Certainly! Let's analyze the reaction between KI (Potassium Iodide) and Pb(NO₃)₂ (Lead(II) Nitrate) and predict the type of reaction and the products formed.
1. Identify the Reactants:
- KI: Potassium Iodide
- Pb(NO₃)₂: Lead(II) Nitrate
2. Determine the Type of Reaction:
This is a double displacement reaction, also known as a precipitation reaction. In a double displacement reaction, the cations and anions of two different compounds swap places, forming two new compounds.
3. Write the Balanced Chemical Equation:
To predict the products of the double displacement reaction, we'll exchange the ions between the reactants:
- Potassium ion (K⁺) from KI
- Iodide ion (I⁻) from KI
- Lead(II) ion (Pb²⁺) from Pb(NO₃)₂
- Nitrate ion (NO₃⁻) from Pb(NO₃)₂
The ions will recombine to form:
- KNO₃: Potassium Nitrate (formed by K⁺ and NO₃⁻)
- PbI₂: Lead(II) Iodide (formed by Pb²⁺ and I⁻)
The balanced chemical equation is:
[tex]\[ 2KI(aq) + Pb(NO₃)₂(aq) \rightarrow 2KNO₃(aq) + PbI₂(s) \][/tex]
4. Identify the Products and State of Matter:
- Potassium Nitrate (KNO₃): This compound remains in the aqueous state (aq) as it is soluble in water.
- Lead(II) Iodide (PbI₂): This compound forms a solid precipitate (s), which is usually a yellow solid, as PbI₂ is insoluble in water.
Therefore, the products are:
- Potassium Nitrate (KNO₃) in aqueous solution.
- Lead(II) Iodide (PbI₂) as a yellow precipitate.
5. Conclusion:
The reaction between KI and Pb(NO₃)₂ is a double displacement reaction resulting in the formation of Potassium Nitrate (KNO₃) in aqueous form and Lead(II) Iodide (PbI₂) as a precipitate.
The expected reaction products are:
- KNO₃ (Potassium Nitrate) in aqueous state (aq)
- PbI₂ (Lead(II) Iodide) as a solid (s)
Specifically, the precipitation product from the reaction is PbI₂ (Lead(II) Iodide).
