The equilibrium constant for
`CN^(-) + CH_(3)COOH hArr HCN + CH_(3)COO^(-)` is: (Given `pK_(b)` for `CN^(-)` = 4.69 and `pK_(a)` for `CH_(3)COOH = 4.75`)

Options

A. `3.7 xx 10^(4)`
B. `2.8 xx 10^(-5)`
C. `1.97 xx 10^(4)`
D. `0.5 xx 10^(-5)`

Answer

To find the equilibrium constant for the reaction:

\[ \text{CN}^- + \text{CH}_3\text{COOH} \rightleftharpoons \text{HCN} + \text{CH}_3\text{COO}^- \]

we will use the provided \( pK_b \) and \( pK_a \) values.

### Step 1: Determine \( K_a \) for acetic acid (\( \text{CH}_3\text{COOH} \))

Given:
- \( pK_a \) for \( \text{CH}_3\text{COOH} = 4.75 \)

Using the relationship:
\[ K_a = 10^{-pK_a} \]

Calculating:
\[ K_a = 10^{-4.75} \approx 1.78 \times 10^{-5} \]

### Step 2: Determine \( K_a \) for hydrocyanic acid (\( \text{HCN} \))

Given:
- \( pK_b \) for \( \text{CN}^- = 4.69 \)

Using the relationship:
\[ pK_a + pK_b = 14 \]
Thus,
\[ pK_a = 14 - pK_b = 14 - 4.69 = 9.31 \]

Now, calculate \( K_a \):
\[ K_a = 10^{-pK_a} = 10^{-9.31} \approx 4.89 \times 10^{-10} \]

### Step 3: Write the expression for the equilibrium constant \( K_{eq} \)

For the reaction:
\[ \text{CN}^- + \text{CH}_3\text{COOH} \rightleftharpoons \text{HCN} + \text{CH}_3\text{COO}^- \]

The equilibrium constant \( K_{eq} \) can be expressed as:
\[ K_{eq} = \frac{[\text{HCN}][\text{CH}_3\text{COO}^-]}{[\text{CN}^-][\text{CH}_3\text{COOH}]} \]

### Step 4: Relate \( K_{eq} \) to \( K_a \) values

From the dissociation of acetic acid and hydrocyanic acid, we can relate:
\[ K_{eq} = \frac{K_{a(\text{CH}_3\text{COOH})}}{K_{a(\text{HCN})}} \]

Substituting the values we calculated:
\[ K_{eq} = \frac{1.78 \times 10^{-5}}{4.89 \times 10^{-10}} \]

### Step 5: Calculate \( K_{eq} \)

Perform the calculation:
\[ K_{eq} \approx \frac{1.78 \times 10^{-5}}{4.89 \times 10^{-10}} \approx 3.63 \times 10^{4} \]

### Conclusion

Thus, the equilibrium constant \( K_{eq} \) for the reaction is approximately:
\[ K_{eq} \approx 3.63 \times 10^{4} \]