If degree of dissociation is `0.01` of decimolar solution of weak acid HA then `pK_(a)` of acid is :

Options

A. 2
B. 3
C. 5
D. 7

Answer

To solve the problem of finding the \( pK_a \) of the weak acid \( HA \) given its degree of dissociation and concentration, we can follow these steps:

### Step 1: Understand the given data
We are given:
- Degree of dissociation (\( \alpha \)) = 0.01
- Concentration of the weak acid (\( C \)) = 0.1 M (decimolar solution)

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

### Step 3: Set up the equilibrium concentrations
At equilibrium, if \( \alpha \) is the degree of dissociation, the concentrations will be:
- Concentration of \( H^+ \) = \( C \cdot \alpha \)
- Concentration of \( A^- \) = \( C \cdot \alpha \)
- Concentration of undissociated \( HA \) = \( C(1 - \alpha) \)

### Step 4: Substitute the values
Given \( C = 0.1 \) M and \( \alpha = 0.01 \):
- Concentration of \( H^+ \) = \( 0.1 \cdot 0.01 = 0.001 \) M
- Concentration of \( A^- \) = \( 0.1 \cdot 0.01 = 0.001 \) M
- Concentration of \( HA \) = \( 0.1(1 - 0.01) = 0.1 \cdot 0.99 = 0.099 \) M

### Step 5: Write the expression for the dissociation constant \( K_a \)
The expression for the dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

Substituting the equilibrium concentrations:
\[ K_a = \frac{(0.001)(0.001)}{0.099} \]

### Step 6: Calculate \( K_a \)
Calculating \( K_a \):
\[ K_a = \frac{0.000001}{0.099} \approx 0.0000101 \]
This can be approximated as:
\[ K_a \approx 1.01 \times 10^{-5} \]

### Step 7: Calculate \( pK_a \)
The \( pK_a \) is calculated using the formula:
\[ pK_a = -\log K_a \]
Substituting the value of \( K_a \):
\[ pK_a = -\log(1.01 \times 10^{-5}) \]

Using logarithmic properties:
\[ pK_a \approx -(-5) = 5 \]

### Final Answer
Thus, the \( pK_a \) of the weak acid \( HA \) is:
\[ \boxed{5} \]

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