Ionic strength of `0.4 M CaCl_(2)` is

Options

A. `1.2`
B. `1.0`
C. `0.9`
D. `0.8`

Explanation

To calculate the ionic strength of a 0.4 M solution of CaCl₂, we can follow these steps:

### Step 1: Identify the ions present in CaCl₂
Calcium chloride (CaCl₂) dissociates in water to produce:
- 1 Calcium ion (Ca²⁺)
- 2 Chloride ions (Cl⁻)

### Step 2: Write down the formula for ionic strength
The ionic strength (I) of a solution is given by 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 (valency) of the ion.

### Step 3: Calculate contributions from each ion
1. **For Calcium ion (Ca²⁺)**:
- Number of ions = 1
- Concentration \( C_{Ca^{2+}} = 0.4 \, M \)
- Valency \( z_{Ca^{2+}} = +2 \)

Contribution to ionic strength:
\[ I_{Ca^{2+}} = C_{Ca^{2+}} \cdot z_{Ca^{2+}}^2 = 0.4 \cdot (2^2) = 0.4 \cdot 4 = 1.6 \]

2. **For Chloride ion (Cl⁻)**:
- Number of ions = 2
- Concentration \( C_{Cl^{-}} = 0.4 \, M \)
- Valency \( z_{Cl^{-}} = -1 \)

Contribution to ionic strength:
\[ I_{Cl^{-}} = 2 \cdot C_{Cl^{-}} \cdot z_{Cl^{-}}^2 = 2 \cdot 0.4 \cdot (-1)^2 = 2 \cdot 0.4 \cdot 1 = 0.8 \]

### Step 4: Sum the contributions
Now, we can sum the contributions from both ions:
\[ I = \frac{1}{2} \left( I_{Ca^{2+}} + I_{Cl^{-}} \right) = \frac{1}{2} \left( 1.6 + 0.8 \right) = \frac{1}{2} \cdot 2.4 = 1.2 \]

### Final Answer
The ionic strength of the 0.4 M CaCl₂ solution is **1.2**.

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