Problem 60 - Chemistry The Central Science In Si Units

Question

Consider a beaker containing a saturated solution of \(\mathrm{Pbl}_{2}\) in equilibrium with undissolved \(\mathrm{Pbl}_{2}(s)\). Now solid \(\mathrm{KI}\) is added to this solution. (a) Will the amount of solid \(\mathrm{PbI}_{2}\) at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of \(\mathrm{Pb}^{2+}\) ions in solution increase or decrease? (c) Will the concentration of I' ions in solution increase or decrease?

Step-by-Step Solution

Answer

(a) Increase, (b) Decrease, (c) Increase.

Step 1: Understanding the Initial Equilibrium

Initially, the system has a saturated solution of \( \mathrm{PbI}_2 \), meaning that the solution is at equilibrium with undissolved \( \mathrm{PbI}_2 \). The equilibrium reaction can be expressed as: \[ \mathrm{PbI}_2(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + 2\mathrm{I}^-(aq) \]. At this point, both lead (\( \mathrm{Pb}^{2+} \)) and iodide (\( \mathrm{I}^- \)) ions are present at their maximum equilibrium concentrations.

Step 2: Adding KI to the Equilibrium

When solid \( \mathrm{KI} \) is added, it dissociates into \( \mathrm{K}^+ \) and \( \mathrm{I}^- \) ions in the solution, increasing the concentration of \( \mathrm{I}^- \) ions significantly.

Step 3: Applying Le Chatelier's Principle

According to Le Chatelier's Principle, increasing the concentration of \( \mathrm{I}^- \) ions will shift the equilibrium to the left to counteract this increase. This means more \( \mathrm{PbI}_2 \) will precipitate out of the solution to absorb the excess \( \mathrm{I}^- \) ions.

Step 4: Analyzing the Effects on Solid PbI2 and Ion Concentrations

As equilibrium shifts left, more \( \mathrm{PbI}_2 \) precipitates, increasing the amount of solid \( \mathrm{PbI}_2 \) at the bottom of the beaker. The concentration of \( \mathrm{Pb}^{2+} \) ions will decrease, as some of these ions are removed from the solution to form solid \( \mathrm{PbI}_2 \). The concentration of \( \mathrm{I}^- \) ions will increase because of the continued addition of \( \mathrm{K}I \).

Step 5: Conclusion

With the addition of \( \mathrm{K}I \), the system tries to re-establish equilibrium by increasing the amount of solid \( \mathrm{PbI}_2 \) at the bottom. The concentration of \( \mathrm{Pb}^{2+} \) in the solution decreases, while the concentration of \( \mathrm{I}^- \) ions increases due to the dissolution of \( \mathrm{K}I \).

Key Concepts

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept in chemistry that helps us predict how a system at equilibrium responds to changes. It's like the system has an "action-reaction" mechanism. If you change something in a chemical reaction that's in equilibrium, the reaction will adjust to compensate.

This principle tells us that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and re-establish equilibrium. For example, if you increase the concentration of a reactant, the equilibrium will shift to reduce that concentration, typically by forming more product.

In the context of the exercise, when additional ( \( ext{I}^- \) ions are added to the saturated solution by dissolving additional \( ext{KI} \), Le Chatelier's Principle predicts the equilibrium will shift to reduce that concentration by forming more solid \( ext{PbI}_2 \). This is why we see more solid ( \( ext{PbI}_2 \) precipitating out of the solution. It’s the system’s way of saying “Let’s take care of this!” and coming back to balance.

Saturated Solution

A saturated solution is one where the maximum amount of a solute has been dissolved in a solvent at a given temperature. In simpler terms, it's like when you've added as much sugar as possible to your cup of tea—no more will dissolve.

In a saturated solution, the dissolved solute and any undissolved solid exist in a balance of dynamic equilibrium. This means that while undissolved and dissolved particles are constantly in motion, their concentrations remain constant over time.

When no more solute can dissolve, any additional solute remains unchanged, in solid form. In the \( \text{PbI}_2 \) system, the solution contains dissolved \( \text{Pb}^{2+} \) and \( ext{I}^- \) ions at equilibrium with un-dissolved \( \text{PbI}_2 \) solid. If we introduce more ions into the system, like those from \( \text{KI} \), it disrupts this delicate balance, prompting Le Chatelier’s Principle to act, causing more of the solid to precipitate.

Ionic Concentration

Ionic concentration refers to the amount of ions present in a solution, often expressed as moles per liter (M). It's a critical aspect when dealing with reactions in aqueous solutions, influencing how the reactions progress and equilibrate.

Changes in ionic concentration can have a significant impact on chemical equilibria. In our example, initially, the concentration of \( \text{Pb}^{2+} \) and \( \text{I}^- \) ions reflects a state of balance in the saturated solution. However, the addition of \( \text{KI} \) increases the concentration of \( \text{I}^- \) ions, disturbing the equilibrium.

Consider it like overcrowding a room: adding more people (ions) causes a shift, in this case, pushing the equilibrium towards forming more solid \( \text{PbI}_2 \), or in simpler terms, removing some ions from the solution as they get incorporated into the solid precipitate. Thus, the concentration of \( \text{Pb}^{2+} \) ions decreases while the \( \text{I}^- \) ions initially increase before reaching new equilibrium dynamics.