`HCOOH` and `CH_(3)COOH` solutions have equal `pH.` If `K_(1)//K_(2)` is `4`, the ratio of their molar concentration will be

Options

A. `0.25`
B. `0.5`
C. `2`
D. `4`

Explanation

To solve the problem, we need to find the ratio of the molar concentrations of formic acid (HCOOH) and acetic acid (CH₃COOH) given that their pH values are equal and the ratio of their dissociation constants \( K_1/K_2 \) is 4.

### Step-by-Step Solution:

1. **Understanding pH Equality**:
Since the pH of both solutions is the same, the concentration of hydrogen ions \([H^+]\) from both acids must be equal. Therefore, we can write:
\[ [H^+]_{HCOOH} = [H^+]_{CH_3COOH} \]

2. **Using the Degree of Ionization**:
Let \( C_1 \) be the concentration of HCOOH and \( C_2 \) be the concentration of CH₃COOH. The degree of ionization (α) for each acid can be expressed as:
\[ [H^+]_{HCOOH} = C_1 \alpha_1 \quad \text{and} \quad [H^+]_{CH_3COOH} = C_2 \alpha_2 \]
Setting these equal gives:
\[ C_1 \alpha_1 = C_2 \alpha_2 \]

3. **Dissociation Constants**:
The dissociation constants for the acids are given by:
\[ K_1 = \frac{[H^+][HCOO^-]}{[HCOOH]} = \frac{C_1 \alpha_1^2}{C_1(1-\alpha_1)} \approx \frac{C_1 \alpha_1^2}{C_1} = K_1 \alpha_1^2 \quad \text{(for weak acids, assuming } \alpha \text{ is small)} \]
\[ K_2 = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} = \frac{C_2 \alpha_2^2}{C_2(1-\alpha_2)} \approx K_2 \alpha_2^2 \]

4. **Relating the Dissociation Constants**:
From the problem, we know that:
\[ \frac{K_1}{K_2} = 4 \]

5. **Finding the Ratio of Concentrations**:
We can express the ratio of the degrees of ionization:
\[ \frac{\alpha_1}{\alpha_2} = \sqrt{\frac{K_1}{K_2}} \cdot \frac{C_2}{C_1} \]
Substituting \( K_1/K_2 = 4 \):
\[ \frac{\alpha_1}{\alpha_2} = 2 \cdot \frac{C_2}{C_1} \]

6. **Substituting Back to Find Concentration Ratio**:
From \( C_1 \alpha_1 = C_2 \alpha_2 \), we can substitute for \(\alpha_1\):
\[ C_1 (2 \cdot \frac{C_2}{C_1}) = C_2 \alpha_2 \]
Rearranging gives:
\[ 2 C_2 = C_2 \cdot \frac{C_2}{C_1} \implies 2 = \frac{C_2}{C_1} \]
Therefore, we find:
\[ \frac{C_1}{C_2} = \frac{1}{2} \]

7. **Final Ratio**:
Since \( \frac{K_1}{K_2} = 4 \), we can also express this as:
\[ \frac{C_2}{C_1} = 4 \implies \frac{C_1}{C_2} = \frac{1}{4} \]

### Conclusion:
The ratio of the molar concentrations of HCOOH to CH₃COOH is:
\[ \frac{C_1}{C_2} = \frac{1}{4} \]