A 0.05 N solution of acetic acid is found to be `1.9%` ionized at `25^(@)C`. Calculate (i) `K_(a)` for acetic acid and (ii) the pH of the solution.

Explanation

To solve the problem, we will calculate (i) the acid dissociation constant \( K_a \) for acetic acid and (ii) the pH of the solution based on the given information.

### Given:
- Normality of acetic acid solution: \( 0.05 \, N \)
- Percentage ionization: \( 1.9\% \)

### Step 1: Calculate the concentration of acetic acid
Since normality (N) for acetic acid (a monoprotic acid) is equal to molarity (M), we can say:
\[ C = 0.05 \, M \]

### Step 2: Calculate the concentration of ionized acetic acid
The percentage ionization tells us how much of the acetic acid is ionized. To find the concentration of ionized acetic acid, we can use:
\[ \text{Ionized concentration} = \text{Total concentration} \times \frac{\text{Percentage ionization}}{100} \]
\[ \text{Ionized concentration} = 0.05 \, M \times \frac{1.9}{100} = 0.00095 \, M \]

### Step 3: Set up the equilibrium expression
The dissociation of acetic acid (\( CH_3COOH \)) can be represented as:
\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]
At equilibrium, the concentrations will be:
- \( [CH_3COOH] = 0.05 - 0.00095 = 0.04905 \, M \)
- \( [CH_3COO^-] = 0.00095 \, M \)
- \( [H^+] = 0.00095 \, M \)

### Step 4: Write the expression for \( K_a \)
The acid dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \]
Substituting the equilibrium concentrations:
\[ K_a = \frac{(0.00095)(0.00095)}{0.04905} \]
\[ K_a = \frac{0.0000009025}{0.04905} \approx 0.0000184 \, \text{or} \, 1.84 \times 10^{-5} \]

### Step 5: Calculate the pH of the solution
The pH is calculated using the concentration of \( H^+ \):
\[ \text{pH} = -\log[H^+] \]
Substituting the value of \( [H^+] \):
\[ \text{pH} = -\log(0.00095) \approx 3.02 \]

### Final Results:
(i) \( K_a \) for acetic acid is approximately \( 1.84 \times 10^{-5} \).

(ii) The pH of the solution is approximately \( 3.02 \).