To solve the problem step by step, we will follow these steps:
### Step 1: Write the balanced chemical equation
The reaction between lead nitrate and chromic sulfate can be represented as:
\[
3 \text{Pb(NO}_3\text{)}_2 + \text{Cr}_2\text{(SO}_4\text{)}_3 \rightarrow 3 \text{PbSO}_4 \downarrow + 2 \text{Cr(NO}_3\text{)}_3
\]
### Step 2: Calculate the number of moles of each reactant
1. **For Lead Nitrate (Pb(NO₃)₂):**
- Volume = 45.0 mL = 0.045 L
- Molarity = 0.25 M
- Moles of Pb(NO₃)₂ = Volume × Molarity
\[
\text{Moles of Pb(NO}_3\text{)}_2 = 0.045 \, \text{L} \times 0.25 \, \text{mol/L} = 0.01125 \, \text{mol} \, (11.25 \, \text{mmol})
\]
2. **For Chromic Sulfate (Cr₂(SO₄)₃):**
- Volume = 25.0 mL = 0.025 L
- Molarity = 0.10 M
- Moles of Cr₂(SO₄)₃ = Volume × Molarity
\[
\text{Moles of Cr}_2\text{(SO}_4\text{)}_3 = 0.025 \, \text{L} \times 0.10 \, \text{mol/L} = 0.0025 \, \text{mol} \, (2.5 \, \text{mmol})
\]
### Step 3: Determine the limiting reagent
From the balanced equation, we see that 3 moles of Pb(NO₃)₂ react with 1 mole of Cr₂(SO₄)₃.
- Moles of Pb(NO₃)₂ required for 2.5 mmol of Cr₂(SO₄)₃:
\[
\text{Required moles of Pb(NO}_3\text{)}_2 = 3 \times 0.0025 \, \text{mol} = 0.0075 \, \text{mol} \, (7.5 \, \text{mmol})
\]
Since we have 11.25 mmol of Pb(NO₃)₂ available, and only 7.5 mmol is needed, Cr₂(SO₄)₃ is the limiting reagent.
### Step 4: Calculate the moles of lead sulfate (PbSO₄) formed
From the balanced equation, 1 mole of Cr₂(SO₄)₃ produces 3 moles of PbSO₄. Therefore, the moles of PbSO₄ formed will be:
\[
\text{Moles of PbSO}_4 = 3 \times \text{Moles of Cr}_2\text{(SO}_4\text{)}_3 = 3 \times 0.0025 \, \text{mol} = 0.0075 \, \text{mol} \, (7.5 \, \text{mmol})
\]
### Step 5: Calculate the remaining moles of Pb(NO₃)₂
Moles of Pb(NO₃)₂ used:
\[
\text{Moles of Pb(NO}_3\text{)}_2 \text{ used} = 7.5 \, \text{mmol}
\]
Remaining moles of Pb(NO₃)₂:
\[
\text{Remaining moles of Pb(NO}_3\text{)}_2 = 11.25 \, \text{mmol} - 7.5 \, \text{mmol} = 3.75 \, \text{mmol}
\]
### Step 6: Calculate the total volume of the solution
Total volume after mixing:
\[
\text{Total Volume} = 45.0 \, \text{mL} + 25.0 \, \text{mL} = 70.0 \, \text{mL} = 0.070 \, \text{L}
\]
### Step 7: Calculate the molar concentration of the remaining species
1. **Molarity of remaining Pb(NO₃)₂:**
\[
\text{Molarity of Pb(NO}_3\text{)}_2 = \frac{3.75 \times 10^{-3} \, \text{mol}}{0.070 \, \text{L}} = 0.0536 \, \text{M}
\]
2. **Moles of Cr(NO₃)₃ formed:**
From the balanced equation, 2 moles of Cr(NO₃)₃ are formed for every mole of Cr₂(SO₄)₃:
\[
\text{Moles of Cr(NO}_3\text{)}_3 = 2 \times 0.0025 \, \text{mol} = 0.0050 \, \text{mol} \, (5.0 \, \text{mmol})
\]
3. **Molarity of Cr(NO₃)₃:**
\[
\text{Molarity of Cr(NO}_3\text{)}_3 = \frac{5.0 \times 10^{-3} \, \text{mol}}{0.070 \, \text{L}} = 0.0714 \, \text{M}
\]
4. **Calculate the concentration of ions left in the solution:**
- Lead ion (Pb²⁺) concentration = 0.0536 M
- Chromium ion (Cr³⁺) concentration = 0.0714 M
- Nitrate ion (NO₃⁻) concentration:
- From Pb(NO₃)₂: 2 × 0.0536 M = 0.1072 M
- From Cr(NO₃)₃: 3 × 0.0714 M = 0.2142 M
- Total NO₃⁻ concentration = 0.1072 M + 0.2142 M = 0.3214 M
### Summary of Results
- Moles of PbSO₄ formed: **7.5 mmol**
- Molar concentration of remaining Pb(NO₃)₂: **0.0536 M**
- Molar concentration of Cr(NO₃)₃: **0.0714 M**
- Molar concentration of Pb²⁺: **0.0536 M**
- Molar concentration of Cr³⁺: **0.0714 M**
- Molar concentration of NO₃⁻: **0.3214 M**
