The total ionic strength (total molarity of all the ions) containing 0.2 M `CuSO_(4) and 0.1M Al_(2)(SO_(4))_(3)` is

Options

A. 0.5 M
B. 0.7 M
C. 0.9 M
D. 1.2 M

Explanation

To calculate the total ionic strength (total molarity of all the ions) of the solution containing 0.2 M CuSO₄ and 0.1 M Al₂(SO₄)₃, we will follow these steps:

### Step 1: Identify the ions and their concentrations

1. **CuSO₄ dissociates into:**
- Cu²⁺ (1 ion)
- SO₄²⁻ (1 ion)

Therefore, from 0.2 M CuSO₄, we get:
- [Cu²⁺] = 0.2 M
- [SO₄²⁻] = 0.2 M

2. **Al₂(SO₄)₃ dissociates into:**
- 2 Al³⁺ (2 ions)
- 3 SO₄²⁻ (3 ions)

Therefore, from 0.1 M Al₂(SO₄)₃, we get:
- [Al³⁺] = 0.1 M × 2 = 0.2 M
- [SO₄²⁻] = 0.1 M × 3 = 0.3 M

### Step 2: Calculate the total concentration of each ion

Now we sum the concentrations of each ion from both compounds:

- **For Cu²⁺:**
- [Cu²⁺] = 0.2 M

- **For Al³⁺:**
- [Al³⁺] = 0.2 M

- **For SO₄²⁻:**
- From CuSO₄: [SO₄²⁻] = 0.2 M
- From Al₂(SO₄)₃: [SO₄²⁻] = 0.3 M
- Total [SO₄²⁻] = 0.2 M + 0.3 M = 0.5 M

### Step 3: Calculate the total ionic strength

The ionic strength (I) can be calculated using the formula:

\[ I = \frac{1}{2} \sum (C_i \cdot z_i^2) \]

Where:
- \( C_i \) is the concentration of each ion.
- \( z_i \) is the charge of each ion.

Now we calculate for each ion:

1. For Cu²⁺:
- \( C = 0.2 \, \text{M}, \, z = 2 \)
- Contribution = \( 0.2 \cdot (2^2) = 0.2 \cdot 4 = 0.8 \)

2. For Al³⁺:
- \( C = 0.2 \, \text{M}, \, z = 3 \)
- Contribution = \( 0.2 \cdot (3^2) = 0.2 \cdot 9 = 1.8 \)

3. For SO₄²⁻:
- \( C = 0.5 \, \text{M}, \, z = 2 \)
- Contribution = \( 0.5 \cdot (2^2) = 0.5 \cdot 4 = 2.0 \)

### Step 4: Sum the contributions and calculate ionic strength

Now, we sum all contributions:

\[ \text{Total Contribution} = 0.8 + 1.8 + 2.0 = 4.6 \]

Now, plug this into the ionic strength formula:

\[ I = \frac{1}{2} \cdot 4.6 = 2.3 \]

### Final Answer

The total ionic strength of the solution is **2.3 M**.

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