Question

According to Appendix I, Cu²⁺ forms two complexes with acetate:

$\begin{array}{c}\mathbf{C}{\mathbf{u}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{C}{\mathbf{H}}_{\mathbf{3}}\mathbf{C}{\mathbf{O}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{\rightleftharpoons }\mathbf{C}\mathbf{u}{\left(C{H}_{3}C{O}_2\right)}^{\mathbf{+}}{\mathbf{\beta }}_{\mathbf{1}}\left(=K_1\right)\\ \mathbf{C}{\mathbf{u}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{2}\mathbf{C}{\mathbf{H}}_{\mathbf{3}}\mathbf{C}{\mathbf{O}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{\rightleftharpoons }\mathbf{C}\mathbf{u}{\left(C{H}_{3}C{O}_2\right)}_{\mathbf{2}}{\mathbf{\beta }}_{\mathbf{2}}\end{array}$

(a) Referring to Box 6-2, find K₂ for the reaction

$\mathit{C}\mathit{u}{\left(C{H}_{3}C{O}_2\right)}^{\mathbf{+}}\mathbf{+}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}{\mathit{O}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{\rightleftharpoons }\mathit{C}\mathit{u}{\left(C{H}_{3}C{O}_2\right)}_{\mathbf{2}}\left(aq\right){\mathit{K}}_{\mathbf{2}}$

(b) Consider 1.00 L of solution prepared by mixing 1.00 × 10^-4 mol Cu(ClO₄)₂ and 0.100 mol CH₃CO₂Na. Use Equation 12-16 to find the fraction of copper in the form Cu²⁺

Short answer

b) The fraction of copper in the form Cu²⁺ will be 0.017

Step-by-step solution

Reaction and their equilibrium constant

Equations given

$\begin{array}{c}C{u}^{2+}+C{H}_{3}C{O}_2^{-}\rightleftharpoons Cu{\left(C{H}_{3}C{O}_2\right)}^{+}{\beta }_1\left(=K_1\right)\\ C{u}^{2+}+2C{H}_{3}C{O}_2^{-}\rightleftharpoons Cu{\left(C{H}_{3}C{O}_2\right)}_2{\beta }_2\\ Cu{\left(C{H}_{3}C{O}_2\right)}^{+}+C{H}_{3}C{O}_2^{-}\rightleftharpoons Cu{\left(C{H}_{3}C{O}_2\right)}_2\left(aq\right)K_2\end{array}$

$\begin{array}{c}{\beta }_1=\frac{\left[Cu{\left(C{H}_{3}C{O}_2\right)}^{+}\right]}{\left[C{H}_{3}C{O}_2^{-}\right]\left[C{u}^{2+}\right]}\\ {\beta }_2=\frac{\left[Cu{\left(C{H}_{3}C{O}_2\right)}_2\right]}{{\left[C{H}_{3}C{O}_2^{-}\right]}^2\left[C{u}^{2+}\right]}\\ K_2=\frac{\left[Cu{\left(C{H}_{3}C{O}_2\right)}_2\right]}{\left[Cu{\left(C{H}_{3}C{O}_2\right)}^{+}\right]\left[C{H}_{3}C{O}_2^{-}\right]}\\ =\frac{\left[Cu{\left(C{H}_{3}C{O}_2\right)}_2\right]}{{\left[C{H}_{3}C{O}_2^{-}\right]}^2\left[C{u}^{2+}\right]}\times \frac{\left[C{H}_{3}C{O}_2^{-}\right]\left[C{u}^{2+}\right]}{\left[Cu{\left(C{H}_{3}C{O}_2\right)}^{+}\right]}\\ ={\beta }_2\times \frac{1}{{\beta }_1}\\ =\frac{{\beta }_2}{{\beta }_1}\end{array}$

Information Given

From Appendix I the following values were obtained for calculation

$\begin{array}{l}\log {\beta }_1=2.23\\ \log {\beta }_2=3.63\end{array}$

As per equation 12-16 we can write

Fraction of free metal ion(copper ion)

${\alpha }_{C{u}^{2+}}=\frac{1}{1+{\beta }_1\left[L\right]+{\beta }_2{\left[L\right]}^2}$

Determine the fraction of copper ion

$\begin{array}{c}{\alpha }_{C{u}^{2+}}=\frac{1}{1+{\beta }_1\left[L\right]+{\beta }_2{\left[L\right]}^2}\\ =\frac{1}{1+{10}^{2.23}\times \left[0.1\right]+{10}^{3.63}\times {\left[0.1\right]}^2}\\ =0.017\end{array}$

The fraction of copper in the form Cu²⁺ will be 0.017