Problem 43 - Chemistry and Chemical Reactivity

Question

When \(1.55 \mathrm{g}\) of solid thallium(I) bromide is added to 1.00 L of water, the salt dissolves to a small extent. $$\operatorname{TiBr}(\mathrm{s}) \rightleftharpoons \mathrm{Tl}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})$$ 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

The value of \(K_{\mathrm{sp}}\) for \(\text{TlBr}\) is \(3.61 \times 10^{-6}\).

Step 1: Understand the Concept of Solubility Product (K_sp)

The solubility product constant, denoted as \(K_{\mathrm{sp}}\), describes the equilibrium between a solid and its ions in a saturated solution. For the dissolution of thallium(I) bromide \(\text{TlBr}\), the equation is \(\text{TlBr} \rightleftharpoons \text{Tl}^+ + \text{Br}^-\). The \(K_{\text{sp}}\) is given by the product of the concentration of the ions: \(K_{\mathrm{sp}} = [\text{Tl}^+][\text{Br}^-]\).

Step 2: Identify Ion Concentrations at Equilibrium

From the problem, we know that the concentration of \(\text{Tl}^+\) and \(\text{Br}^-\) ions at equilibrium in the solution are both \(1.9 \times 10^{-3} \text{ M}\).

Step 3: Calculate the Solubility Product (K_sp)

Substitute the equilibrium concentrations of the ions into the \(K_{\mathrm{sp}}\) expression: \[K_{\mathrm{sp}} = [\text{Tl}^+][\text{Br}^-] = (1.9 \times 10^{-3})(1.9 \times 10^{-3}) = 3.61 \times 10^{-6} \]

Key Concepts

Thallium(I) Bromide

Thallium(I) bromide, often represented as TlBr, is an intriguing compound due to its limited solubility in water. When added to water, it forms a dynamic equilibrium between the solid form and its ions in the solution. This process can be described by the chemical equation:

• \[\text{TlBr(s)} \rightleftharpoons \text{Tl}^+\text{(aq)} + \text{Br}^-\text{(aq)}\]

This equation shows the dissolution and formation of thallium and bromide ions in water.
Thallium is a soft, malleable metal, and in this case, it's combined with bromide, a halide ion.
The fact that only a small amount of TlBr dissolves indicates that the solubility is quite low, making the study of its solubility product (\(K_{\text{sp}}\)) particularly useful.

Ion Concentration

Ion concentration refers to the amount of dissolved ions in a solution, which plays a key role in understanding solubility.
In the equilibrium of TlBr, the concentration of thallium ions (\(\text{Tl}^+\)) and bromide ions (\(\text{Br}^-\)) are equal at \(1.9 \times 10^{-3} \text{ M}\).

• These equal concentrations result from the 1:1 stoichiometry of the dissolution reaction.
• Knowing these ion concentrations allows us to calculate the solubility product, \(K_{\text{sp}}\).

The relationship between ion concentration and \(K_{\text{sp}}\) is crucial in predicting how much of a compound can dissolve before reaching saturation.
This balance helps chemists understand saturation and potential precipitation.

Chemical Equilibrium

Chemical equilibrium in solubility refers to the point where the rate of dissolution of a solid equals the rate at which it precipitates out of solution.
This is depicted by the reversible equation for TlBr:

• \[\text{TlBr(s)} \rightleftharpoons \text{Tl}^+\text{(aq)} + \text{Br}^-\text{(aq)}\]

In this state, concentrations of ions remain constant over time.
The solubility product \(K_{\text{sp}}\) helps quantify this equilibrium:

• \[K_{\mathrm{sp}} = [\text{Tl}^+][\text{Br}^-]\]
• For TlBr, it was calculated as \(3.61 \times 10^{-6}\), reflecting how little the salt dissolves.

Understanding this concept is vital for controlling reactions and processes in chemical manufacturing, environmental science, and pharmacology, where solubility is a factor.