The pH of 0.1 M acetic acid is 3, the dissociation constant of acid will be

Options

A. `1.0 xx 10^(-4)`
B. `1.0 xx 10^(-5)`
C. `1.0 xx 10^(-3)`
D. `1.0 xx 10^(-8)`

Explanation

To find the dissociation constant (Ka) of acetic acid given that the pH of a 0.1 M solution is 3, we can follow these steps:

### Step 1: Determine the concentration of H⁺ ions
Given that the pH of the solution is 3, we can calculate the concentration of hydrogen ions (H⁺) using the formula:
\[ \text{pH} = -\log[H^+] \]
Thus,
\[ [H^+] = 10^{-\text{pH}} = 10^{-3} \, \text{M} \]

### Step 2: Set up the dissociation equation
Acetic acid (CH₃COOH) dissociates in water as follows:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]
Let α be the degree of dissociation. Initially, we have:
- Concentration of CH₃COOH = 0.1 M
- Change in concentration due to dissociation = -0.1α for CH₃COOH, +0.1α for CH₃COO⁻ and H⁺.

At equilibrium:
- [CH₃COOH] = 0.1 - 0.1α
- [CH₃COO⁻] = 0.1α
- [H⁺] = 0.1α

### Step 3: Relate H⁺ concentration to α
From Step 1, we have:
\[ [H^+] = 0.1α \]
Setting this equal to the concentration we calculated from pH:
\[ 0.1α = 10^{-3} \]
Thus,
\[ α = \frac{10^{-3}}{0.1} = 10^{-2} \]

### Step 4: Substitute α into the expression for Ka
The dissociation constant (Ka) is given by:
\[ K_a = \frac{[\text{CH}_3\text{COO}^-][H^+]}{[\text{CH}_3\text{COOH}]} \]
Substituting the equilibrium concentrations:
\[ K_a = \frac{(0.1α)(0.1α)}{(0.1 - 0.1α)} \]
Substituting α = 10^{-2}:
\[ K_a = \frac{(0.1 \times 10^{-2})(0.1 \times 10^{-2})}{(0.1 - 0.1 \times 10^{-2})} \]
Calculating the values:
\[ K_a = \frac{(0.1 \times 10^{-2})^2}{0.1(1 - 10^{-2})} = \frac{10^{-4}}{0.1(0.99)} \approx \frac{10^{-4}}{0.099} \approx 10^{-3.99} \approx 10^{-4} \]

### Step 5: Final calculation of Ka
Thus, we can approximate:
\[ K_a \approx 10^{-5} \]

### Conclusion
The dissociation constant (Ka) of acetic acid is approximately \(10^{-5}\).

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