If the ionic product of Ni(OH)2 is 1 .9 x 10-15, then the molar solubility of Ni(OH)2 in 1.0 M NaOH is

NaOH⇌Na++OH−                    CM   CMNi(OH)2⇌Ni2++2OH−       X            X         X Total OH−=x+C ∴Ksp=Ni2+OH−2=x(x+C)2 =xx2+2Cx+C2  or Ksp=xC2 (neglecting higher powers of x )  x=KspC2=1.9×10−15(1)2=1.9×10−15M