Balanced Chemical Equation:
AgBr(s) ⇌ Ag+(aq) + Br-(aq)
The solubility product constant is defined by:
$$K_{sp} = [\text{Ag}^+] [\text{Br}^-]$$
Where [Ag^+] and [Br^-] represent the equilibrium concentrations of silver and bromide ions, respectively.
Given \(K_{sp} = 5 \times 10^{-13}\) at \(25 \degree C\), we can set up the following ICE (Initial, Change, Equilibrium) table to determine the equilibrium concentrations of silver and bromide ions:
ICE table:
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| AgBr(s) | - | - | - |
| Ag+(aq) | 0 | +x | x |
| Br-(aq) | 0 | +x | x |
Since AgBr dissolves \(1:1\), the change in the concentration of Ag+ and Br- ions will be equal, represented as \(+\text x\) and \(-\text x\).
Substituting the equilibrium concentrations into the \(K_{sp}\) expression:
$$5 \times 10^{-13} = [\text{x}][\text{x}]$$
Solving for \([\text{x}]\), we obtain:
$$x = [\text{Ag}^+] = [\text{Br}^-] = \sqrt{5 \times 10^{-13} \ M}$$
Therefore, the equilibrium concentrations of silver and bromide ions are:
$$[\text{Ag}^+] = [\text{Br}^-] = \sqrt{5 \times 10^{-13} \ M} \approx 2.24 \times 10^{-7}\ M$$
