A monoprotic acid in `0.1 M` solution has `K_(a)=1.0xx10^(-5)`. The degree of dissociation for acid is

Options

A. `1.0%`
B. `99.9%`
C. `0.1%`
D. `99%`

Explanation

To find the degree of dissociation (α) of a monoprotic acid in a 0.1 M solution with a dissociation constant (K_a) of 1.0 x 10^(-5), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Dissociation of the Acid:**
A monoprotic acid (HA) dissociates in water as follows:
\[ HA \rightleftharpoons H^+ + A^- \]
Initially, we have a concentration of 0.1 M for the acid.

2. **Setting Up the Initial Concentrations:**
Let the initial concentration of the acid (HA) be \( C = 0.1 \, \text{M} \).
At the start (before dissociation):
- \([HA] = C = 0.1 \, \text{M}\)
- \([H^+] = 0\)
- \([A^-] = 0\)

3. **Defining the Change in Concentrations:**
If \( \alpha \) is the degree of dissociation, then at equilibrium:
- \([HA] = C(1 - \alpha)\)
- \([H^+] = C\alpha\)
- \([A^-] = C\alpha\)

4. **Writing the Expression for \( K_a \):**
The expression for the dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
Substituting the equilibrium concentrations into this expression:
\[ K_a = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} = \frac{C^2\alpha^2}{C(1 - \alpha)} \]
Simplifying this gives:
\[ K_a = \frac{C\alpha^2}{1 - \alpha} \]

5. **Substituting Known Values:**
We know \( K_a = 1.0 \times 10^{-5} \) and \( C = 0.1 \):
\[ 1.0 \times 10^{-5} = \frac{0.1 \alpha^2}{1 - \alpha} \]

6. **Assuming \( \alpha \) is Small:**
Since \( K_a \) is small, we can assume that \( \alpha \) is much smaller than 1, which allows us to approximate \( 1 - \alpha \approx 1 \):
\[ 1.0 \times 10^{-5} \approx 0.1 \alpha^2 \]

7. **Solving for \( \alpha^2 \):**
Rearranging gives:
\[ \alpha^2 = \frac{1.0 \times 10^{-5}}{0.1} = 1.0 \times 10^{-4} \]

8. **Finding \( \alpha \):**
Taking the square root:
\[ \alpha = \sqrt{1.0 \times 10^{-4}} = 1.0 \times 10^{-2} \]

9. **Converting to Percentage:**
To express the degree of dissociation as a percentage:
\[ \alpha \text{ (in percent)} = \alpha \times 100 = (1.0 \times 10^{-2}) \times 100 = 1\% \]

### Final Answer:
The degree of dissociation of the monoprotic acid in a 0.1 M solution is **1%**.