The toxic compound 2,4-dinitrophenol has `K_(a)=10^(-4).` In an experiment, a buffer solution of 2,4-dinitrophenol was prepared with the pH adjusted to 5. Calculate the ratio of the concentrations of the dissociated ion to the undissociated acid:

Options

A. `0.01`
B. `0.1`
C. 10
D. 100

Answer

To solve the problem, we need to find the ratio of the concentrations of the dissociated ion (salt) to the undissociated acid for 2,4-dinitrophenol, given its dissociation constant \( K_a \) and the pH of the solution.

### Step-by-Step Solution:

1. **Identify Given Values**:
- \( K_a = 10^{-4} \)
- \( \text{pH} = 5 \)

2. **Calculate \( pK_a \)**:
- The relationship between \( K_a \) and \( pK_a \) is given by:
\[ pK_a = -\log(K_a) \]
- Substituting the value of \( K_a \):
\[ pK_a = -\log(10^{-4}) = 4 \]

3. **Use the Henderson-Hasselbalch Equation**:
- The Henderson-Hasselbalch equation relates pH, \( pK_a \), and the ratio of the concentrations of the salt (dissociated ion) and the acid (undissociated):
\[ \text{pH} = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
- Plugging in the known values:
\[ 5 = 4 + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

4. **Rearranging the Equation**:
- To isolate the logarithmic term, subtract \( pK_a \) from both sides:
\[ 5 - 4 = \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
\[ 1 = \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

5. **Exponentiating Both Sides**:
- To eliminate the logarithm, we exponentiate both sides:
\[ 10^1 = \frac{[\text{Salt}]}{[\text{Acid}]} \]
\[ \frac{[\text{Salt}]}{[\text{Acid}]} = 10 \]

6. **Conclusion**:
- The ratio of the concentrations of the dissociated ion (salt) to the undissociated acid is:
\[ \frac{[\text{Salt}]}{[\text{Acid}]} = 10 \]

### Final Answer:
The ratio of the concentrations of the dissociated ion to the undissociated acid is **10**.

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