Calculate the degree of ionisation of 0.4 M acetic acid in water. Dissociation constant of acetic acid is `1.8 xx 10^(-5)`. What will be `H^+` concentration?

Answer

To calculate the degree of ionization of 0.4 M acetic acid in water and the concentration of \( H^+ \) ions, we can follow these steps:

### Step 1: Write the dissociation equation for acetic acid
Acetic acid (\( CH_3COOH \)) dissociates in water as follows:
\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]

### Step 2: Define the degree of ionization
Let \( \alpha \) be the degree of ionization. This means that if we start with a concentration \( C \) of acetic acid, the concentration of dissociated acetic acid at equilibrium will be:
- Concentration of \( CH_3COOH \) at equilibrium: \( C(1 - \alpha) \)
- Concentration of \( CH_3COO^- \) at equilibrium: \( C\alpha \)
- Concentration of \( H^+ \) at equilibrium: \( C\alpha \)

### Step 3: Set up the expression for the dissociation constant \( K_a \)
The dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \]
Substituting the equilibrium concentrations, we get:
\[ K_a = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} = \frac{C^2\alpha^2}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha} \]

### Step 4: Substitute the known values
Given:
- \( C = 0.4 \, M \)
- \( K_a = 1.8 \times 10^{-5} \)

Substituting these values into the equation:
\[ 1.8 \times 10^{-5} = \frac{0.4\alpha^2}{1 - \alpha} \]

### Step 5: Assume \( \alpha \) is small
Since \( K_a \) is small, we can assume \( \alpha \) is small compared to 1, hence \( 1 - \alpha \approx 1 \). Thus, the equation simplifies to:
\[ 1.8 \times 10^{-5} \approx 0.4\alpha^2 \]

### Step 6: Solve for \( \alpha \)
Rearranging the equation gives:
\[ \alpha^2 = \frac{1.8 \times 10^{-5}}{0.4} \]
Calculating the right side:
\[ \alpha^2 = 4.5 \times 10^{-5} \]
Taking the square root:
\[ \alpha = \sqrt{4.5 \times 10^{-5}} \approx 6.71 \times 10^{-3} \]

### Step 7: Calculate the concentration of \( H^+ \)
The concentration of \( H^+ \) ions is given by:
\[ [H^+] = C\alpha = 0.4 \times 6.71 \times 10^{-3} \approx 2.68 \times 10^{-3} \, M \]

### Final Results
- Degree of ionization \( \alpha \approx 6.71 \times 10^{-3} \)
- Concentration of \( H^+ \approx 2.68 \times 10^{-3} \, M \)