The dissociation constants for acetic acid and HCN at `25^(@)C` are `1.5xx10^(-5)` and `4.5xx10^(-10)` , respectively. The equilibrium constant for the equilibirum `CN^(-) + CH_(3)COOHhArr HCN + CH_(3)COO^(-)` would be

Options

A. `3.0xx10^5`
B. `3.0xx10^-5`
C. `3.0xx10^-4`
D. `3.0xx10^(4)`

Explanation

To find the equilibrium constant for the reaction \( CN^- + CH_3COOH \rightleftharpoons HCN + CH_3COO^- \), we can relate it to the dissociation constants of acetic acid and HCN.

### Step 1: Write the dissociation equations and their constants

1. **Dissociation of Acetic Acid:**
\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]
The dissociation constant \( K_{a1} \) for acetic acid is given by:
\[ K_{a1} = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} = 1.5 \times 10^{-5} \]

2. **Dissociation of HCN:**
\[ HCN \rightleftharpoons CN^- + H^+ \]
The dissociation constant \( K_{a2} \) for HCN is given by:
\[ K_{a2} = \frac{[CN^-][H^+]}{[HCN]} = 4.5 \times 10^{-10} \]

### Step 2: Write the equilibrium constant for the desired reaction

The equilibrium constant \( K \) for the reaction:
\[ CN^- + CH_3COOH \rightleftharpoons HCN + CH_3COO^- \]
can be expressed in terms of \( K_{a1} \) and \( K_{a2} \).

### Step 3: Relate the equilibrium constant to the dissociation constants

Using the relationship between the equilibrium constants, we can express \( K \) as:
\[ K = \frac{K_{a1}}{K_{a2}} \]

### Step 4: Substitute the known values

Substituting the known values of \( K_{a1} \) and \( K_{a2} \):
\[ K = \frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}} \]

### Step 5: Calculate the value of \( K \)

Now, perform the calculation:
\[ K = \frac{1.5}{4.5} \times 10^{(-5) - (-10)} = \frac{1.5}{4.5} \times 10^{5} = \frac{1}{3} \times 10^{5} \]
\[ K = 0.333 \times 10^{5} = 3.33 \times 10^{4} \]

### Final Answer
The equilibrium constant \( K \) for the reaction \( CN^- + CH_3COOH \rightleftharpoons HCN + CH_3COO^- \) is approximately \( 3.33 \times 10^{4} \).

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