Acetic acid tends to form dimer due to formation of intermolcular hydrogen bonding.
`2CH_(2)COOHhArr(CH_(3)COOH)_(2)`
The equilibrium constant for this reaction is `1.5xx10^(2)M^(-1)`in benzene solution and `3.6xx10^(-2)` in water. In benzene, monomer does not dissociate but it water, monomer dissociation simultaneously with acid dissociation constant `2.0xx10^(-5)`M. Dimer does not dissociate in benzene as well as water.
The molar ratio of dimer to monmer for 0.1M acetic acid in water (neglecting the dissciation of acetic acid in water ) is equal to :

Options

A. `250:1`
B. `1:250`
C. `9:2500`
D. `2500:9`

Answer

To solve the problem, we need to find the molar ratio of dimer to monomer for a 0.1 M solution of acetic acid in water, while neglecting the dissociation of acetic acid.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction for the formation of the dimer from acetic acid can be written as:
\[ 2 \text{CH}_3\text{COOH} \rightleftharpoons \text{(CH}_3\text{COOH)}_2 \]
Here, two moles of acetic acid combine to form one mole of dimer.

2. **Equilibrium Constant in Water**:
The equilibrium constant \( K \) for the dimerization in water is given as:
\[ K = \frac{[\text{(CH}_3\text{COOH)}_2]}{[\text{CH}_3\text{COOH}]^2} = 3.6 \times 10^{-2} \text{ M}^{-1} \]

3. **Initial Concentration**:
The initial concentration of acetic acid is given as:
\[ [\text{CH}_3\text{COOH}] = 0.1 \text{ M} \]

4. **Setting Up the Equation**:
Let \( x \) be the concentration of the dimer formed at equilibrium. The concentration of the monomer at equilibrium will then be:
\[ [\text{CH}_3\text{COOH}] = 0.1 - 2x \]
Substituting into the equilibrium expression:
\[ K = \frac{x}{(0.1 - 2x)^2} \]
Given \( K = 3.6 \times 10^{-2} \), we can write:
\[ 3.6 \times 10^{-2} = \frac{x}{(0.1 - 2x)^2} \]

5. **Neglecting Dimerization**:
Since \( x \) is expected to be small compared to 0.1 M, we can approximate:
\[ 0.1 - 2x \approx 0.1 \]
Thus, the equation simplifies to:
\[ 3.6 \times 10^{-2} = \frac{x}{(0.1)^2} \]

6. **Solving for \( x \)**:
Rearranging gives:
\[ x = 3.6 \times 10^{-2} \times (0.1)^2 = 3.6 \times 10^{-2} \times 0.01 = 3.6 \times 10^{-4} \text{ M} \]

7. **Calculating the Molar Ratio**:
The molar ratio of dimer to monomer is given by:
\[ \text{Molar Ratio} = \frac{[\text{(CH}_3\text{COOH)}_2]}{[\text{CH}_3\text{COOH}]} \]
Substituting the values:
\[ \text{Molar Ratio} = \frac{3.6 \times 10^{-4}}{0.1} = \frac{3.6 \times 10^{-4}}{1 \times 10^{-1}} = 3.6 \times 10^{-3} \]
To express this in a more manageable form:
\[ \text{Molar Ratio} = \frac{3.6}{1000} = \frac{9}{2500} \]

8. **Final Answer**:
Therefore, the molar ratio of dimer to monomer for 0.1 M acetic acid in water is:
\[ \text{Molar Ratio} = 9 : 2500 \]