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 pH of 0.1M acetic acid solution in water, considering the simultaneous dimerisation and dissociation of acid is :

Options

A. 1
B. `2.85`
C. `5.7`
D. `3.42`

Answer

To find the pH of a 0.1 M acetic acid solution in water, considering the simultaneous dimerization and dissociation of the acid, we can follow these steps:

### Step 1: Understand the equilibrium reactions
Acetic acid (CH₃COOH) can form a dimer in water:
\[ 2 \text{CH}_3\text{COOH} \rightleftharpoons (\text{CH}_3\text{COOH})_2 \]

Additionally, acetic acid can dissociate:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

### Step 2: Write the equilibrium expressions
The equilibrium constant for dimerization in water is given as \( K_d = 3.6 \times 10^{-2} \, \text{M}^{-1} \). The dissociation constant \( K_a \) for acetic acid is given as \( K_a = 2.0 \times 10^{-5} \, \text{M} \).

### Step 3: Set up the expressions for concentrations
Let \( x \) be the concentration of acetic acid that dissociates. The initial concentration of acetic acid is 0.1 M. Therefore, at equilibrium:
- Concentration of CH₃COOH = \( 0.1 - x \)
- Concentration of CH₃COO⁻ = \( x \)
- Concentration of H⁺ = \( x \)

### Step 4: Apply the dissociation constant
Using the expression for \( K_a \):
\[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} = \frac{x^2}{0.1 - x} \]
Substituting \( K_a = 2.0 \times 10^{-5} \):
\[ 2.0 \times 10^{-5} = \frac{x^2}{0.1 - x} \]

### Step 5: Assume \( x \) is small
Assuming \( x \) is small compared to 0.1, we can simplify:
\[ 2.0 \times 10^{-5} \approx \frac{x^2}{0.1} \]
Thus,
\[ x^2 = 2.0 \times 10^{-5} \times 0.1 = 2.0 \times 10^{-6} \]
\[ x = \sqrt{2.0 \times 10^{-6}} \approx 1.414 \times 10^{-3} \, \text{M} \]

### Step 6: Calculate pH
Now that we have the concentration of H⁺ ions:
\[ \text{pH} = -\log [\text{H}^+] = -\log(1.414 \times 10^{-3}) \]
Calculating this gives:
\[ \text{pH} \approx 3 - 0.1473 \approx 2.8527 \]

### Final Answer
Thus, the pH of the 0.1 M acetic acid solution in water is approximately **2.85**.

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