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

Options

A. `2.8`
B. `1.2`
C. `1.0`
D. `1.8`

Explanation

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

### Step 1: Understand the Dissociation of CaCl₂
Calcium chloride (CaCl₂) dissociates in water into its constituent ions:
\[ \text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2 \text{Cl}^- \]

### Step 2: Determine the Concentration of Ions
From the dissociation, we can see that:
- 1 mole of CaCl₂ produces 1 mole of Ca²⁺ ions.
- 1 mole of CaCl₂ produces 2 moles of Cl⁻ ions.

Given that the concentration of CaCl₂ is 0.4 M:
- The concentration of Ca²⁺ ions = 0.4 M
- The concentration of Cl⁻ ions = 2 × 0.4 M = 0.8 M

### Step 3: Calculate the Ionic Strength
The ionic strength (I) of a solution is calculated using the formula:
\[ I = \frac{1}{2} \sum c_i z_i^2 \]
where \( c_i \) is the concentration of each ion and \( z_i \) is the charge of each ion.

For our solution:
- For Ca²⁺: \( c_{\text{Ca}^{2+}} = 0.4 \, \text{M} \) and \( z_{\text{Ca}^{2+}} = +2 \)
- For Cl⁻: \( c_{\text{Cl}^-} = 0.8 \, \text{M} \) and \( z_{\text{Cl}^-} = -1 \)

Now we can plug in the values:
\[ I = \frac{1}{2} \left( c_{\text{Ca}^{2+}} z_{\text{Ca}^{2+}}^2 + c_{\text{Cl}^-} z_{\text{Cl}^-}^2 \right) \]
\[ I = \frac{1}{2} \left( 0.4 \times (2)^2 + 0.8 \times (-1)^2 \right) \]
\[ I = \frac{1}{2} \left( 0.4 \times 4 + 0.8 \times 1 \right) \]
\[ I = \frac{1}{2} \left( 1.6 + 0.8 \right) \]
\[ I = \frac{1}{2} \times 2.4 \]
\[ I = 1.2 \]

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

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