For the equilibrium `:`
`I: C_(6)H_(5)COOH+H_(2)OhArr C_(6)H_(5)COO^(-)+H_(3)O^(+),K_(1)=6.30xx10^(-5)`
`II: C_(6)H_(5)COOH+OH^(-)hArrC_(6)H_(5)COO^(-)+H_(2)O, K_(2)=6.30xx10^(9)`
`III: H_(2)O+H_(2)OhArr H_(3)O^(+)+OH^(-), K_(3)=?`
`K_(3)` using above equilibria is` :`

Options

A. `(6.30)^(2)xx10^(4)`
B. `1.0xx10^(-14)`
C. `1xx10^(14)`
D. `(6.30)^(-2)xx10^(-4)`

Explanation

To find the equilibrium constant \( K_3 \) for the reaction:

\[ \text{H}_2\text{O} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^- \]

we will use the first two equilibria provided.

### Step-by-Step Solution:

1. **Identify the Reactions and Their Equilibrium Constants**:
- Reaction I:
\[ \text{C}_6\text{H}_5\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{C}_6\text{H}_5\text{COO}^- + \text{H}_3\text{O}^+ \quad K_1 = 6.30 \times 10^{-5} \]
- Reaction II:
\[ \text{C}_6\text{H}_5\text{COOH} + \text{OH}^- \rightleftharpoons \text{C}_6\text{H}_5\text{COO}^- + \text{H}_2\text{O} \quad K_2 = 6.30 \times 10^{9} \]

2. **Rearranging Reaction II**:
- To relate it to the third reaction, we need to reverse Reaction II:
\[ \text{C}_6\text{H}_5\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{C}_6\text{H}_5\text{COOH} + \text{OH}^- \]
- When we reverse a reaction, the equilibrium constant becomes the reciprocal:
\[ K_{II}' = \frac{1}{K_2} = \frac{1}{6.30 \times 10^{9}} \]

3. **Adding the Reactions**:
- Now we will add Reaction I and the reversed Reaction II:
- From Reaction I:
\[ \text{C}_6\text{H}_5\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{C}_6\text{H}_5\text{COO}^- + \text{H}_3\text{O}^+ \]
- From the reversed Reaction II:
\[ \text{C}_6\text{H}_5\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{C}_6\text{H}_5\text{COOH} + \text{OH}^- \]

- When we add these two reactions, the terms \( \text{C}_6\text{H}_5\text{COOH} \) and \( \text{C}_6\text{H}_5\text{COO}^- \) will cancel out:
\[ \text{H}_2\text{O} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^- \]

4. **Finding the Equilibrium Constant \( K_3 \)**:
- The equilibrium constant for the overall reaction is the product of the equilibrium constants of the individual reactions:
\[ K_3 = K_1 \times K_{II}' \]
- Substituting the values:
\[ K_3 = K_1 \times \frac{1}{K_2} = (6.30 \times 10^{-5}) \times \left(\frac{1}{6.30 \times 10^{9}}\right) \]
- Simplifying:
\[ K_3 = \frac{6.30 \times 10^{-5}}{6.30 \times 10^{9}} = 10^{-14} \]

### Final Answer:
Thus, the value of \( K_3 \) is:
\[ K_3 = 1.0 \times 10^{-14} \]

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