To determine the type of reaction between potassium iodide (KI) and lead(II) nitrate [Pb(NO₃)₂], follow these steps:
1. Identify the Reactants:
- Potassium iodide (KI) is a compound that, when dissolved in water, dissociates into potassium ions (K⁺) and iodide ions (I⁻).
- Lead(II) nitrate [Pb(NO₃)₂] is a compound that, when dissolved in water, dissociates into lead(II) ions (Pb²⁺) and nitrate ions (NO₃⁻).
2. Write the Balanced Chemical Equation:
- The chemical reaction involves the mixing of aqueous solutions of KI and Pb(NO₃)₂.
[tex]\[ 2\text{KI} (aq) + \text{Pb}(\text{NO}_3)_2 (aq) \rightarrow 2\text{KNO}_3 (aq) + \text{PbI}_2 (s) \][/tex]
3. Reaction Type Analysis:
- This reaction can be classified as a double displacement reaction. In a double displacement reaction, the cations and anions of two different compounds switch places, forming two new compounds.
- The cations ([tex]\(K^+\)[/tex] from KI and [tex]\(Pb^{2+}\)[/tex] from Pb(NO₃)₂) and the anions ([tex]\(I^-\)[/tex] from KI and [tex]\(NO₃^-\)[/tex] from Pb(NO₃)₂) exchange partners to form new compounds.
4. Observation of Products:
- Potassium nitrate (KNO₃) remains in aqueous solution.
- Lead(II) iodide (PbI₂) precipitates out of the solution as a solid due to its low solubility in water.
The precipitate formation indicates that the reaction is a double displacement reaction leading to precipitation.
Result:
The type of reaction between potassium iodide (KI) and lead(II) nitrate [Pb(NO₃)₂] is a double displacement reaction leading to precipitation.
