Find out the solubility of Ni(OH)2 in 0.1M NaOH. Given that the ionic product of Ni(OH)2 is 2x10-15
0.1 M NaOH=0.1 M OH− ions NiOH2→Ni2++2OH−KSP=S×10−12S=2×10−1510−2=2×10−13M