Chemistry - Electrochemistry Question with Solution | TestHub

The molar conductivity at infinite dilution for Co₂(Fe(CN)₆) is:

$${\Lambda }_m^o=2{\lambda }_{C{o}^{2+}}^o+{\lambda }_{[Fe(CN{)}_6{]}^{4-}}^o$$

$${\Lambda }_m^o=2(88)+444.0=176+444.0=620.0Sc{m}^{2}mo{l}^{-1}$$

The conductivity of the Co₂(Fe(CN)₆) solution is:

$${\kappa }_{solution}={\kappa }_{saturated}-{\kappa }_{water}=2.22\times 10^{-6}-3.6\times 10^{-7}=1.86\times 10^{-6}Sc{m}^{-1}$$

The solubility (s) of the salt is calculated using:

$${\Lambda }_m^o=\frac{1000{\kappa }_{solution}}{s}$$

$$s=\frac{1000{\kappa }_{solution}}{{\Lambda }_m^o}=\frac{1000\times 1.86\times 10^{-6}}{620.0}=\frac{1.86\times 10^{-3}}{620.0}=3.0\times 10^{-6}mol{L}^{-1}$$

The question asks for the solubility in M³, which implies solubility product $K_{sp}$.

For Co₂(Fe(CN)₆), the dissociation is:

Co₂(Fe(CN)₆)(s) $\rightleftharpoons$ 2Co²⁺(aq) + [Fe(CN)₆]⁴⁻(aq)

If 's' is the solubility, then $[C{o}^{2+}]=2s$ and $[Fe(CN{)}_6^{4-}]=s$.

The solubility product $K_{sp}$ is:

$$K_{sp}=[C{o}^{2+}{]}^2[Fe(CN{)}_6^{4-}]=(2s{)}^2(s)=4s^3$$

$$K_{sp}=4\times (3.0\times 10^{-6}{)}^3=4\times (27\times 10^{-18})=108\times 10^{-18}=1.08\times 10^{-16}{M}^3$$

The solubility is $1.08\times 10^{-16}$ M³.

Comparing with $x\times 10^{-y}$ M³, we have $x=1.08$ and $y=16$.

The required answer is $(x+y)=(1.08+16)=17.08$.

Rounding to the nearest whole number, the answer is 17.