At certain temperature, dissociation constant of formic acid and acetic acid are `1.8xx10^(-4)` and `1.8xx10^(-5)` respectively. At what concentration of acetic solution, the `H_3O^(+)` ion concentration is same as that in 0.001 M formic acid solution

Options

A. 0.01 M
B. 0.001 M
C. 0.1 M
D. 0.0001 M

Explanation

To solve the problem, we need to find the concentration of acetic acid solution (C2) such that the concentration of hydronium ions (H3O+) is the same as that in a 0.001 M formic acid solution.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Dissociation constant of formic acid (Ka1) = \(1.8 \times 10^{-4}\)
- Dissociation constant of acetic acid (Ka2) = \(1.8 \times 10^{-5}\)
- Concentration of formic acid (C1) = 0.001 M

2. **Write the Expression for Hydronium Ion Concentration:**
The concentration of hydronium ions (H3O+) produced from a weak acid can be expressed as:
\[ [H_3O^+] = \sqrt{K_a \cdot C} \]
where \(K_a\) is the dissociation constant and \(C\) is the concentration of the acid.

3. **Set Up the Equation:**
Since the concentration of H3O+ is the same in both solutions, we can equate the expressions for the two acids:
\[ \sqrt{K_{a1} \cdot C_1} = \sqrt{K_{a2} \cdot C_2} \]
Squaring both sides to eliminate the square roots gives:
\[ K_{a1} \cdot C_1 = K_{a2} \cdot C_2 \]

4. **Substitute the Known Values:**
Substitute the values of \(K_{a1}\), \(C_1\), and \(K_{a2}\) into the equation:
\[ (1.8 \times 10^{-4}) \cdot (0.001) = (1.8 \times 10^{-5}) \cdot C_2 \]

5. **Calculate C2:**
Rearranging the equation to solve for \(C_2\):
\[ C_2 = \frac{(1.8 \times 10^{-4}) \cdot (0.001)}{(1.8 \times 10^{-5})} \]
Simplifying the right side:
\[ C_2 = \frac{1.8 \times 10^{-7}}{1.8 \times 10^{-5}} = 0.01 \, M \]

6. **Conclusion:**
The concentration of acetic acid solution (C2) required for the hydronium ion concentration to be the same as that in a 0.001 M formic acid solution is **0.01 M**.