The equivalent conductance of `0.001` M acetic acid is `50 " ohm"^(-1)cm^(2)eq^(-1)`. The maximum value of equivalent conductance is 250 `"ohm"^(-1) cm^(2) eq^(-1)`.What is its degree of ionization?

Options

A. `0.2%`
B. `2%`
C. `20%`
D. `40%`

Answer

To find the degree of ionization of acetic acid, we can use the formula for degree of ionization (α):

\[ \alpha = \frac{\Lambda_m}{\Lambda_m^0} \]

where:
- \(\Lambda_m\) is the equivalent conductance at the given concentration,
- \(\Lambda_m^0\) is the maximum equivalent conductance.

**Step-by-step solution:**

1. **Identify the given values:**
- Equivalent conductance (\(\Lambda_m\)) = 50 ohm\(^{-1}\)cm\(^2\)eq\(^{-1}\)
- Maximum equivalent conductance (\(\Lambda_m^0\)) = 250 ohm\(^{-1}\)cm\(^2\)eq\(^{-1}\)

2. **Substitute the values into the formula:**
\[ \alpha = \frac{50}{250} \]

3. **Calculate the degree of ionization:**
\[ \alpha = \frac{50}{250} = \frac{1}{5} = 0.2 \]

4. **Convert the degree of ionization to percentage:**
\[ \text{Percentage of ionization} = \alpha \times 100 = 0.2 \times 100 = 20\% \]

5. **Final answer:**
The degree of ionization of acetic acid is 20%.

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