Question

Which of the following options are correct for solubility and solubility product of
$C{O}_2\left[Fe{\left(CN\right)}_6\right]$ in water at ${25}^{\circ }$. Conductivity of a saturated solution of $C{O}_2\left[Fe{\left(CN\right)}_6\right]$ is $2.06\times {10}^{-6}{\Omega }^{-1}{\text{cm}}^{-1}$ and that of water used is $4.1\times {10}^{-7}{\Omega }^{-1}{\text{cm}}^{-1}.$
The ionic molar conductivities of $C{o}^{2+}\text{and}Fe(CN{)}_6^{4-}$ are $86.0{\Omega }^{-1}{\text{cm}}^2{\text{mol}}^{-1}\text{and}444.0{\Omega }^{-1}{\text{cm}}^2{\text{mol}}^{-1},$ respectively.

• A. $K_{sp}=7.69\times {10}^{-17}$
• B. $K_{sp}=7.6872\times {10}^{-14}$
• C. $\text{Solubility =}2.6\times {10}^{-6}\text{mol/L}$
• D. $\text{Solubility =}2.6\times {10}^{-5}\text{mol/L}$

Answer 1

Answer: (A), (C)

The correct options are
A $K_{sp}=7.69\times {10}^{-17}$
C $\text{Solubility =}2.6\times {10}^{-6}\text{mol/L}$

$Molarconductivity{\lambda }_m=\frac{\kappa \times 1000}{M}=\frac{[K_{soln}-K_w]\times 1000}{M}$
Since the complex is dissociating, we can consider solubility and molarity are equal.
By substituting the given values we get,
$86\times 2+444=\frac{(2.06\times {10}^{-6}-4.1\times {10}^{-7})\times 1000}{s}$
$s=2.6\times {10}^{-6}mol/L$
Solubility product =$4s^3$
$=4\times (2.6\times {10}^{-6}{)}^3$
=$K_{sp}=7.69\times {10}^{-17}$