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 acid dissociation constant (Ka) for HOCl from the given equilibrium constant (K) for the reaction:

\[ \text{OCl}^- (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{HOCl} (aq) + \text{OH}^- (aq) \]

we can use the relationship between the ionization constants of water (Kw), the acid dissociation constant (Ka), and the base dissociation constant (Kb).

### Step-by-Step Solution:

1. **Identify the given values:**
- The equilibrium constant for the reaction (K) is given as \( K = 3.6 \times 10^{-7} \).
- The ion product of water (Kw) at 25°C is \( K_w = 1.0 \times 10^{-14} \).

2. **Use the relationship between Kw, Ka, and Kb:**
\[ K_w = K_a \times K_b \]
Here, Kb is the base dissociation constant for OCl⁻, which can be derived from the given equilibrium constant (K).

3. **Rearranging the equation to find Ka:**
\[ K_a = \frac{K_w}{K_b} \]

4. **Substituting the known values:**
Since we need to find Kb from the given equilibrium constant K:
\[ K_b = K = 3.6 \times 10^{-7} \]
Now substituting into the equation for Ka:
\[ K_a = \frac{1.0 \times 10^{-14}}{3.6 \times 10^{-7}} \]

5. **Calculating Ka:**
\[ K_a = \frac{1.0 \times 10^{-14}}{3.6 \times 10^{-7}} \approx 2.777 \times 10^{-8} \]

6. **Final Answer:**
\[ K_a \approx 2.777 \times 10^{-8} \]