Question: Calculate the equilibrium concentration of Ni²⁺ in a 1.0 M solution\(\left( {Ni{{\left( {N{H_3}} \right)}_6}} \right){\left( {N{O_3}} \right)_2}\).

Let us calculate the equilibrium concentration ofNi²⁺in a 1.0 M solution\(\left( {{\rm{Ni}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_6}} \right){\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\)

\(\left( {{\rm{Ni}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_6}} \right){\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\)dissociates completely into 1.0 M\({\left( {{\rm{Ni}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_6}} \right)^{2 + }}\)and 2.0 M NO₃^-

The reaction ofdissociation of\(\left( {{\rm{Ni}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_6}} \right)\)

• The formulation constant of\({\left( {{\rm{Ni}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_6}} \right)^{2 + }}\)is\({K_f} = 2 \cdot {10^8}\)
• Hence, the dissociation constant is\({K_d} = \frac{1}{{{K_f}}} = \frac{1}{{2 \cdot {{10}^8}}} = 5 \cdot {10^{ - 9}}\)

calculate the equilibrium concentration of Ni²⁺

\(\begin{array}{*{20}{c}}{{K_d} = \frac{{\left( {N{i^{2 + }}} \right) \cdot {{\left( {N{H_3}} \right)}^6}}}{{\left( {{{\left( {Ni{{\left( {N{H_3}} \right)}_6}} \right)}^{2 + }}} \right)}}}\\{5 \cdot {{10}^{ - 9}} = \frac{{x \cdot {{(6x)}^6}}}{{(1.0 - x)}}}\end{array}\)

Since K_d is smaller than 10^-4,

We will assume that \(1.0 - {\rm{x}} \approx 1.0\)

\(\begin{array}{*{20}{c}}{5 \cdot {{10}^{ - 9}} = \frac{{x \cdot {{(6x)}^6}}}{{1.0}}}\\{46656{x^7} = 5 \cdot {{10}^{ - 9}}}\\{x = 0.014{\rm{M}}}\\{\left( {N{i^{2 + }}} \right) = 0.014{\rm{M}}}\end{array}\)