Silver bromide dissolves in aqueous sodium thiosulfate, Na2...

Silver bromide dissolves in aqueous sodium thiosulfate, ${Na}_{2} {S}_{2} {O}_{3}$ yielding the complex ion ${Ag}({S}_{2} {O}_{3})_{2}^{3-} :$ $$\operatorname{AgBr}(s)+2 {S}_{2} {O}_{3}^{2-(a q)} \rightleftharpoons {Ag}({S}_{2} {O}_{3})_{2}^{3-}(a q)+{Br}^{-(a q)} \quad K=?$$ Calculate the equilibrium constant for the dissolution reaction, and calculate the molar solubility of AgBr in $0.10 {M} {Na}_{2} {S}_{2} {O}_{3} . K_{{sp}}=5.4 \times 10^{-13}$ for AgBr and $K_{{f}}=4.7 \times 10^{13}$ for ${Ag}({S}_{2} {O}_{3})_{2}^{3-} .$

Silver bromide dissolves in aqueous sodium thiosulfate, ${Na}_{2} {S}_{2} {O}_{3}$ yielding the complex ion ${Ag}({S}_{2} {O}_{3})_{2}^{3-} :$
$$\operatorname{AgBr}(s)+2 {S}_{2} {O}_{3}^{2-(a q)} \rightleftharpoons {Ag}({S}_{2} {O}_{3})_{2}^{3-}(a q)+{Br}^{-(a q)} \quad K=?$$
Calculate the equilibrium constant for the dissolution reaction, and calculate the molar solubility of AgBr in $0.10 {M} {Na}_{2} {S}_{2} {O}_{3} . K_{{sp}}=5.4 \times 10^{-13}$ for AgBr and $K_{{f}}=4.7 \times 10^{13}$ for ${Ag}({S}_{2} {O}_{3})_{2}^{3-} .$

Key Concepts

Solubility Product (Ksp)

The solubility product constant is the equilibrium constant that describes the dissolution of a sparingly soluble solid salt into its constituent ions in solution. It is defined as the product of the ion concentrations, each raised to the power of its stoichiometric coefficient, and is a measure of how little of the salt dissolves under saturation conditions.

Formation (Stability) Constant (Kf)

The formation constant (or stability constant) quantifies the equilibrium favoring the formation of a complex ion from a metal ion and one or more ligands. A high Kf value indicates a very stable complex, meaning the ligand strongly binds the metal ion, which can significantly alter the solubility behavior of the metal salt.

Complex Ion Formation and Its Effect on Solubility

When a ligand is present that can complex with the metal ion, the formation of a complex ion effectively removes free metal ion from the solution. This shifts the dissolution equilibrium of a sparingly soluble salt (as dictated by the solubility product) to dissolve further, thereby increasing the salt's apparent solubility.

Overall Equilibrium Constant Calculation

For a reaction that occurs in consecutive steps, such as the dissolution of a salt followed by the complexation of its metal ion, the overall equilibrium constant is calculated as the product of the equilibrium constants for each individual step. This method allows for the analysis of multiple equilibria acting in concert.

Molar Solubility in the Presence of a Reactive Common Ion

In systems where a common ion participates in a secondary reaction (such as complex ion formation) in addition to the dissolution equilibrium, the molar solubility must be determined by combining mass balance with the modified equilibrium expression. The consumption of a common ion by a complexation reaction affects the concentration of free ions, thus altering the solubility compared to a simple Ksp scenario.

Transcript

Please give Ace some feedback