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

Calculate the molar solubility of Ni(OH)_2 in 0.10 M NaOH. The ionic p

1.	Calculate the molar solubility of Ni(OH)_2 in 0.10 M NaOH. The ionic product of Ni(OH)_2 is 2.0xx10^(-15) .
Answer» Solution :0.1 M NaOH is STRONG base , so COMPLETE ionised Thus NaOH `to Na_((aq))^(+) + OH_((aq))^(-)` `therefore [NaOH]=[OH^-]=0.1`M , NaOH Suppose solubility of `Ni(OH)_2 ="S mol L"^(-1)` So, At equilibrium in this salt `[Ni^(2+)] = "S mol L"^(-1)` `[OH^-] = "2 S mol L"^(-1)` produce `{:(Ni(OH)_(2(s)) hArr, Ni_((aq))^(2+) + , 2OH_((aq))^(-)),("At equilibrium ", SM , 2SM):}` CONCENTRATION So, total `[OH^-]`=(0.1+2S)M (ADDITION of `HO^-`of NaOH and `Ni(OH)_2` ) but the VALUE of S is much less , So ineligible in compare to 0.1 in solution. `[OH^-]=(0.1 +2S) approx "0.1 mol L"^(-1)` `K_(sp) =[Ni^(2+)] [OH^-]^2 = 2xx10^(-15)` `therefore (S)(0.1)^2 =2xx10^(-15)` `therefore S=2.0 xx 10^(-13) "mol L"^(-1) = [Ni^(2+)]`