The solubility of `PbI_(2)` is `0.0013`M. Then the solubility product of `PbI_(2)` is

The solubility product of CaF_2 is

Calculate the molar solubility of Ni(OH)_2 in 0.10 M NaOH. The solubility product of Ni(OH)_2 is 2.0xx10^(-15) .

Calculate the molar solubility of Ni(OH)_2 in 0.10 M NaOH. The solubility product of Ni(OH)_2 is 2.0 xx 10^(-15) .

The solubility proudct of a salt having formula M_(2)X_(3) is 2.2 xx 10^(-20) . If the solubility of an another salt having formula M_(2)X is twice the molar solubility of M_(2)X_(3) , the solubility product of M_(2)X is

The solubility proudct of a salt having formula M_(2)X_(3) is 2.2 xx 10^(-20) . If the solubility of an another salt having formula M_(2)X is twice the molar solubility of M_(2)X_(3) , the solubility product of M_(2)X is

The solubility product of PbI_2 is 8.0 xx 10^(-9) . The solubility of lead iodide in 0.1 molar solution of lead nitrate is x xx 10^(-6) mol /L. The value of x is __________. (Rounded off to the nearest integer ). [Given : sqrt(2) = 1.41]

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