Question

Without performing detailed calculations, show that significant disproportionation of AuCl occurs if you attempt to make a saturated aqueous solution. Use data from Table 23.7 and \(K_{\mathrm{sp}}(\mathrm{AuCl})=\) \(2.0 \times 10^{-13}\).

Short answer

Disproportionation of AuCl will occur if the concentration of Cl- ion in the solution is greater than the molar solubility calculated using \(K_{sp}\) equation.

Step-by-step solution

- Understanding Disproportionation

First, understand the concept of Disproportionation. It's a type of redox reaction where one species is both reduced and oxidized.

- Identifying the Reaction

Next, identify that the AuCl will disporportionate into Au and AuCl3 - \(2AuCl(s) \rightarrow Au(s) + AuCl3(aq)\). This is your disproportionation reaction for AuCl.

- Calculating Molarity

Derive the molar solubility 's' using the given \(K_{sp}\) for AuCl. For AuCl, \(K_{sp} = [Au^{+}][Cl^{-}]\), and since AuCl dissociates into Au+ and Cl-, the concentrations of Au+ and Cl- are 's'. So, \(K_{sp} = s^2\). Solving for 's' gives us \(s = \sqrt{K_{sp}}\). Use \(K_{sp} = 2 x 10^{-13}\) to calculate 's'.

- Examining the Table 23.7

Utilize Table 23.7 to understand the effect of chlorine ion concentration on the reaction. The table will provide values for standard potential values needed for later steps.

- Conclusion

Finally, compare 's' with concentrations from the table, if the concentration of Cl- ion is much higher than the molar solubility 's', then disproportionation is favored and will occur.

Key concepts

Redox Reactions

Redox reactions, or oxidation-reduction reactions, are processes where electrons are transferred between two substances. The substance that loses electrons is oxidized, while the substance that gains electrons is reduced. This is often remembered by the mnemonic 'LEO goes GER' - Lose Electrons Oxidation, Gain Electrons Reduction.

Disproportionation is a special kind of redox reaction where a single substance undergoes both oxidation and reduction. For example, if you have a compound like AuCl, it can break down and form two new substances, Au (gold metal) and AuCl3 (gold(III) chloride). During this reaction, the single reactant, AuCl, is being both oxidized (to form AuCl3) and reduced (to form Au). This type of reaction showcases the dual characteristic of the reactant as it acts as both a reducing agent and an oxidizing agent.

Solubility Product Constant Ksp

The solubility product constant, typically abbreviated as Ksp, describes the equilibrium between a solid and its constituent ions in a solution. When a solid dissolves, it dissociates into ions, and at equilibrium, the product of the concentrations of these ions, each raised to the power of its stoichiometric coefficient, equals the Ksp.

For the compound AuCl, which dissociates into Au+ and Cl- ions, the equation for Ksp is \(K_{sp} = [Au^{+}][Cl^{-}]\). This mathematical representation tells us how much of the solid can dissolve before the system becomes saturated. By understanding Ksp, we are able to predict whether a precipitate will form in a solution or if a solid will dissolve, helping us identify the conditions under which a reaction like disproportionation is likely to occur.

Molar Solubility

Molar solubility refers to the number of moles of a solute that can be dissolved per liter of solution before the solution becomes saturated. You can think of it as the solute's capacity to dissolve. It is derived from the solubility product constant (Ksp) and gives us insight into the extent of a substance's solubility in a specific solvent. For AuCl, we calculate molar solubility by taking the square root of Ksp, since \(K_{sp} = s^2\) when the solute dissociates into one mole of each ion.

Knowing the molar solubility helps in predicting whether a reaction, like disproportionation of AuCl, could occur significantly in a solution. If a calculated molar solubility is very small, for example, it means the compound is not very soluble. In a disproportionation reaction, if the ions produced exceed this molar solubility, the shift towards products would be so great that the reaction would proceed readily to form a precipitate or result in other products as the system seeks to re-establish equilibrium.