Answer :
When potassium iodide (KI) reacts with lead(II) nitrate [tex]\(\text{Pb}\left(\text{NO}_3\right)_2\)[/tex], a double displacement reaction is expected to occur. In this type of reaction, the cations and anions of two different molecules switch places, forming two new compounds. Here's the step-by-step process:
1. Identify Reactants and Their Ions:
- Potassium iodide (KI) dissociates in water to form potassium ions ([tex]\(\text{K}^+\)[/tex]) and iodide ions ([tex]\(\text{I}^-\)[/tex]).
- Lead(II) nitrate [tex]\(\text{Pb}\left(\text{NO}_3\right)_2\)[/tex] dissociates in water to form lead ions ([tex]\(\text{Pb}^{2+}\)[/tex]) and nitrate ions ([tex]\(\text{NO}_3^-\)[/tex]).
2. Write the Dissociation Equations:
- [tex]\(\text{KI}\rightarrow \text{K}^+ + \text{I}^-\)[/tex]
- [tex]\(\text{Pb}\left(\text{NO}_3\right)_2 \rightarrow \text{Pb}^{2+} + 2\text{NO}_3^-\)[/tex]
3. Predict the Products:
- The cations ([tex]\(\text{K}^+\)[/tex] and [tex]\(\text{Pb}^{2+}\)[/tex]) will pair with the anions ([tex]\(\text{I}^-\)[/tex] and [tex]\(\text{NO}_3^-\)[/tex]) of the other compound to form new products.
- Potassium ions ([tex]\(\text{K}^+\)[/tex]) pair with nitrate ions ([tex]\(\text{NO}_3^-\)[/tex]) to form potassium nitrate ([tex]\(\text{KNO}_3\)[/tex]).
- Lead ions ([tex]\(\text{Pb}^{2+}\)[/tex]) pair with iodide ions ([tex]\(\text{I}^-\)[/tex]) to form lead(II) iodide ([tex]\(\text{PbI}_2\)[/tex]).
4. Write the Balanced Chemical Equation:
- [tex]\(\text{2KI} + \text{Pb}\left(\text{NO}_3\right)_2 \rightarrow 2\text{KNO}_3 + \text{PbI}_2\)[/tex]
5. Describe the Reaction:
- Potassium nitrate ([tex]\(\text{KNO}_3\)[/tex]) is soluble in water and remains in solution.
- Lead(II) iodide ([tex]\(\text{PbI}_2\)[/tex]) is insoluble in water and precipitates out as a yellow solid.
The overall chemical reaction is:
[tex]\[ \text{2KI}_{(aq)} + \text{Pb}\left(\text{NO}_3\right)_2_{(aq)} \rightarrow 2\text{KNO}_3_{(aq)} + \text{PbI}_2_{(s)} \][/tex]
Thus, the expected reaction between potassium iodide and lead(II) nitrate is a double displacement reaction that produces potassium nitrate in solution and a yellow precipitate of lead(II) iodide.
1. Identify Reactants and Their Ions:
- Potassium iodide (KI) dissociates in water to form potassium ions ([tex]\(\text{K}^+\)[/tex]) and iodide ions ([tex]\(\text{I}^-\)[/tex]).
- Lead(II) nitrate [tex]\(\text{Pb}\left(\text{NO}_3\right)_2\)[/tex] dissociates in water to form lead ions ([tex]\(\text{Pb}^{2+}\)[/tex]) and nitrate ions ([tex]\(\text{NO}_3^-\)[/tex]).
2. Write the Dissociation Equations:
- [tex]\(\text{KI}\rightarrow \text{K}^+ + \text{I}^-\)[/tex]
- [tex]\(\text{Pb}\left(\text{NO}_3\right)_2 \rightarrow \text{Pb}^{2+} + 2\text{NO}_3^-\)[/tex]
3. Predict the Products:
- The cations ([tex]\(\text{K}^+\)[/tex] and [tex]\(\text{Pb}^{2+}\)[/tex]) will pair with the anions ([tex]\(\text{I}^-\)[/tex] and [tex]\(\text{NO}_3^-\)[/tex]) of the other compound to form new products.
- Potassium ions ([tex]\(\text{K}^+\)[/tex]) pair with nitrate ions ([tex]\(\text{NO}_3^-\)[/tex]) to form potassium nitrate ([tex]\(\text{KNO}_3\)[/tex]).
- Lead ions ([tex]\(\text{Pb}^{2+}\)[/tex]) pair with iodide ions ([tex]\(\text{I}^-\)[/tex]) to form lead(II) iodide ([tex]\(\text{PbI}_2\)[/tex]).
4. Write the Balanced Chemical Equation:
- [tex]\(\text{2KI} + \text{Pb}\left(\text{NO}_3\right)_2 \rightarrow 2\text{KNO}_3 + \text{PbI}_2\)[/tex]
5. Describe the Reaction:
- Potassium nitrate ([tex]\(\text{KNO}_3\)[/tex]) is soluble in water and remains in solution.
- Lead(II) iodide ([tex]\(\text{PbI}_2\)[/tex]) is insoluble in water and precipitates out as a yellow solid.
The overall chemical reaction is:
[tex]\[ \text{2KI}_{(aq)} + \text{Pb}\left(\text{NO}_3\right)_2_{(aq)} \rightarrow 2\text{KNO}_3_{(aq)} + \text{PbI}_2_{(s)} \][/tex]
Thus, the expected reaction between potassium iodide and lead(II) nitrate is a double displacement reaction that produces potassium nitrate in solution and a yellow precipitate of lead(II) iodide.
