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. `6.63 xx 10^(-10)`
B. ` 6.63 xx 10^(10)`
C. `1.8 xx 10^(9)`
D. `1.8 xx 10^(-9)`

Answer

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 that 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 \( K_1 = 1.8 \times 10^{-5} \), we can follow these steps:

### Step 1: Write the Equilibrium Constant Expression for Both Reactions

For the first reaction, the equilibrium constant \( K_1 \) can be expressed as:

\[ K_1 = \frac{[\text{CH}_3\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{COOH}]} \]

For the second reaction, the equilibrium constant \( K_2 \) can be expressed as:

\[ K_2 = \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}][\text{OH}^-]} \]

### Step 2: Relate the Two Equilibrium Constants

We can relate \( K_1 \) and \( K_2 \) by manipulating the equations. We know that:

\[ K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 10^{-14} \]

Now, if we divide \( K_1 \) by \( K_2 \):

\[ \frac{K_1}{K_2} = \frac{[\text{CH}_3\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{COOH}]} \div \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}][\text{OH}^-]} \]

This simplifies to:

\[ \frac{K_1}{K_2} = \frac{[\text{H}_3\text{O}^+]}{[\text{OH}^-]} \]

### Step 3: Substitute \( K_w \)

From the relationship \( K_w = [\text{H}_3\text{O}^+][\text{OH}^-] \), we can express \( [\text{H}_3\text{O}^+] \) in terms of \( K_w \) and \( [\text{OH}^-] \):

\[ [\text{H}_3\text{O}^+] = \frac{K_w}{[\text{OH}^-]} \]

Substituting this into our equation gives:

\[ \frac{K_1}{K_2} = \frac{K_w}{[\text{OH}^-]^2} \]

### Step 4: Solve for \( K_2 \)

Rearranging gives:

\[ K_2 = \frac{K_1 \cdot [\text{OH}^-]^2}{K_w} \]

Since we are looking for \( K_2 \) and we know \( K_w = 10^{-14} \) and \( K_1 = 1.8 \times 10^{-5} \), we can substitute the values:

\[ K_2 = \frac{1.8 \times 10^{-5} \cdot [\text{OH}^-]^2}{10^{-14}} \]

### Step 5: Calculate \( K_2 \)

Assuming \( [\text{OH}^-] \) is at standard conditions (which is typically \( 1 \times 10^{-7} \) M for pure water), we can calculate:

\[ K_2 = \frac{1.8 \times 10^{-5} \cdot (1 \times 10^{-7})^2}{10^{-14}} \]
\[ K_2 = \frac{1.8 \times 10^{-5} \cdot 1 \times 10^{-14}}{10^{-14}} = 1.8 \times 10^{9} \]

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:

\[ K_2 = 1.8 \times 10^{9} \]

### Final Answer

The equilibrium constant \( K_2 \) is \( 1.8 \times 10^{9} \).

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