Using Table \(20.4\), calculate, for the disproportionation of \(\mathrm{Au}^{+}\), (a) \(K\). (b) the concentration of \(\mathrm{Au}^{+}\) in equilibrium with \(0.10 \mathrm{M} \mathrm{Au}^{3+}\).

The disproportionation of \(\mathrm{Au}^{+}\) ions can be described by the following balanced chemical equation: \begin{equation} 3\mathrm{Au}^{+} \rightarrow \mathrm{Au} + 2\mathrm{Au}^{3+} \end{equation} This equation tells us that three \(\mathrm{Au}^{+}\) ions will react to form one \(\mathrm{Au}\) atom and two \(\mathrm{Au}^{3+}\) ions.

According to the Nernst Equation, we can relate the standard cell potential (\(E^\circ\)) and the equilibrium constant (K) as follows: \begin{equation} E^\circ = \frac{0.0592 \,V}{n} \log K \end{equation} Here, \(n\) is the number of electron transfer involved in the reaction, and in our case, n = 2, since two electrons are transferred when one \(\mathrm{Au}^{3+}\) ion is produced from one \(\mathrm{Au}^{+}\) ion. The standard potential (\(E^\circ\)) can be found using the Table \(20.4\) provided. We will find the difference in the standard potentials for Au(III) and Au(I) to get the overall standard potential for this reaction.

From Table \(20.4\), we know the standard reduction potential for the following half-cell reactions: \begin{equation} \mathrm{Au}^{3+} + 3\mathrm e^- \rightarrow \mathrm{Au} \quad E^\circ_{Au(III) \rightarrow Au} = 1.498 \, V \end{equation} \begin{equation} \mathrm{Au}^{+} + \mathrm e^- \rightarrow \mathrm{Au} \quad E^\circ_{Au(I) \rightarrow Au} = 1.692 \, V \end{equation} To calculate the overall standard potential for the disproportionation reaction, take the difference between these two standard potentials: \begin{equation} E^\circ = E^\circ_{Au(III) \rightarrow Au} - E^\circ_{Au(I) \rightarrow Au} = 1.498 - 1.692 = -0.194 \, V \end{equation}

Now, use the value of \(E^\circ\) and the Nernst equation to calculate the equilibrium constant (K): \begin{equation} K = 10^{\frac{-n E^\circ}{0.0592 \,V}} = 10^{\frac{-2 (-0.194 \,V)}{0.0592 \,V}} \approx 162 \end{equation} So, the equilibrium constant (K) for the disproportionation reaction is approximately 162.

We can now apply the equilibrium expression for the reaction: \begin{equation} K = \frac{[\mathrm{Au}^{3+}]^2}{[\mathrm{Au}^{+}]^3} \end{equation} Substituting the provided concentration of \(\mathrm{Au}^{3+}\) and the value of K, we have: \begin{equation} 162 = \frac{(0.10)^2}{[\mathrm{Au}^{+}]^3} \end{equation} To solve for the concentration of \(\mathrm{Au}^{+}\), we will rearrange the equation and take the cube root: \begin{equation} [\mathrm{Au}^{+}] = \sqrt[3]{\frac{(0.10)^2}{162}} \approx 0.017 \, M \end{equation} So, the concentration of \(\mathrm{Au}^{+}\) ions in equilibrium with \(0.10 \mathrm{M}\) of \(\mathrm{Au}^{3+}\) ions is approximately \(0.017 \mathrm{M}\). To summarize: (a) The equilibrium constant (K) is approximately 162. (b) The concentration of \(\mathrm{Au}^{+}\) at equilibrium with \(0.10 \mathrm{M}\) of \(\mathrm{Au}^{3+}\) ions is approximately \(0.017 \mathrm{M}\).