. 12 (109)] = 10x1.5410-10% (340001