Question: Using the dissociation constant, \({K_d} = 1 \times 1{0^{ - 44}}\), calculate the equilibrium concentrations of \(F{e^{3 + }}\;and\;C{N^ - }\) in a \(0.333M\) solution of \(Fe(CN)_6^{3 - }\).

The product of the reactant concentration divided by the product concentration yields the dissociation constant.

\({K_d} = \)Product of concentration of reactants\( \div \)concentration of products

Consider the reaction,

Since, it is given that, the initial concentration of\({\rm{Fe}}({\rm{CN}})_6^{3 - }{\rm{\;is\;}}0.333{\rm{M}}\)and the dissociation constant is\({K_d} = 1 \cdot {10^{ - 44}}\).

Calculate the equilibrium concentrations of \({\rm{F}}{{\rm{e}}^{3 + }}{\rm{\;and\;C}}{{\rm{N}}^ - }\),

Find the equilibrium concentrations using the formula,

\({K_d} = \frac{{\left( {F{e^{3 + }}} \right) \cdot {{\left( {C{N^ - }} \right)}^6}}}{{\left( {Fe(CN)_6^{3 - }} \right)}}\)

Substitute the values,

\(1 \cdot {10^{ - 44}} = \frac{{x \cdot {{(6x)}^6}}}{{0.333 - x}}\)

Since,\({K_d}\)is smaller than\({10^{ - 4}}\),assume that\(0.333 - {\rm{x}} \approx 0.333\)

\(\begin{array}{*{20}{c}}{1 \cdot {{10}^{ - 44}} = \frac{{x \cdot {{(6x)}^6}}}{{0.333}}}\\{46656{x^7} = 3.33 \cdot {{10}^{ - 45}}}\end{array}\)

\(\begin{array}{*{20}{c}}{{x^7} = 7.14 \cdot {{10}^{ - 50}}}\\{x = 9.53 \cdot {{10}^{ - 8}}{\rm{M}}}\end{array}\)

Hence, the equilibrium constant can be calculated as,

\(\begin{array}{*{20}{c}}{\left( {F{e^{3 + }}} \right) = x = 9.53 \cdot {{10}^{ - 8}}{\rm{M}}}\\{\left( {C{N^ - }} \right) = 6x = 5.72 \cdot {{10}^{ - 7}}{\rm{M}}}\end{array}\)

Therefore, the equilibrium concentrations of \({\rm{F}}{{\rm{e}}^{3 + }}{\rm{\;and\;C}}{{\rm{N}}^ - }\) in a \(0.333{\rm{M}}\) solution of \(Fe({\rm{CN}})_6^{3 - }\)is\(\left( {F{e^{3 + }}} \right) = 9.53 \cdot {10^{ - 8}}{\rm{M}}\)and\(\left( {C{N^ - }} \right) = 5.72 \cdot {10^{ - 7}}{\rm{M}}\).