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, we will follow these steps:

### Step 1: Determine the moles of each reactant
1. **Calculate the moles of Ag₂SO₄:**
\[ \text{Moles of } Ag₂SO₄ = \text{Volume (L)} \times \text{Molarity (M)} = 2 \, \text{L} \times 2 \, \text{M} = 4 \, \text{moles} \]

2. **Calculate the moles of NaCl:**
\[ \text{Moles of } NaCl = \text{Volume (L)} \times \text{Molarity (M)} = 4 \, \text{L} \times 1 \, \text{M} = 4 \, \text{moles} \]

### Step 2: Determine the reaction and the moles of precipitate formed
The reaction between silver sulfate and sodium chloride can be represented as:
\[ Ag₂SO₄ + 2NaCl \rightarrow 2AgCl + Na₂SO₄ \]
From the balanced equation, we see that 1 mole of Ag₂SO₄ reacts with 2 moles of NaCl to produce 2 moles of AgCl.

- **From 4 moles of Ag₂SO₄, we can produce:**
\[ \text{Moles of } AgCl = 2 \times \text{Moles of } Ag₂SO₄ = 2 \times 4 = 8 \, \text{moles} \]
- **However, we only have 4 moles of NaCl available. Therefore, we can only react:**
\[ \text{Moles of } AgCl = 2 \times \text{Moles of } NaCl = 2 \times 4 = 8 \, \text{moles} \]

Since we have enough NaCl to react with all the Ag₂SO₄, we will produce 8 moles of AgCl.

### Step 3: Calculate the total volume of the solution
The total volume after mixing is:
\[ \text{Total Volume} = 2 \, \text{L} + 4 \, \text{L} = 6 \, \text{L} \]

### Step 4: Calculate the concentrations of the remaining ions
After the complete precipitation of AgCl, the remaining ions in the solution are Na⁺ and SO₄²⁻.

1. **Calculate the concentration of Na⁺:**
- Moles of Na⁺ = Moles of NaCl = 4 moles
- Concentration of Na⁺ = \(\frac{\text{Moles}}{\text{Total Volume}} = \frac{4 \, \text{moles}}{6 \, \text{L}} = \frac{2}{3} \, \text{M} \approx 0.67 \, \text{M}\)

2. **Calculate the concentration of SO₄²⁻:**
- Moles of SO₄²⁻ = Moles of Ag₂SO₄ = 4 moles
- Concentration of SO₄²⁻ = \(\frac{\text{Moles}}{\text{Total Volume}} = \frac{4 \, \text{moles}}{6 \, \text{L}} = \frac{2}{3} \, \text{M} \approx 0.67 \, \text{M}\)

### Step 5: Sum of the molar concentrations of all ions
The total molar concentration of all ions in the solution is:
\[ \text{Total Concentration} = [Na^+] + [SO₄^{2-}] = 0.67 \, \text{M} + 0.67 \, \text{M} = 1.34 \, \text{M} \]

### Final Answer
The sum of the molar concentration of all the ions is **1.34 M**.

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