Question

Calculate the solubility product of $C{o}_2\left[Fe(CN{)}_6\right]$ in water at $25^{\circ }C$ Given, conductivity of saturated solutions of $C{o}_2\left[Fe(CN{)}_6\right]$ is $2.06\times 10^{-6}{\Omega }^{-1}c{m}^{-1}$ and that of water used is $4.1\times 10^{-7}{\Omega }^{-1}c{m}^{-1}$. The ionic molar conductivities of $C{o}^{2+}$ and ${\left[Fe(CN{)}_6\right]}^{4-}$ are $86.0\Omega c{m}^{2}mo{l}^{-1}$ and $444.0{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$, respectively.

• A. $\text{7}\text{.87}\times {\text{10}}^{-\text{17}}$
• B. $7.87\times 10^{-6}$
• C. $7.87\times 10^{-8}$
• D. $7.87\times 10^{-9}$

Answer 1

Answer: (A)

${\kappa }_{C{o}_2\left[Fe(CN{)}_6\right]}=2.06\times 10^{-6}-0.41\times 10^{-6}=1.65\times 10^{-6}$

${\Lambda }_{mC{o}_2\left[Fe(CN{)}_6\right]}=2{\Lambda }_{m}C{o}^{2+}+{\Lambda }_m{\left[Fe(CN{)}_6\right]}^{4-}$

$=2\times 86+444=616Sc{m}^{2}mo{l}^{-1}$

${\Lambda }_m=1000\frac{\kappa }{c}$

$c=2.7\times 10^{-6}M$

$K_{spC{o}_2\left[Fe(CN{)}_6\right]}=4c^3=7.87\times 10^{-17}{M}^3$