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 acid \( HA \), we can follow these steps:

### Step 1: Understand the given information
We know that the acid \( HA \) dissociates in water as follows:
\[ HA \rightleftharpoons H^+ + A^- \]
The pH of a 1.0 M solution of this acid is given as 5.

### Step 2: Calculate the concentration of \( H^+ \)
The pH is related to the concentration of hydrogen ions \( [H^+] \) by the formula:
\[ \text{pH} = -\log[H^+] \]
Given that the pH is 5, we can calculate \( [H^+] \):
\[ [H^+] = 10^{-\text{pH}} = 10^{-5} \, \text{M} \]

### Step 3: Determine the concentration of \( A^- \)
Since the dissociation of \( HA \) produces equal amounts of \( H^+ \) and \( A^- \), we have:
\[ [A^-] = [H^+] = 10^{-5} \, \text{M} \]

### Step 4: Calculate the concentration of undissociated \( HA \)
Initially, we had a concentration of \( HA \) as 1.0 M. The amount that dissociates is equal to the concentration of \( H^+ \) produced, which is \( 10^{-5} \, \text{M} \). Therefore, the concentration of undissociated \( HA \) at equilibrium is:
\[ [HA] = 1.0 - [H^+] = 1.0 - 10^{-5} \approx 1.0 \, \text{M} \]
(Note: The change is negligible due to the small value of \( 10^{-5} \) compared to 1.0 M.)

### Step 5: Write the expression for the dissociation constant \( K_a \)
The dissociation constant \( K_a \) is given by the formula:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
Substituting the values we have:
\[ K_a = \frac{(10^{-5})(10^{-5})}{1.0} = \frac{10^{-10}}{1.0} = 10^{-10} \]

### Step 6: Final answer
Thus, the dissociation constant \( K_a \) for the acid \( HA \) is:
\[ K_a = 1.0 \times 10^{-10} \]