The copper(I) ion forms a complex ion with \(\mathrm{CN}^{-}\) according to the following equation: \(\mathrm{Cu}^{+}(a q)+3 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Cu}(\mathrm{CN})_{3}^{2-}(a q) \quad K_{\mathrm{f}}=1.0 \times 10^{11}\) a. Calculate the solubility of \(\operatorname{CuBr}(s)\left(K_{\text {sp }}=1.0 \times 10^{-5}\right)\) in \(1.0 \mathrm{~L}\) of \(1.0 \mathrm{M} \mathrm{NaCN}\).b. Calculate the concentration of \(\mathrm{Br}^{-}\) at equilibrium. c. Calculate the concentration of \(\mathrm{CN}^{-}\) at equilibrium.

Firstly, we should write the dissociation equilibrium equation for the given salt \(\mathrm{CuBr}(s)\), which is: \[ \mathrm{CuBr}(s) \rightleftharpoons \mathrm{Cu}^{+}(a q) + \mathrm{Br}^{-}(a q) \]

Now, we will create an Initial, Change, and Equilibrium (ICE) table to express the change in concentrations from the initial to equilibrium states. The initial concentration of \(\mathrm{CN}^{-}\) is \(1.0~\mathrm{M}\) while initial concentrations of \(\mathrm{Cu}^{+}\), \(\mathrm{Cu}(\mathrm{CN})_{3}^{2-}\), and \(\mathrm{Br}^{-}\) are \(0\): | | \(\mathrm{Cu}^{+}(a q)\) | \(\mathrm{CN}^{-}(a q)\) | \(\mathrm{Cu}(\mathrm{CN})_{3}^{2-}(a q)\) | \(\mathrm{Br}^{-}(a q)\) | |--------|---------------|--------------|----------------------|-------------| | Initial | 0 | 1.0 | 0 | 0 | | Change | +x | -3x | +x | +x | | Equilibrium | x | 1.0 - 3x | x | x |

First, we will use the dissociation equilibrium equation for \(\mathrm{CuBr}\) and the \(K_{sp}\) value (\(1.0 \times 10^{-5}\)). From the ICE table, we know the equilibrium concentrations of \(\mathrm{Cu}^{+}\) and \(\mathrm{Br}^{-}\) are both equal to x: \[ K_{sp} = [\mathrm{Cu}^+][\mathrm{Br}^-] = x^2 \] Next, we will set up the expression using the complex formation equilibrium equation with cyanide ions and the given \(K_f\) value (\(1.0 \times 10^{11}\)). The equilibrium concentrations have been determined from the ICE table: \[ K_f = \frac{[\mathrm{Cu(CN)_3}^{2-}]}{[\mathrm{Cu}^+][\mathrm{CN}^-]^3} = \frac{x}{x(1.0 - 3x)^3} \]

By relating the two constants \(K_{sp}\) and \(K_f\), we can solve for x: \[ K_{sp} = K_f \cdot [\mathrm{Cu}^{+}][\mathrm{CN}^{-}]^3 \\ x^2 = (1.0 \times 10^{11}) x(1.0 - 3x)^3 \] Now, you can make a reasonable approximation by considering that the change (3x) in the \(\mathrm{CN}^{-}\) concentration is negligible since it is considerably smaller than the initial concentration (1.0 M). With that assumption, the expression above simplifies to: \[ x^2 \approx (1.0 \times 10^{11}) x(1.0)^3 \\ x \approx \frac{1}{10^{11}} \]

Now that we have the value of x, we can determine the solubility of \(\mathrm{CuBr}\): Solubility of \(\mathrm{CuBr} = x = \frac{1}{10^{11}}\) mol/L

From the ICE table, we know that the concentration of \(\mathrm{Br}^-\) at equilibrium is equal to x. So, \[ [\mathrm{Br}^-]_{\text{eq}} = x = \frac{1}{10^{11}}~\text{mol/L} \]

Similarly, we can calculate the concentration of \(\mathrm{CN}^-\) ions at equilibrium by substituting the value of x: \[ [\mathrm{CN}^-]_{\text{eq}} = 1.0 - 3x = 1.0 - 3 \left(\frac{1}{10^{11}}\right) \approx 1.0~\text{mol/L} \] So, the solubility and concentrations at equilibrium are: - Solubility of \(\mathrm{CuBr} \approx \frac{1}{10^{11}}~\text{mol/L}\) - \([\mathrm{Br}^{-}]_{\text{eq}} \approx \frac{1}{10^{11}}~\text{mol/L}\) - \([\mathrm{CN}^{-}]_{\text{eq}} \approx 1.0~\text{mol/L}\)