Calculate the molar solubility of Ni (OH)_(2) in 0.10 M NaOH. The ionic product of Ni(OH)_(2) and 2.0xx10^(-15).

Calculate the molar solubility of Ni (OH)_(2) in 0.10 M NaOH. The ioni

1.	Calculate the molar solubility of Ni (OH)_(2) in 0.10 M NaOH. The ionic product of Ni(OH)_(2) and 2.0xx10^(-15).
Answer» Solution :SUPPOSE the solubility of `Ni(OH)_(2)` in 0.10 M NaOH = s mol `L^(-1)` `Ni(OH)_(2)` in the solution dissociates as :`Ni(OH)_(2) hArr Ni^(2+)+2OH^(-)`. Thus, s mol `L^(-1)` of `Ni(OH)_(2)` in the solution gives s mol `L^(-1) ` of `Ni^(2+)` ion and 2 s mol `L^(-1)` of `OH^(-)` IONS. But `OH^(-)` ions are also PRODUCED from NaOH . As NaOH dissociates completely , `OH^(-)` from NaOH = 0.1 M . HENCE, `[Ni^(2+)]=s "mol" L^(-1) and [OH^(-)]=2 s + 0.1 "mol " L^(-1) ~~ 0.1 "mol" L^(-1)` (`:'` 2sisvery small compared to 0.1 ) `K_(sp) ` of `Ni(OH)_(2) = [Ni^(2+)][OH^(-)]^(2)` `:. s (0.1)^(2) = 2.0 XX 10^(-15 ) or s = 2.0 xx 10^(-13) "mol" L^(-1)`