Given `HF + H_2O hArr H_3O^(+) + F^(-) : K_a`
` " " F^(-) +H_2O hArr HF +OH^(-) ,K_b `
Which of the following relations is correct ?

Options

A. `K_b=K_w `
B. ` K_a= (1)/(K_w)`
C. ` K_a xx K_b= K_w`
D. ` K_a//K_b= K_w`

Answer

To solve the problem, we need to analyze the two given equilibrium reactions and their corresponding equilibrium constants \( K_a \) and \( K_b \).

### Step 1: Write the Equilibrium Reactions
1. The first equilibrium reaction is:
\[ HF + H_2O \rightleftharpoons H_3O^+ + F^- \]
The equilibrium constant for this reaction is denoted as \( K_a \).

2. The second equilibrium reaction is:
\[ F^- + H_2O \rightleftharpoons HF + OH^- \]
The equilibrium constant for this reaction is denoted as \( K_b \).

### Step 2: Write the Expressions for \( K_a \) and \( K_b \)
- For the first reaction, the expression for \( K_a \) is:
\[ K_a = \frac{[H_3O^+][F^-]}{[HF]} \]

- For the second reaction, the expression for \( K_b \) is:
\[ K_b = \frac{[HF][OH^-]}{[F^-]} \]

### Step 3: Combine the Two Reactions
If we add the two reactions together, we can derive a relationship involving \( K_a \) and \( K_b \).

- Adding the two reactions:
\[ (HF + H_2O \rightleftharpoons H_3O^+ + F^-) + (F^- + H_2O \rightleftharpoons HF + OH^-) \]
results in:
\[ 2H_2O \rightleftharpoons H_3O^+ + OH^- \]

### Step 4: Write the Expression for \( K_w \)
The equilibrium constant for the ionization of water, \( K_w \), is given by:
\[ K_w = [H_3O^+][OH^-] \]

### Step 5: Relate \( K_a \), \( K_b \), and \( K_w \)
From the combined reactions, we can express \( K_w \) in terms of \( K_a \) and \( K_b \):
- Multiplying the expressions for \( K_a \) and \( K_b \):
\[ K_a \cdot K_b = \left( \frac{[H_3O^+][F^-]}{[HF]} \right) \cdot \left( \frac{[HF][OH^-]}{[F^-]} \right) \]
Simplifying this gives:
\[ K_a \cdot K_b = [H_3O^+][OH^-] \]
Thus, we have:
\[ K_a \cdot K_b = K_w \]

### Conclusion
The correct relation among the options provided is:
\[ K_a \cdot K_b = K_w \]
Thus, the answer is option C.

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