If the ionic product of `Ni(OH)_(2) is 1.9 xx 10^(-15)`, the molar solubility of `Ni(OH)_(2)`, in 1.0 M NAOH is

If the ionic product of Ni (OH)_2 is 1.9xx10^(-15) , the molar solubility of Ni (OH)_2 in 1.0 M NaOH is

If the ionic product of Ni (OH)_2 is 1.9xx10^(-15) , the molar solubility of Ni (OH)_2 in 1.0 M NaOH is

The solubility product of Ni(OH)_(2) is 2.0xx10^(-15) . The molar solubility of Ni(OH)_(2) in 0.1 M NaOH solution is

The solubility product of Ni(OH)_(2) is 2.0xx10^(-15) . The molar solubility of Ni(OH)_(2) in 0.1 M NaOH solution is

If the solubility product of Ni(OH)_2 is 4.0 xx 10^(-15) the solubility (in mol L^(-1) ) is

The solubility product of Mg(OH)_2 is 4 xx 10^(-12) (mol L^(-1))^3 . Solubility of Mg(OH)_2 is

The K_(sp) of M(OH)_(2)" is "5 times 10^(-10)M^(3) . The molar solubility of M(OH)_(2) in a 0.1 M NaOH solution is

the solubility product of Al(OH)_3 is 1xx10^-36 . Calculate the solubility of Al(OH)_3.