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

### Step 1: Determine the final concentrations after mixing
When equal volumes of the two solutions are mixed, the concentrations will be halved because the total volume doubles.

- For NaCl:
\[ \text{Initial concentration} = 0.01 \, \text{M} \\ \text{Final concentration} = \frac{0.01}{2} = 0.005 \, \text{M} \]

- For AlCl₃:
\[ \text{Initial concentration} = 0.02 \, \text{M} \\ \text{Final concentration} = \frac{0.02}{2} = 0.01 \, \text{M} \]

### Step 2: Identify the ions produced
- NaCl dissociates into:
\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]
Thus, the concentrations of the ions are:
- \([\text{Na}^+] = 0.005 \, \text{M}\)
- \([\text{Cl}^-] = 0.005 \, \text{M}\)

- AlCl₃ dissociates into:
\[ \text{AlCl}_3 \rightarrow \text{Al}^{3+} + 3\text{Cl}^- \]
Thus, the concentrations of the ions are:
- \([\text{Al}^{3+}] = 0.01 \, \text{M}\)
- \([\text{Cl}^-] = 3 \times 0.01 = 0.03 \, \text{M}\)

### Step 3: Calculate the total concentration of chloride ions
The total concentration of chloride ions is the sum from both sources:
\[ [\text{Cl}^-]_{\text{total}} = 0.005 + 0.03 = 0.035 \, \text{M} \]

### Step 4: Calculate the ionic strength (μ)
The formula for ionic strength is given by:
\[ \mu = \frac{1}{2} \sum c_i z_i^2 \]
where \(c_i\) is the concentration of each ion and \(z_i\) is the charge of the ion.

Now we will calculate the contributions from each ion:

1. For \(\text{Na}^+\):
\[ \mu_{\text{Na}^+} = 0.005 \times (1^2) = 0.005 \]

2. For \(\text{Al}^{3+}\):
\[ \mu_{\text{Al}^{3+}} = 0.01 \times (3^2) = 0.01 \times 9 = 0.09 \]

3. For \(\text{Cl}^-\):
\[ \mu_{\text{Cl}^-} = 0.035 \times (1^2) = 0.035 \]

### Step 5: Combine the contributions to find the total ionic strength
Now, we sum the contributions:
\[ \mu = \frac{1}{2} \left( 0.005 + 0.09 + 0.035 \right) \]
\[ \mu = \frac{1}{2} \left( 0.13 \right) = 0.065 \]

### Final Answer
The ionic strength of the solution is:
\[ \mu = 0.065 \]