Ni(OH)2 ⇌ Ni2+ + 2OH−
Let the solubility of Ni(OH)2 be equal to S. Dissolution of S mol/L of Ni(OH)2 provides S mol/L of Ni2+ and 2S moVL of OH−, but the total concentration of OH− = (0.10 + 2S) mol/L because the solution already contains 0.10 mol/L of OH− from NaOH.
Ksp= 2.0 × 10−15
Ksp = [Ni2+][OH−]2
2.0 × 10−15 = (S)(0.10 + 2s)2
AS Ksp is small, 2S ⊲⊲ 0.10
∴ 0.10 + 2S ≃ 0.10
Hence, 2 × 10−15 = (S) (0.10)2
S= \(\frac{2\times10^{-15}}{0.01}\)
S= 2 × 10−13 = [Ni2+]
