Find out the percentage dissociation of an acid having conc. Of 10 M and dissociation constant `1.0xx10^(-3)`.

Options

A. 0.1
B. `0.5`
C. `1.0`
D. `2.0`

Answer

To find the percentage dissociation of an acid with a concentration of 10 M and a dissociation constant \( K_a = 1.0 \times 10^{-3} \), we can follow these steps:

### Step 1: Set up the equation for dissociation
Let the acid dissociate as follows:
\[ HA \rightleftharpoons H^+ + A^- \]

Let the initial concentration of the acid \( [HA] = C = 10 \, \text{M} \). If \( x \) is the amount that dissociates, then at equilibrium:
- Concentration of \( HA = C - x = 10 - x \)
- Concentration of \( H^+ = x \)
- Concentration of \( A^- = x \)

### Step 2: Write the expression for the dissociation constant
The dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{x \cdot x}{10 - x} = \frac{x^2}{10 - x} \]

### Step 3: Substitute the known value of \( K_a \)
We know that \( K_a = 1.0 \times 10^{-3} \). Therefore, we can write:
\[ 1.0 \times 10^{-3} = \frac{x^2}{10 - x} \]

### Step 4: Assume \( x \) is small compared to \( C \)
Since \( K_a \) is small, we can assume that \( x \) is small compared to 10 M, so \( 10 - x \approx 10 \). This simplifies our equation to:
\[ 1.0 \times 10^{-3} = \frac{x^2}{10} \]

### Step 5: Solve for \( x \)
Rearranging gives:
\[ x^2 = 1.0 \times 10^{-3} \times 10 = 1.0 \times 10^{-2} \]
\[ x = \sqrt{1.0 \times 10^{-2}} = 0.1 \, \text{M} \]

### Step 6: Calculate the percentage dissociation
The percentage dissociation \( \alpha \) is given by:
\[ \alpha = \frac{x}{C} \times 100 = \frac{0.1}{10} \times 100 = 1\% \]

### Final Answer
The percentage dissociation of the acid is **1%**.

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