The ionization constant of acetic acid `1.74xx10^(-5)`. Calculate the degree of dissociation of acetic acid in its `0.05 M` solution. Calculate the concentration of acetate ion in the solution and its `pH`.

Explanation

To solve the problem, we need to calculate the degree of dissociation (α) of acetic acid in a 0.05 M solution, the concentration of acetate ions, and the pH of the solution.

### Step 1: Write the dissociation reaction and expression for the ionization constant (Ka)

The dissociation of acetic acid (CH₃COOH) can be represented as:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

The ionization constant (Ka) is given by:
\[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \]

### Step 2: Set up the initial concentrations

Let:
- Initial concentration of acetic acid, \( C = 0.05 \, M \)
- Degree of dissociation, \( \alpha \)

At equilibrium, the concentrations will be:
- \([\text{CH}_3\text{COOH}] = C(1 - \alpha) = 0.05(1 - \alpha)\)
- \([\text{CH}_3\text{COO}^-] = [\text{H}^+] = C\alpha = 0.05\alpha\)

### Step 3: Substitute into the Ka expression

Substituting these values into the Ka expression:
\[ K_a = \frac{(0.05\alpha)(0.05\alpha)}{0.05(1 - \alpha)} \]

This simplifies to:
\[ K_a = \frac{0.0025\alpha^2}{0.05(1 - \alpha)} \]

### Step 4: Substitute the known value of Ka

Given \( K_a = 1.74 \times 10^{-5} \):
\[ 1.74 \times 10^{-5} = \frac{0.0025\alpha^2}{0.05(1 - \alpha)} \]

### Step 5: Simplify the equation

Multiplying both sides by \( 0.05(1 - \alpha) \):
\[ 1.74 \times 10^{-5} \times 0.05(1 - \alpha) = 0.0025\alpha^2 \]

This simplifies to:
\[ 8.7 \times 10^{-7}(1 - \alpha) = 0.0025\alpha^2 \]

### Step 6: Assume \( \alpha \) is small

Since \( K_a \) is small, we can assume \( 1 - \alpha \approx 1 \):
\[ 8.7 \times 10^{-7} \approx 0.0025\alpha^2 \]

### Step 7: Solve for α

Rearranging gives:
\[ \alpha^2 = \frac{8.7 \times 10^{-7}}{0.0025} \]
\[ \alpha^2 = 3.48 \times 10^{-4} \]
\[ \alpha = \sqrt{3.48 \times 10^{-4}} \]
\[ \alpha \approx 0.01865 \]

### Step 8: Calculate the concentration of acetate ions

The concentration of acetate ions \([\text{CH}_3\text{COO}^-]\) is given by:
\[ [\text{CH}_3\text{COO}^-] = C\alpha = 0.05 \times 0.01865 \]
\[ [\text{CH}_3\text{COO}^-] \approx 9.33 \times 10^{-4} \, M \]

### Step 9: Calculate the pH of the solution

The concentration of hydrogen ions \([\text{H}^+]\) is equal to the concentration of acetate ions:
\[ [\text{H}^+] = 9.33 \times 10^{-4} \, M \]

Now, calculate the pH:
\[ \text{pH} = -\log{[H^+]} = -\log{(9.33 \times 10^{-4})} \]
\[ \text{pH} \approx 3.03 \]

### Final Results

- Degree of dissociation (α) ≈ 0.01865
- Concentration of acetate ions ≈ \( 9.33 \times 10^{-4} \, M \)
- pH ≈ 3.03