Problem 39 - Chemistry and Chemical Reactivity

Question

When 1.55 g of solid thallium (I) bromide is added to \(1.00 \mathrm{L}\) of water, the salt dissolves to a small extent. $$\operatorname{TIBr}(\mathrm{s}) \rightleftarrows \mathrm{Tl}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})$$ The thallium(I) and bromide ions in equilibrium with TIBr each have a concentration of \(1.9 \times 10^{-3} \mathrm{M} .\) What is the value of \(K_{\mathrm{sp}}\) for TIBr?

Step-by-Step Solution

Answer

\( K_{\text{sp}} = 3.61 \times 10^{-6} \)

Step 1: Understanding the Dissolution Process

The dissolution of thallium(I) bromide (TlBr) in water dissociates into thallium ions (\( \text{Tl}^+ \)) and bromide ions (\( \text{Br}^- \)). Both ions have a concentration of \(1.9 \times 10^{-3} \, \text{M}\) at equilibrium.

Step 2: Writing the Solubility Product Expression

The solubility product \( K_{\text{sp}} \) for the reaction \( \text{TlBr(s)} \rightleftharpoons \text{Tl}^+(\text{aq}) + \text{Br}^-(\text{aq}) \) is given by the product of the concentrations of the ions: \[ K_{\text{sp}} = [\text{Tl}^+][\text{Br}^-] \]

Step 3: Substituting Concentrations into \(K_{\text{sp}}\) Formula

Since both ion concentrations are \(1.9 \times 10^{-3} \, \text{M}\), substitute these into the \( K_{\text{sp}} \) expression:\[ K_{\text{sp}} = (1.9 \times 10^{-3})(1.9 \times 10^{-3}) \]

Step 4: Calculating the Value of \(K_{\text{sp}}\)

Perform the multiplication: \[ K_{\text{sp}} = (1.9 \times 10^{-3})^2 = 3.61 \times 10^{-6} \]

Step 5: Conclusion

The solubility product constant \( K_{\text{sp}} \) for TlBr is \( 3.61 \times 10^{-6} \).

Key Concepts

Thallium Bromide Dissolution

Thallium bromide (TlBr) is a sparingly soluble ionic compound. When added to water, it dissociates into its constituent ions: thallium ions (\( \text{Tl}^+ \)) and bromide ions (\( \text{Br}^- \)). This dissolution is governed by chemical equilibrium as represented by the equation:\[\operatorname{TIBr}(\mathrm{s}) \rightleftarrows \mathrm{Tl}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})\]This means that at a certain point, the dissolution process achieves a balance where the rate of dissolution equals the rate at which ions recombine to form solid TlBr again. This balance defines the equilibrium concentrations of the ions in solution. In this particular scenario, each ion reaches a concentration of \(1.9 \times 10^{-3} \text{M}\) in the solution. Understanding the dissolution process and equilibrium is crucial for calculating the solubility product constant, which gives us insight into how much of the compound dissolves in water.

Equilibrium Concentrations

Equilibrium concentrations refer to the steady-state concentrations of ions in a solution when a dissolving solid and its ions are at equilibrium. When thallium bromide dissolves slightly in water, it forms an equilibrium where the ions present in solution have fixed concentrations.

• At equilibrium, the concentration of \( \text{Tl}^+ \) is \(1.9 \times 10^{-3} \text{M}\).
• The concentration of \( \text{Br}^- \) is also \(1.9 \times 10^{-3} \text{M}\).

These consistent concentrations occur because the initial amount of TlBr is enough to establish an equilibrium but not enough to fully saturate the solution. Once the equilibrium is set in motion, the solution effectively contains equal concentrations of thallium and bromide ions as determined by the limited solubility of TlBr.

Ion Concentration Calculation

Calculating ion concentrations at equilibrium is an integral part of determining the solubility product. Consider that when thallium bromide dissolves in water, each molecule of TlBr produces one thallium ion and one bromide ion. Given the equilibrium concentration for both ions,\[[\text{Tl}^+] = 1.9 \times 10^{-3} \text{M}, \quad [\text{Br}^-] = 1.9 \times 10^{-3} \text{M}.\]The solubility product constant (\( K_{\text{sp}} \)) is directly associated with these concentrations. To find \( K_{\text{sp}} \), multiply these ion concentrations:\[K_{\text{sp}} = [\text{Tl}^+][\text{Br}^-] = (1.9 \times 10^{-3})(1.9 \times 10^{-3}).\]This multiplication yields the \( K_{\text{sp}} \) value, which quantifies the extent to which the thallium bromide dissolves under these conditions. This straightforward calculation reflects the standard procedure for solving equilibrium concentration problems.

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that describes the state where the concentrations of reactants and products remain constant over time. This occurs when the forward and reverse reactions happen at the same rate. In the case of thallium bromide, when it dissolves in water, a dynamic equilibrium is established:\[\operatorname{TIBr}(\mathrm{s}) \rightleftarrows \mathrm{Tl}^{+}(\mathrm{aq}) + \mathrm{Br}^{-}(\mathrm{aq})\]\Even though it seems externally still, at the microscopic level, ions constantly dissolve and recombine at equal rates.

• The system achieves balance, not because reactions stop, but because their rates equalize.
• The dissolving process results in a steady concentration of ions.

Understanding chemical equilibrium helps illustrate why calculated ion concentrations accurately represent the dissolution extent. For sparingly soluble salts like TlBr, achieving equilibrium allows us to predict how much dissolves at a given condition using the solubility product constant, \( K_{\text{sp}} \). This insight is pivotal for applications in chemistry and environmental sciences.