2.) Suppose we have a solution PbI, and it is clear. PЫ, (8) > Рb* (аq) + 21 (аq) 2.a) If you add Pb(NO,),(aq) to the solution what direction does the reaction shift to (the left, right, or no shift in the reaction). 2.b) What does addition of KBr(aq) shift the equilibrium towards (the left, right, or no shift in the reaction).

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**Equilibrium and Reaction Shifts**

**2.) Suppose we have a solution PbI₂ and it is clear.**

\[ \text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq) \]

**2.a) If you add Pb(NO₃)₂(aq) to the solution, what direction does the reaction shift to (the left, right, or no shift in the reaction)?**

**2.b) What does addition of KBr(aq) shift the equilibrium towards (the left, right, or no shift in the reaction)?**

### Explanation:

The equilibrium expression shows the dissociation of lead(II) iodide (\( \text{PbI}_2 \)) in aqueous solution to form lead ions (\( \text{Pb}^{2+} \)) and iodide ions (\( \text{I}^- \)).

- **Part 2.a:** Adding \( \text{Pb(NO}_3)_2 \), which contains \( \text{Pb}^{2+} \) ions, increases the concentration of \( \text{Pb}^{2+} \) in the solution. According to Le Chatelier's Principle, the reaction will shift to the left to counter the increase in product concentration by forming more solid PbI₂.

- **Part 2.b:** Adding \( \text{KBr} \), which may precipitate other ions (e.g., by forming insoluble compounds with \( \text{Pb}^{2+} \)), may not directly affect the defined equilibrium unless it changes ion concentrations involved in the dissolution of PbI₂. In this context, adding a compound not containing shared ions would more likely cause no shift in the dissolution equilibrium of PbI₂.
Transcribed Image Text:**Equilibrium and Reaction Shifts** **2.) Suppose we have a solution PbI₂ and it is clear.** \[ \text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq) \] **2.a) If you add Pb(NO₃)₂(aq) to the solution, what direction does the reaction shift to (the left, right, or no shift in the reaction)?** **2.b) What does addition of KBr(aq) shift the equilibrium towards (the left, right, or no shift in the reaction)?** ### Explanation: The equilibrium expression shows the dissociation of lead(II) iodide (\( \text{PbI}_2 \)) in aqueous solution to form lead ions (\( \text{Pb}^{2+} \)) and iodide ions (\( \text{I}^- \)). - **Part 2.a:** Adding \( \text{Pb(NO}_3)_2 \), which contains \( \text{Pb}^{2+} \) ions, increases the concentration of \( \text{Pb}^{2+} \) in the solution. According to Le Chatelier's Principle, the reaction will shift to the left to counter the increase in product concentration by forming more solid PbI₂. - **Part 2.b:** Adding \( \text{KBr} \), which may precipitate other ions (e.g., by forming insoluble compounds with \( \text{Pb}^{2+} \)), may not directly affect the defined equilibrium unless it changes ion concentrations involved in the dissolution of PbI₂. In this context, adding a compound not containing shared ions would more likely cause no shift in the dissolution equilibrium of PbI₂.
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