Question: Using the dissociation constant, \({K_d} = 2.2 \times 1{0^{ - 34}}\), calculate the equilibrium concentrations of\(C{o^{3 + }}\;and\;N{H_3}\)in a\(0.500 - M\;solution of\;Co\left( {N{H_3}} \right)_6^{3 + }\).

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, the initial concentration of\({\rm{Co}}\left( {{\rm{N}}{{\rm{H}}_3}} \right)_6^{3 + }\)is\(0.500{\rm{M}}\)and the dissociation constant is\({K_{\rm{d}}} = 2.2 \times {10^{ - 34}}\),

Calculate the equilibrium concentration of\({\rm{C}}{{\rm{o}}^{3 + }}{\rm{\;and\;N}}{{\rm{H}}_3}\),

Use the formula to solve,

\({K_d} = \frac{{\left( {C{o^{3 + }}} \right) \cdot {{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}^6}}}{{\left( {Co\left( {{\rm{N}}{{\rm{H}}_3}} \right)_6^{3 + }} \right)}}\)

Substitute the values,

\(2.2 \cdot {10^{ - 34}} = \frac{{x \cdot {{(6x)}^6}}}{{0.500 - x}}\)

Since,\({K_d}\)is smaller than\({10^{ - 4}}\), assume that\(0.500 - {\rm{x}} \approx 0.500\)

\(2.2 \cdot {10^{ - 34}} = \frac{{x \cdot {{(6x)}^6}}}{{0.500}}\)

\(46656 \cdot {x^7} = 1.1 \cdot {10^{ - 34}}\)

\(\begin{array}{*{20}{c}}{{x^7} = 2.36 \cdot {{10}^{ - 39}}}\\{x = 3.03 \cdot {{10}^{ - 6}}{\rm{M}}}\end{array}\)

Hence,

\(\begin{array}{*{20}{c}}{\left( {C{o^{3 + }}} \right) = x = 3.03 \cdot {{10}^{ - 6}}{\rm{M}}}\\{\left( {N{H_3}} \right) = 6x = 1.82 \cdot {{10}^{ - 5}}{\rm{M}}}\end{array}\)

Therefore, the equilibrium concentrations of \({\rm{C}}{{\rm{o}}^{3 + }}{\rm{\;and\;N}}{{\rm{H}}_3}\)are,

\(\begin{array}{*{20}{c}}{\left( {{\rm{C}}{{\rm{o}}^{3 + }}} \right) = 3.03 \cdot {{10}^{ - 6}}{\rm{M}}}\\{\left( {N{H_3}} \right) = 1.82 \cdot {{10}^{ - 5}}{\rm{M}}}\end{array}\)