An acid `HA` ionizes as `HAhArrH^(+)+A^(-)` The `pH` of `1.0 M` solution is `5`. Its dissociation constant would be

Options

A. `1xx10^(-10)`
B. 5
C. `5xx10^(-8)`
D. `1xx10^(-5)`

Explanation

To find the dissociation constant \( K_a \) of the weak acid \( HA \), we can follow these steps:

### Step 1: Understand the Ionization of the Acid
The ionization of the acid \( HA \) can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
This means that for every mole of \( HA \) that dissociates, one mole of \( H^+ \) and one mole of \( A^- \) are produced.

### Step 2: Determine the Initial Concentration and Change
Given that the initial concentration of \( HA \) is \( 1.0 \, M \) and it dissociates to produce \( H^+ \) and \( A^- \):
- Initial concentrations:
- \([HA] = 1.0 \, M\)
- \([H^+] = 0\)
- \([A^-] = 0\)

- Change in concentrations at equilibrium:
- \([HA] = 1 - \alpha\)
- \([H^+] = \alpha\)
- \([A^-] = \alpha\)

### Step 3: Calculate the Concentration of \( H^+ \)
The pH of the solution is given as \( 5 \). We can calculate the concentration of \( H^+ \) ions using the formula:
\[ \text{pH} = -\log[H^+] \]
From this, we can find:
\[ [H^+] = 10^{-5} \, M \]

### Step 4: Relate \( H^+ \) Concentration to Degree of Dissociation
Since the concentration of \( H^+ \) at equilibrium is equal to \( \alpha \):
\[ \alpha = [H^+] = 10^{-5} \]

### Step 5: Write the Expression for the Dissociation Constant \( K_a \)
The dissociation constant \( K_a \) can be expressed as:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
Substituting the equilibrium concentrations:
\[ K_a = \frac{\alpha \cdot \alpha}{1 - \alpha} = \frac{\alpha^2}{1 - \alpha} \]

### Step 6: Simplify the Expression
Since \( \alpha \) is very small compared to \( 1 \) (i.e., \( \alpha \ll 1 \)), we can approximate:
\[ 1 - \alpha \approx 1 \]
Thus, the equation simplifies to:
\[ K_a \approx \alpha^2 \]

### Step 7: Substitute the Value of \( \alpha \)
Now substituting \( \alpha = 10^{-5} \):
\[ K_a \approx (10^{-5})^2 = 10^{-10} \]

### Final Answer
The dissociation constant \( K_a \) for the acid \( HA \) is:
\[ K_a = 1 \times 10^{-10} \]

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