The equilibrium constant for the reaction given is `3.6xx10^(-7)OCl^(-)(aq)+H_2O(l)hArrHOCl(aq)+OH^(-)(aq).`
What is Ka for HOCl ?

Options

A. `2.77xx10^(-8)`
B. `3.6xx10^(-7)`
C. `6xx10^(-4)`
D. `2.8 xx10^(-6)`

Answer

To find the dissociation constant \( K_a \) for hypochlorous acid (HOCl), we can use the relationship between the ion product of water (\( K_w \)), the acid dissociation constant (\( K_a \)), and the base dissociation constant (\( K_b \)).

### Step-by-step Solution:

1. **Identify the given data**:
- The equilibrium constant \( K_b \) for the reaction involving the hypochlorite ion (\( OCl^- \)) is given as:
\[ K_b = 3.6 \times 10^{-7} \]
- The ion product of water \( K_w \) at 25°C is:
\[ K_w = 1.0 \times 10^{-14} \]

2. **Use the relationship between \( K_a \), \( K_b \), and \( K_w \)**:
The relationship is given by:
\[ K_w = K_a \times K_b \]
Rearranging this equation to solve for \( K_a \):
\[ K_a = \frac{K_w}{K_b} \]

3. **Substitute the known values**:
Now, substituting the values of \( K_w \) and \( K_b \) into the equation:
\[ K_a = \frac{1.0 \times 10^{-14}}{3.6 \times 10^{-7}} \]

4. **Perform the calculation**:
\[ K_a = \frac{1.0}{3.6} \times 10^{-14 + 7} = \frac{1.0}{3.6} \times 10^{-7} \]
\[ K_a \approx 0.2778 \times 10^{-7} \approx 2.78 \times 10^{-8} \]

5. **Final result**:
Thus, the value of \( K_a \) for HOCl is approximately:
\[ K_a \approx 2.8 \times 10^{-8} \]