Calculate the solubility product of `Co_(2)[Fe(CN)_(6)]` in water at `25^(@)C`.
Given, conductivity of saturated solutions of `Co_(2)[Fe(CN)_(6)]` is `2.06 xx 10^(-6) Omega^(-1)cm^(-1)` and that of water used is `4.1 xx 10^(-7) Omega^(-1)cm^(-1)`. The ionic molar conductivities of `Co^(2+)` and `[Fe(CN)_(6)]^(4)` are `86.0 Omega cm^(2)mol^(-1)` and `444.0 Omega^(-1) cm^(2)mol^(-1)`, respectively.

Explanation

Since conductivity is additive property ,
`kappa` (complex salt) = `kappa` (solution ) `- kappa` (water)
`= 2.06 xx 10^(-6) - 4.0 xx 10^(-7)`
`= 16.6 xx 10^(-7) ohm^(-1) cm^(-1)`
`Lambda_(m)` (complex salt) = `2 xx lambda_(m) (Co^(2+)) + lambda_(m) ([Fe(CN)_(6)]^(4-))`
`= 2 xx 86 + 444.0 = 616 ohm^(-1) cm^(2) mol^(-1)`
`Lambda_(m) = (kappa xx 1000)/(M)`
when M is the solubility of saturated solution
`616 = (16.6 xx 10^(-7) xx 1000)/(M)`
or `M = (16.6 xx 10^(-7) xx 1000)/(616) = 2.69 xx 10^(-6) mol L^(-1)`
`Co_(2) [Fe (CN)_(6)] hArr 2 Co^(2+) + 2 [Fe (CN)_(6)]^(4-)`
If s is the solubility
`[Co^(2+)] = 2 s , [Fe (CN)_(6) ]^(4-) = s`
`K_(sp) = (2s)^2 (s) = 4 s^(3)`
`= 4 xx (2.69 xx 10^(-6))^(3) = 7.79 xx 10^(-17) mol^(3) L^(-3)`