Question

Determine the pH required for the onset of precipitation of Ni(OH)2 from 0.010M NiSO4(aq). (Ksp =[Ni2+][OH−] 2= 6.5 × 10^−18)

Answer 1

NiSO4(aq)+2H2O(l) -----> Ni(OH)2 (s) + H2SO4 (aq)

                                     Ni(OH)2 (s) <=> Ni+2 (aq) + 2OH- (aq)

at t=0                                 0.01                  0                0            (since for solids we wont consider the concentration) change)

at t=equilibrium                                            s                2s

expression for Ksp = s x (2s)² = 4s³

Given Ksp (solubility product ) for this reaction is Ksp =[Ni2+][OH−] 2= 6.5 × 10^−18

therefore Ksp : 4s³ = 6.5 × 10^−18

by simplifying s = 1.18 × 10^−6

therefore [OH-] = 2s = 2x 1.18× 10^−6

                           = 2.36 × 10^−6

pOH = -log[2.36 × 10^−6]

       = 5.63

pH = 14-pOH

      = 8.37

for the onset of precipitation of Ni(OH)2,

ionic product should be more than solubility product.

i.e., [Ni2+][OH−] 2 ionic product > [Ni2+][OH−] 2 solubility product

hence pOH should be less than 5.63

pH should be more than 8.37