2 M of 100 mL `Na_2 SO_4` is mixed with 3 M of 100 mL NaCl solution and 1 M of 200 mL `CaCl_2` solution . Then the ratio of the concentration of cation and anion is

Options

A. `1:1`
B. `2:1`
C. `2:3`
D. `1:2`

Answer

To solve the problem, we need to find the total concentration of cations and anions after mixing the given solutions. Let's break it down step by step.

### Step 1: Calculate the number of moles of each solution.

1. **Sodium Sulfate (Na₂SO₄)**:
- Molarity (M) = 2 M
- Volume (V) = 100 mL = 0.1 L
- Number of moles (n) = M × V = 2 mol/L × 0.1 L = 0.2 moles

2. **Sodium Chloride (NaCl)**:
- Molarity (M) = 3 M
- Volume (V) = 100 mL = 0.1 L
- Number of moles (n) = M × V = 3 mol/L × 0.1 L = 0.3 moles

3. **Calcium Chloride (CaCl₂)**:
- Molarity (M) = 1 M
- Volume (V) = 200 mL = 0.2 L
- Number of moles (n) = M × V = 1 mol/L × 0.2 L = 0.2 moles

### Step 2: Determine the dissociation of each compound.

1. **Dissociation of Na₂SO₄**:
- Na₂SO₄ → 2 Na⁺ + SO₄²⁻
- From 0.2 moles of Na₂SO₄, we get:
- Na⁺ = 2 × 0.2 = 0.4 moles
- SO₄²⁻ = 0.2 moles

2. **Dissociation of NaCl**:
- NaCl → Na⁺ + Cl⁻
- From 0.3 moles of NaCl, we get:
- Na⁺ = 0.3 moles
- Cl⁻ = 0.3 moles

3. **Dissociation of CaCl₂**:
- CaCl₂ → Ca²⁺ + 2 Cl⁻
- From 0.2 moles of CaCl₂, we get:
- Ca²⁺ = 0.2 moles
- Cl⁻ = 2 × 0.2 = 0.4 moles

### Step 3: Calculate the total moles of cations and anions.

1. **Total cations**:
- Na⁺ from Na₂SO₄ = 0.4 moles
- Na⁺ from NaCl = 0.3 moles
- Ca²⁺ from CaCl₂ = 0.2 moles
- Total cations = 0.4 + 0.3 + 0.2 = 0.9 moles

2. **Total anions**:
- SO₄²⁻ from Na₂SO₄ = 0.2 moles
- Cl⁻ from NaCl = 0.3 moles
- Cl⁻ from CaCl₂ = 0.4 moles
- Total anions = 0.2 + 0.3 + 0.4 = 0.9 moles

### Step 4: Calculate the ratio of cations to anions.

- Ratio of cations to anions = Total cations / Total anions = 0.9 moles / 0.9 moles = 1

### Final Answer:
The ratio of the concentration of cations to anions is **1:1**.

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