Question

Nanotechnology has become an important field, with applications ranging from high-density data storage to the design of "nano machines.” One common building block of nanostructured architectures is manganese oxide nanoparticles. The particles can be formed from manganese oxalate nanorods, the formation of which can be described as follows:

$$\begin{gathered} \mathit{M}{\mathit{n}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}{\mathit{C}}_{\mathbf{2}}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathit{M}\mathit{n}{\mathit{C}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}{\mathit{\sigma }}_{\mathbf{X}}^{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{K}}_{\mathbf{1}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}} \\ \mathit{M}\mathit{n}{\mathit{C}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}{\mathit{C}}_{\mathbf{2}}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathit{M}\mathit{n}{\mathbf{(}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}{\mathbf{)}}_{\mathbf{2}}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{K}}_{\mathbf{2}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}} \end{gathered}$$

Calculate the value for the overall formation constant forrole="math" localid="1663739242405" ${\mathbf{\text{Mn(C}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{4}}}{{\mathbf{\text{)}}}_{\mathbf{\text{2}}}}^{\mathbf{\text{2 -}}}$:

$\mathit{K}\mathbf{=}\frac{\left[Mn{{\left(C_2{O}_4\right)}_2}^{2-}\right]}{\left[M{n}^{2+}\right]{\left[C_2{O_4}^{2-}\right]}^{\mathbf{2}}}$

Short answer

The value of the overall formation constant is$6.24\times {10}^5.$

Step-by-step solution

The equilibrium constant for both reactions

The expressions can be denoted asfollows:

$$\begin{gathered} K_1=\frac{\left[Mn{C}_2{O}_4\right]}{\left[M{n}^{2+}\right]\left[C_2{O_4}^{2-}\right]}......\left(1\right) \\ {K}_2=\frac{\left[{{\left(Mn{C}_2{O}_4\right)}_2}^{2-}\right]}{\left[Mn{C}_2{O}_4\right]\left[C_2{O_4}^{2-}\right]}......\left(2\right) \end{gathered}$$

The overall formation constant

$$\begin{gathered} K_1\times K_2=\frac{\left[Mn{C}_2{O}_4\right]}{\left[M{n}^{2+}\right]\left[C_2{O_4}^{2-}\right]}\times \frac{\left[{{\left(Mn{C}_2{O}_4\right)}_2}^{2-}\right]}{\left[Mn{C}_2{O}_4\right]\left[C_2{O_4}^{2-}\right]} \\ {K}_1\times K_2=\frac{\left[{{\left(Mn{C}_2{O}_4\right)}_2}^{2-}\right]}{\left[M{n}^{2+}\right]{\left[C_2{O_4}^{2-}\right]}^2} \\ K=\frac{\left[{{\left(Mn{C}_2{O}_4\right)}_2}^{2-}\right]}{\left[M{n}^{2+}\right]{\left[C_2{O_4}^{2-}\right]}^2} \end{gathered}$$

The overall formation constant for${\text{Mn(C}}_{\text{2}}{\text{O}}_{\text{4}}{{\text{)}}_{\text{2}}}^{\text{2 -}}$ is

$$\begin{gathered} K=K{}_1\times K_2 \\ =(7.9\times {10}^3)(7.9\times {10}^1) \\ =6.24\times {10}^5 \end{gathered}$$