Check my work please
The solubility product of nickel (II) hydroxide at 25°C is 1.6 x 10-16.
Calculate the molar solubility of nickel (II) hydroxide in 0.100 M sodium hydroxide.
Ni(OH)2 <=> Ni2+ + 2OH–
I = Ni2+ is 0 & 2OH– is 0.1
Change = Ni2+ is X and 2OH- is 2x
Equili = Ni2+ is X & 2OH- is (0.1+2x)
0.10/ Ksp = 0.10/1.6 x 10-16