The weak acid, HA has a Ka of 1×`10^(−3)` . If 0.1 mol of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closet to:

Options

A. 1%
B. 0.1%
C. 10%
D. 100%

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the initial concentration of the acid
Given that 0.1 moles of the weak acid HA is dissolved in 1 liter of water, the initial concentration of HA is:
\[ \text{Concentration of HA} = \frac{0.1 \text{ moles}}{1 \text{ liter}} = 0.1 \, M \]

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

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

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

### Step 5: Simplify the equation
This simplifies to:
\[ K_a = \frac{0.01\alpha^2}{0.1(1 - \alpha)} = \frac{0.1\alpha^2}{1 - \alpha} \]

### Step 6: Substitute the value of \(K_a\)
We know that \(K_a = 1 \times 10^{-3}\). Thus, we can set up the equation:
\[ 1 \times 10^{-3} = \frac{0.1\alpha^2}{1 - \alpha} \]

### Step 7: Assume \(\alpha\) is small
Since \(K_a\) is small, we can assume that \(\alpha\) is much smaller than 1, which allows us to simplify \(1 - \alpha \approx 1\):
\[ 1 \times 10^{-3} = 0.1\alpha^2 \]

### Step 8: Solve for \(\alpha^2\)
Rearranging gives:
\[ \alpha^2 = \frac{1 \times 10^{-3}}{0.1} = 1 \times 10^{-2} \]

### Step 9: Solve for \(\alpha\)
Taking the square root:
\[ \alpha = \sqrt{1 \times 10^{-2}} = 0.1 \]

### Step 10: Calculate the percentage of dissociation
The percentage of acid dissociated is given by:
\[ \text{Percentage of dissociation} = \alpha \times 100 = 0.1 \times 100 = 10\% \]

### Final Answer
The percentage of acid dissociated at equilibrium is closest to **10%**.

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