Assuming complete precipitation of `AgCl`, calculate the sum of the molar concentration of all the ions if 2 lit of 2 M `Ag_(2)SO_(4)` is mixed
with 4 lit of 1 M NaCl solution is

Options

A. 4 M
B. 2 M
C. 3 M
D. 2.5 M

Answer

To solve the problem step by step, we will follow the outlined process of calculating the molar concentration of ions after the complete precipitation of AgCl when mixing Ag₂SO₄ and NaCl solutions.

### Step 1: Write the balanced chemical equation
When silver sulfate (Ag₂SO₄) reacts with sodium chloride (NaCl), the reaction can be represented as follows:

\[ \text{Ag}_2\text{SO}_4 + 2 \text{NaCl} \rightarrow 2 \text{AgCl} + \text{Na}_2\text{SO}_4 \]

### Step 2: Identify the initial concentrations and volumes
- Volume of Ag₂SO₄ solution (V₁) = 2 L
- Molarity of Ag₂SO₄ (M₁) = 2 M
- Volume of NaCl solution (V₂) = 4 L
- Molarity of NaCl (M₂) = 1 M

### Step 3: Calculate the number of moles of each reactant
1. **Moles of Ag₂SO₄:**
\[ \text{Moles of Ag}_2\text{SO}_4 = M_1 \times V_1 = 2 \, \text{mol/L} \times 2 \, \text{L} = 4 \, \text{mol} \]

2. **Moles of NaCl:**
\[ \text{Moles of NaCl} = M_2 \times V_2 = 1 \, \text{mol/L} \times 4 \, \text{L} = 4 \, \text{mol} \]

### Step 4: Determine the limiting reactant
From the balanced equation, we see that 1 mole of Ag₂SO₄ reacts with 2 moles of NaCl. Therefore, for 4 moles of Ag₂SO₄, we need:
\[ 4 \, \text{mol Ag}_2\text{SO}_4 \times 2 \, \text{mol NaCl/mol Ag}_2\text{SO}_4 = 8 \, \text{mol NaCl} \]
Since we only have 4 moles of NaCl, NaCl is the limiting reactant.

### Step 5: Calculate the moles of AgCl produced
From the balanced equation, 2 moles of NaCl produce 2 moles of AgCl. Therefore, 4 moles of NaCl will produce:
\[ \text{Moles of AgCl} = \frac{4 \, \text{mol NaCl}}{2} \times 2 = 4 \, \text{mol AgCl} \]

### Step 6: Calculate the total moles of ions in the solution after precipitation
After precipitation, the remaining ions in the solution will be:
- From Na₂SO₄: 2 moles of Na⁺ and 1 mole of SO₄²⁻ (from 4 moles of Ag₂SO₄)
- From the precipitated AgCl: 4 moles of Ag⁺ and 4 moles of Cl⁻ (from 4 moles of NaCl)

Thus, the total moles of ions:
\[ \text{Total moles of ions} = 2 \, \text{mol Na}^+ + 1 \, \text{mol SO}_4^{2-} + 4 \, \text{mol Ag}^+ + 4 \, \text{mol Cl}^- = 11 \, \text{mol ions} \]

### Step 7: Calculate the total volume of the solution
The total volume of the mixed solution:
\[ \text{Total Volume} = V_1 + V_2 = 2 \, \text{L} + 4 \, \text{L} = 6 \, \text{L} \]

### Step 8: Calculate the molar concentration of all ions
The total molar concentration of all ions (C) is given by:
\[ C = \frac{\text{Total moles of ions}}{\text{Total volume}} = \frac{11 \, \text{mol}}{6 \, \text{L}} \approx 1.83 \, \text{M} \]

### Final Answer
The sum of the molar concentration of all the ions in the solution is approximately **1.83 M**.