If equilibrium constant of `CH_3COOH + H_2O Leftrightarrow CH_3COO^(-) + H_3O+ is 1.8 xx 10^(-5)`, equilibrium constant for
`CH_(3)COOH+OH^(-) Leftrightarrow CH_(3)COO^(-)+H_(2)O` is

Options

A. `1.8 xx 10^(-9)`
B. `1.8 xx 10^(9)`
C. `5.55 xx 10^(-9)`
D. `5.55 xx 10^(10`)

Answer

To solve the problem, we need to find the equilibrium constant for the reaction:

\[ \text{CH}_3\text{COOH} + \text{OH}^- \leftrightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \]

Given the equilibrium constant for the reaction:

\[ \text{CH}_3\text{COOH} + \text{H}_2\text{O} \leftrightarrow \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+ \]

is \( K_1 = 1.8 \times 10^{-5} \).

### Step-by-Step Solution:

1. **Write the expression for \( K_1 \)**:
The equilibrium constant \( K_1 \) for the first reaction can be expressed as:
\[ K_1 = \frac{[\text{CH}_3\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{COOH}][\text{H}_2\text{O}]} \]
Since the concentration of pure liquids (like water) does not appear in the expression, we can simplify it to:
\[ K_1 = \frac{[\text{CH}_3\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{COOH}]} \]

2. **Write the expression for \( K_2 \)**:
The equilibrium constant \( K_2 \) for the second reaction can be expressed as:
\[ K_2 = \frac{[\text{CH}_3\text{COO}^-][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{OH}^-]} \]

3. **Relate \( K_1 \) and \( K_2 \)**:
We can relate the two equilibrium constants using the ion product of water \( K_w \):
\[ K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1.0 \times 10^{-14} \text{ at } 25^\circ C \]
From the reactions, we can write:
\[ K_1 \cdot K_2 = K_w \]

4. **Substitute the known values**:
We know \( K_1 = 1.8 \times 10^{-5} \) and \( K_w = 1.0 \times 10^{-14} \). Thus:
\[ K_2 = \frac{K_w}{K_1} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \]

5. **Calculate \( K_2 \)**:
Performing the calculation:
\[ K_2 = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10} \]

6. **Final Calculation**:
To express it in scientific notation:
\[ K_2 = \frac{1.0}{1.8} \times 10^{-14 + 5} = 0.555 \times 10^{-9} \approx 5.56 \times 10^{-10} \]

Thus, the equilibrium constant for the reaction \( \text{CH}_3\text{COOH} + \text{OH}^- \leftrightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \) is approximately \( 5.56 \times 10^{-10} \).