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. `3.3xx10^(-5)`
B. `3.3xx10^(-4)`
C. `3.3xx10^(4)`
D. `3.3xx10^(5)`

Answer

To find the equilibrium constant for the reaction \( CN^- + CH_3COOH \rightleftharpoons HCN + CH_3COO^- \), we will use the dissociation constants of acetic acid and HCN provided in the question.

### Step-by-step Solution:

1. **Write the Dissociation Reactions**:
- For acetic acid (\( CH_3COOH \)):
\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \quad (K_a1 = 1.5 \times 10^{-5}) \]
- For hydrocyanic acid (HCN):
\[ HCN \rightleftharpoons H^+ + CN^- \quad (K_a2 = 4.5 \times 10^{-10}) \]

2. **Write the Equilibrium Expression for the Given Reaction**:
The equilibrium expression for the reaction we are interested in is:
\[ K = \frac{[HCN][CH_3COO^-]}{[CN^-][CH_3COOH]} \]

3. **Relate the Given Reactions to the Desired Reaction**:
To relate the dissociation constants to the equilibrium constant of the desired reaction, we can manipulate the reactions:
- The first reaction (acetic acid dissociation) is already in the correct form.
- The second reaction (HCN dissociation) needs to be reversed:
\[ H^+ + CN^- \rightleftharpoons HCN \quad (K = \frac{1}{K_a2} = \frac{1}{4.5 \times 10^{-10}}) \]

4. **Combine the Equilibrium Constants**:
The equilibrium constant for the desired reaction can be expressed as:
\[ K = K_a1 \times \frac{1}{K_a2} \]
Substituting the values:
\[ K = \frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}} \]

5. **Calculate the Value of K**:
\[ K = \frac{1.5}{4.5} \times 10^{(-5) - (-10)} = 0.333 \times 10^{5} = 3.33 \times 10^{4} \]

6. **Final Result**:
Therefore, the equilibrium constant for the reaction \( CN^- + CH_3COOH \rightleftharpoons HCN + CH_3COO^- \) is:
\[ K = 3.3 \times 10^{4} \]