Ionic strength of a solution made by mixing equal volumes of `0.01 M NaCl` and `0.02 M AlCl_(3)`

Options

A. `0.065`
B. `0.13`
C. `0.0325`
D. `0.0216`

Explanation

To calculate the ionic strength of a solution made by mixing equal volumes of `0.01 M NaCl` and `0.02 M AlCl₃`, we can follow these steps:

### Step 1: Calculate the concentrations after mixing
When equal volumes of two solutions are mixed, the concentrations are halved.

- For NaCl:
- Initial concentration = 0.01 M
- After mixing (halved) = 0.01 M / 2 = 0.005 M

- For AlCl₃:
- Initial concentration = 0.02 M
- After mixing (halved) = 0.02 M / 2 = 0.01 M

### Step 2: Determine the dissociation of the salts
Next, we need to consider how each salt dissociates in solution.

- NaCl dissociates into:
- Na⁺ (1 ion) and Cl⁻ (1 ion)
- From 0.005 M NaCl, we get:
- [Na⁺] = 0.005 M
- [Cl⁻] = 0.005 M

- AlCl₃ dissociates into:
- Al³⁺ (1 ion) and 3 Cl⁻ (3 ions)
- From 0.01 M AlCl₃, we get:
- [Al³⁺] = 0.01 M
- [Cl⁻] = 3 × 0.01 M = 0.03 M

### Step 3: Calculate the total concentration of ions
Now, we will sum up the concentrations of each ion:

- Total concentration of Na⁺ = 0.005 M
- Total concentration of Al³⁺ = 0.01 M
- Total concentration of Cl⁻ = 0.005 M (from NaCl) + 0.03 M (from AlCl₃) = 0.035 M

### Step 4: Apply the formula for ionic strength
The ionic strength (I) of a solution is given by 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.

Now we can substitute the values:

- For Na⁺:
- Concentration = 0.005 M, Charge = +1
- Contribution to ionic strength = \( 0.005 \times (1)^2 = 0.005 \)

- For Al³⁺:
- Concentration = 0.01 M, Charge = +3
- Contribution to ionic strength = \( 0.01 \times (3)^2 = 0.01 \times 9 = 0.09 \)

- For Cl⁻:
- Concentration = 0.035 M, Charge = -1
- Contribution to ionic strength = \( 0.035 \times (1)^2 = 0.035 \)

### Step 5: Calculate the total ionic strength
Now we can sum these contributions:
\[ I = \frac{1}{2} \left( 0.005 + 0.09 + 0.035 \right) \]
\[ I = \frac{1}{2} \left( 0.13 \right) = 0.065 \]

### Final Answer
The ionic strength of the solution is **0.065 M**.

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