Calculate the molar solubility of Ni(OH)2 in 0.10 M NaOH. The ionic product of Ni(OH)2 is 2.0×10-15.

Let the solubility of Ni(OH)2 = S, dissolution of S mol/L of Ni(OH)2 provides S mol/L of Ni2+and 2S mol/L of OH-, but the total concentration of OH-   = (0.10 + 2S ) mol/L  because the solution already contains 0.10 mol/L of O¯H from NaOH.Ksp=2.0×10-15=Ni2+OH-2                                =(S)(0.10+2S)2 As Ksp is small,   2S<0.10,  thus, (0.10+2S)≈0.10Hence, 2.0×10-15=S(0.10)2S=2.0×10-13M=Ni2+