A weak acid, HA, has a `K_(a)` of `1.00xx10^(-5)`. If `0.100` mol of the acid is dissolved in 1 L of water, the percentage of the acid dissociated at equilibrium is the close to

Options

A. `99.0%`
B. `1.00%`
C. `99.9%`
D. `0.100%`

Explanation

To solve the problem of determining the percentage of a weak acid (HA) that dissociates at equilibrium, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given information:**
- The dissociation constant \( K_a \) of the weak acid \( HA \) is \( 1.00 \times 10^{-5} \).
- The initial concentration of the acid \( [HA] \) is \( 0.100 \, \text{mol/L} \).

2. **Set up the dissociation equation:**
- The dissociation of the weak acid can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]

3. **Define the degree of dissociation:**
- Let \( \alpha \) be the degree of dissociation of the acid. At equilibrium, the concentrations will be:
- \( [HA] = 0.1(1 - \alpha) \)
- \( [H^+] = 0.1\alpha \)
- \( [A^-] = 0.1\alpha \)

4. **Write the expression for the acid dissociation constant \( K_a \):**
- The expression for \( K_a \) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
- Substituting the equilibrium concentrations into the expression:
\[ K_a = \frac{(0.1\alpha)(0.1\alpha)}{0.1(1 - \alpha)} \]

5. **Simplify the equation:**
- This simplifies to:
\[ K_a = \frac{0.01\alpha^2}{0.1(1 - \alpha)} \]
- Further simplifying gives:
\[ K_a = \frac{0.1\alpha^2}{1 - \alpha} \]

6. **Assume \( \alpha \) is small:**
- Since \( K_a \) is small, we can assume \( \alpha \) is much less than 1, which allows us to approximate \( 1 - \alpha \approx 1 \):
\[ K_a \approx 0.1\alpha^2 \]

7. **Substitute the value of \( K_a \):**
- Now substituting \( K_a = 1.00 \times 10^{-5} \):
\[ 1.00 \times 10^{-5} = 0.1\alpha^2 \]

8. **Solve for \( \alpha \):**
- Rearranging gives:
\[ \alpha^2 = \frac{1.00 \times 10^{-5}}{0.1} = 1.00 \times 10^{-4} \]
- Taking the square root:
\[ \alpha = \sqrt{1.00 \times 10^{-4}} = 0.01 \]

9. **Calculate the percentage dissociated:**
- The percentage of the acid that has dissociated is given by:
\[ \text{Percentage dissociated} = \alpha \times 100 = 0.01 \times 100 = 1\% \]

### Conclusion:
The percentage of the acid dissociated at equilibrium is approximately **1%**.

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