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 will follow these steps:

### Step 1: Write the dissociation equation
The dissociation of the acid \( HA \) can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]

### Step 2: Define the initial concentration and change in concentration
Let the initial concentration of the acid \( [HA] \) be \( C = 10 \, \text{M} \). At equilibrium, if \( \alpha \) is the degree of dissociation, the concentrations will be:
- \( [HA] = C(1 - \alpha) = 10(1 - \alpha) \)
- \( [H^+] = [A^-] = C\alpha = 10\alpha \)

### Step 3: Write the expression for the dissociation constant
The expression for the dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
Substituting the equilibrium concentrations, we get:
\[ K_a = \frac{(10\alpha)(10\alpha)}{10(1 - \alpha)} \]

### Step 4: Simplify the equation
This simplifies to:
\[ K_a = \frac{100\alpha^2}{10(1 - \alpha)} \]
\[ K_a = \frac{10\alpha^2}{1 - \alpha} \]

### Step 5: Substitute the known values
Substituting \( K_a = 1.0 \times 10^{-3} \):
\[ 1.0 \times 10^{-3} = \frac{10\alpha^2}{1 - \alpha} \]

### Step 6: Assume \( \alpha \) is small
Assuming \( \alpha \) is small (which is valid for strong concentrations), we can approximate \( 1 - \alpha \approx 1 \):
\[ 1.0 \times 10^{-3} = 10\alpha^2 \]
\[ \alpha^2 = \frac{1.0 \times 10^{-3}}{10} \]
\[ \alpha^2 = 1.0 \times 10^{-4} \]

### Step 7: Solve for \( \alpha \)
Taking the square root of both sides:
\[ \alpha = \sqrt{1.0 \times 10^{-4}} = 0.01 \]

### Step 8: Calculate the percentage dissociation
To find the percentage dissociation, we multiply \( \alpha \) by 100:
\[ \text{Percentage dissociation} = \alpha \times 100 = 0.01 \times 100 = 1\% \]

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

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