Answer :
To determine the products of the reaction between lead(II) nitrate [tex]\(\text{Pb(NO}_3\text{)}_2(aq)\)[/tex] and potassium sulfate [tex]\(\text{K}_2\left( \text{SO}_4 \right)(aq)\)[/tex], we'll follow these steps:
### Step 1: Identify the Ions in Solution
When lead(II) nitrate [tex]\(\text{Pb(NO}_3\text{)}_2\)[/tex] and potassium sulfate [tex]\(\text{K}_2\left( \text{SO}_4 \right)\)[/tex] are dissolved in water, they dissociate into their respective ions:
- [tex]\(\text{Pb(NO}_3\text{)}_2(aq) \rightarrow \text{Pb}^{2+}(aq) + 2\text{NO}_3^- (aq)\)[/tex]
- [tex]\(\text{K}_2\left( \text{SO}_4 \right)(aq) \rightarrow 2\text{K}^+(aq) + \text{SO}_4^{2-}(aq)\)[/tex]
### Step 2: Exchange of Ions
When these ions are mixed, they can potentially recombine to form new compounds. The possible combinations are:
- [tex]\(\text{Pb}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4\)[/tex]
- [tex]\(\text{K}^+(aq) + \text{NO}_3^-(aq) \rightarrow \text{KNO}_3\)[/tex]
### Step 3: Determine Solubility
Using the solubility rules:
- [tex]\(\text{PbSO}_4\)[/tex] (Lead(II) sulfate) is insoluble and will precipitate out of the solution.
- [tex]\(\text{KNO}_3\)[/tex] (Potassium nitrate) is soluble and will remain in aqueous form.
### Step 4: Write the Balanced Reaction
Since [tex]\(\text{PbSO}_4\)[/tex] will precipitate and [tex]\(\text{KNO}_3\)[/tex] will stay dissolved, the products are:
[tex]\[ \text{Pb(NO}_3\text{)}_2(aq) + \text{K}_2\left(\text{SO}_4\right)(aq) \rightarrow \text{PbSO}_4 (s) + 2\text{KNO}_3 (aq) \][/tex]
(Note: the [tex]\( 2 \text{KNO}_3 \)[/tex] is necessary to balance the equation, but it’s not part of the multiple-choice options.)
### Step 5: Select the Correct Option
Of the given multiple-choice answers, the correct one must include [tex]\(\text{PbSO}_4(s)\)[/tex] as a solid and [tex]\(\text{KNO}_3(aq)\)[/tex] as the remaining product in the aqueous form.
Thus, the correct answer is:
A. [tex]\(\text{PbSO}_4(s) + \text{KNO}_3(aq)\)[/tex]
So the products of the unbalanced reaction are:
[tex]\[ \text{PbSO}_4(s) + \text{KNO}_3(aq) \][/tex]
### Step 1: Identify the Ions in Solution
When lead(II) nitrate [tex]\(\text{Pb(NO}_3\text{)}_2\)[/tex] and potassium sulfate [tex]\(\text{K}_2\left( \text{SO}_4 \right)\)[/tex] are dissolved in water, they dissociate into their respective ions:
- [tex]\(\text{Pb(NO}_3\text{)}_2(aq) \rightarrow \text{Pb}^{2+}(aq) + 2\text{NO}_3^- (aq)\)[/tex]
- [tex]\(\text{K}_2\left( \text{SO}_4 \right)(aq) \rightarrow 2\text{K}^+(aq) + \text{SO}_4^{2-}(aq)\)[/tex]
### Step 2: Exchange of Ions
When these ions are mixed, they can potentially recombine to form new compounds. The possible combinations are:
- [tex]\(\text{Pb}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4\)[/tex]
- [tex]\(\text{K}^+(aq) + \text{NO}_3^-(aq) \rightarrow \text{KNO}_3\)[/tex]
### Step 3: Determine Solubility
Using the solubility rules:
- [tex]\(\text{PbSO}_4\)[/tex] (Lead(II) sulfate) is insoluble and will precipitate out of the solution.
- [tex]\(\text{KNO}_3\)[/tex] (Potassium nitrate) is soluble and will remain in aqueous form.
### Step 4: Write the Balanced Reaction
Since [tex]\(\text{PbSO}_4\)[/tex] will precipitate and [tex]\(\text{KNO}_3\)[/tex] will stay dissolved, the products are:
[tex]\[ \text{Pb(NO}_3\text{)}_2(aq) + \text{K}_2\left(\text{SO}_4\right)(aq) \rightarrow \text{PbSO}_4 (s) + 2\text{KNO}_3 (aq) \][/tex]
(Note: the [tex]\( 2 \text{KNO}_3 \)[/tex] is necessary to balance the equation, but it’s not part of the multiple-choice options.)
### Step 5: Select the Correct Option
Of the given multiple-choice answers, the correct one must include [tex]\(\text{PbSO}_4(s)\)[/tex] as a solid and [tex]\(\text{KNO}_3(aq)\)[/tex] as the remaining product in the aqueous form.
Thus, the correct answer is:
A. [tex]\(\text{PbSO}_4(s) + \text{KNO}_3(aq)\)[/tex]
So the products of the unbalanced reaction are:
[tex]\[ \text{PbSO}_4(s) + \text{KNO}_3(aq) \][/tex]