The weak acid, HA has a Ka of 1×`10^(−1)` . 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. 100%
D. .01%

Explanation

To solve the problem, we need to find the percentage of dissociation of the weak acid HA with a given dissociation constant (Ka) and initial concentration. Here’s the step-by-step solution:

### Step 1: Write the dissociation equation
The weak acid HA dissociates in water as follows:
\[ HA \rightleftharpoons H^+ + A^- \]

### Step 2: Set up the initial concentrations
When 0.1 moles of HA is dissolved in 1 liter of water, the initial concentration of HA is:
\[ [HA]_{initial} = 0.1 \, M \]
At the start, the concentrations of \( H^+ \) and \( A^- \) are both 0.

### Step 3: Define the change in concentration at equilibrium
Let \( \alpha \) be the degree of dissociation. At equilibrium, the concentrations will be:
- \( [HA] = 0.1(1 - \alpha) \)
- \( [H^+] = 0.1\alpha \)
- \( [A^-] = 0.1\alpha \)

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

### Step 5: Substitute the value of Ka
Given that \( K_a = 1 \times 10^{-1} \):
\[ 1 \times 10^{-1} = \frac{(0.1\alpha)(0.1\alpha)}{0.1(1 - \alpha)} \]

### Step 6: Simplify the equation
This simplifies to:
\[ 1 \times 10^{-1} = \frac{0.01\alpha^2}{0.1(1 - \alpha)} \]
\[ 1 \times 10^{-1} = \frac{0.1\alpha^2}{1 - \alpha} \]

### Step 7: Cross-multiply and solve for α
Cross-multiplying gives:
\[ 0.1(1 - \alpha) = 0.1\alpha^2 \]
Dividing both sides by 0.1:
\[ 1 - \alpha = \alpha^2 \]
Rearranging gives:
\[ \alpha^2 + \alpha - 1 = 0 \]

### Step 8: Use the quadratic formula
Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 1, c = -1 \):
\[ \alpha = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ \alpha = \frac{-1 \pm \sqrt{1 + 4}}{2} \]
\[ \alpha = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 9: Calculate the positive root
Taking the positive root:
\[ \alpha = \frac{-1 + \sqrt{5}}{2} \approx 0.618 \]

### Step 10: Calculate the percentage dissociated
The percentage of acid dissociated is:
\[ \text{Percentage dissociated} = \alpha \times 100 \approx 0.618 \times 100 \approx 61.8\% \]

### Final Answer
The percentage of acid dissociated at equilibrium is closest to **62%**.

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