91Ó°ÊÓ

Let us calculate the Co²⁺ equilibrium concentration when 0.100 moles of [Co (NH₃)₆] (NO₃)₂ is added to a solution with 0.025 M NH₃. Assume the volume is 1.00 L.

The reaction of formation of\(\left( {{\rm{Co}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_6}} \right)\)

• The constant of formation of \(\left( {{\rm{Co}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_6}} \right){\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}{\rm{\;is\;}}{K_f} = 1.3 \cdot {10^5}\)
• Initial concentration of \({\rm{N}}{{\rm{H}}_3}{\rm{\;is\;}}0.025{\rm{M}}\)
• Initial concentration of\(\left( {{\rm{Co}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_6}} \right){\rm{\;is\;}}0.100{\rm{M}}\).

First, let us calculate the equilibrium concentration of [CO²⁺]

\(\begin{array}{*{20}{c}}{{K_f} = \frac{{\left( {{{\left( {{\rm{Co}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_6}} \right)}^{2 + }}} \right)}}{{\left( {{\rm{C}}{{\rm{o}}^{2 + }}} \right) \cdot {{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}^6}}}}\\{1.3 \cdot {{10}^5} = \frac{{0.100 - x}}{{x \cdot {{(0.025 + 6x)}^6}}}}\\{{\rm{\;By solving this equation, we get\;}}}\\{x = 0.0242{\rm{M}}}\\{\left( {{\rm{C}}{{\rm{o}}^{2 + }}} \right) = 0.0242{\rm{M}}}\end{array}\)