Question

question in aqueous solution the $\ce{co^{2+}}$ ion forms a complex with six ammonia molecules. write the formation constant expression for the equilibrium between the hydrated metal ion and the aqueous complex. under that, write the balanced chemical equation for the last step in the formation of the complex.

Explanation:

Step1: Identify the complex formation

The hydrated cobalt(II) ion is $\ce{[Co(H_{2}O)_{6}]^{2+}}$, and it reacts with $\ce{NH_{3}}$ to form the complex $\ce{[Co(NH_{3})_{6}]^{2+}}$. The last step in the formation (when all six $\ce{H_{2}O}$ are replaced by $\ce{NH_{3}}$) would be the reaction of the pentaammine complex with one more $\ce{NH_{3}}$. But more simply, the overall last step (assuming step - by - step replacement, the sixth step) is: $\ce{[Co(NH_{3})_{5}(H_{2}O)]^{2+} + NH_{3}
ightleftharpoons [Co(NH_{3})_{6}]^{2+} + H_{2}O}$. However, if we consider the formation from the fully hydrated ion, the reaction is $\ce{[Co(H_{2}O)_{6}]^{2+} + 6NH_{3}
ightleftharpoons [Co(NH_{3})_{6}]^{2+} + 6H_{2}O}$. The formation constant expression $K_f$ for the reaction $\ce{[Co(H_{2}O)_{6}]^{2+}(aq) + 6NH_{3}(aq)
ightleftharpoons [Co(NH_{3})_{6}]^{2+}(aq) + 6H_{2}O(l)}$ is $K_f=\frac{[\ce{Co(NH_{3})_{6}^{2+}}]}{[\ce{Co(H_{2}O)_{6}^{2+}}][\ce{NH_{3}}]^{6}}$ (since the concentration of liquid water is considered 1 and omitted). The balanced chemical equation for the last step (if we consider the step - wise formation, the sixth coordination step) is $\ce{[Co(NH_{3})_{5}(H_{2}O)]^{2+}(aq) + NH_{3}(aq)
ightleftharpoons [Co(NH_{3})_{6}]^{2+}(aq) + H_{2}O(l)}$.

Step2: Write the formation constant expression for the equilibrium

For the reaction $\ce{[Co(NH_{3})_{5}(H_{2}O)]^{2+}(aq) + NH_{3}(aq)
ightleftharpoons [Co(NH_{3})_{6}]^{2+}(aq) + H_{2}O(l)}$, the formation constant (for this step) $K_{f6}=\frac{[\ce{Co(NH_{3})_{6}^{2+}}]}{[\ce{Co(NH_{3})_{5}(H_{2}O)^{2+}}][\ce{NH_{3}}]}$. If we consider the overall formation from $\ce{[Co(H_{2}O)_{6}]^{2+}}$, the reaction is $\ce{[Co(H_{2}O)_{6}]^{2+}(aq) + 6NH_{3}(aq)
ightleftharpoons [Co(NH_{3})_{6}]^{2+}(aq) + 6H_{2}O(l)}$, and the formation constant $K_f=\frac{[\ce{Co(NH_{3})_{6}^{2+}}]}{[\ce{Co(H_{2}O)_{6}^{2+}}][\ce{NH_{3}}]^{6}}$.

Answer:

The balanced chemical equation for the last step in the formation of $\ce{[Co(NH_{3})_{6}]^{2+}}$ from the hydrated ion (considering step - wise, the sixth step) is $\boldsymbol{\ce{[Co(NH_{3})_{5}(H_{2}O)]^{2+}(aq) + NH_{3}(aq)
ightleftharpoons [Co(NH_{3})_{6}]^{2+}(aq) + H_{2}O(l)}}$. The formation constant expression for the equilibrium $\ce{[Co(H_{2}O)_{6}]^{2+}(aq) + 6NH_{3}(aq)
ightleftharpoons [Co(NH_{3})_{6}]^{2+}(aq) + 6H_{2}O(l)}$ is $\boldsymbol{K_f=\frac{[\ce{Co(NH_{3})_{6}^{2+}}]}{[\ce{Co(H_{2}O)_{6}^{2+}}][\ce{NH_{3}}]^{6}}}$.