The mass of acetic acid present in `500 ml` of solution in which it is `1%` ionides (Ka of ` CH_(3)COOH = 1.8 xx 10^(-4)`)

Options

A. `5.4`
B. `12.6`
C. `6.4`
D. `10.8`

Explanation

To solve the problem of finding the mass of acetic acid in a 500 ml solution that is 1% ionized, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a solution of acetic acid (CH₃COOH) with a volume of 500 ml.
- The solution is 1% ionized, meaning that 1% of the acetic acid molecules have dissociated into ions.
- The dissociation constant (Ka) for acetic acid is given as \( 1.8 \times 10^{-5} \).

2. **Convert the Degree of Ionization**:
- The degree of ionization (α) is given as 1%, which can be expressed as a decimal:
\[ \alpha = \frac{1}{100} = 0.01 \]

3. **Using the Dissociation Constant**:
- For weak acids, the dissociation constant \( K_a \) can be expressed as:
\[ K_a = C \alpha^2 \quad \text{(neglecting α in the term (1 - α))} \]
- Rearranging this gives us the concentration (C):
\[ C = \frac{K_a}{\alpha^2} \]

4. **Calculate α²**:
- Calculate \( \alpha^2 \):
\[ \alpha^2 = (0.01)^2 = 0.0001 = 1 \times 10^{-4} \]

5. **Substituting Values into the Concentration Formula**:
- Now substitute the values of \( K_a \) and \( \alpha^2 \):
\[ C = \frac{1.8 \times 10^{-5}}{1 \times 10^{-4}} = 0.18 \, \text{M} \]

6. **Calculating the Mass of Acetic Acid**:
- Normality (N) is defined as:
\[ N = \frac{\text{mass of solute (g)}}{\text{equivalent weight (g/equiv)} \times \text{volume of solution (L)}} \]
- The equivalent weight of acetic acid (CH₃COOH) is 60 g/mol (since it donates one H⁺ ion).
- The volume of the solution in liters is:
\[ 500 \, \text{ml} = 0.5 \, \text{L} \]
- Rearranging the normality equation to find the mass:
\[ \text{mass} = N \times \text{equivalent weight} \times \text{volume} \]
- Substitute the values:
\[ \text{mass} = 0.18 \, \text{N} \times 60 \, \text{g/equiv} \times 0.5 \, \text{L} \]
\[ \text{mass} = 0.18 \times 60 \times 0.5 = 5.4 \, \text{g} \]

7. **Final Answer**:
- The mass of acetic acid present in the solution is **5.4 g**.