In decinormal solution, `CH_(3)COOH` acid is ionised to the extent of `1.3%`. If `log 1.3 = 0.11,` what is the `pH` of the solution?

Options

A. `3.89`
B. `2.89`
C. `4.89`
D. Unpredictable

Explanation

To find the pH of a decinormal solution of acetic acid (CH₃COOH) that is ionized to an extent of 1.3%, we can follow these steps:

### Step 1: Understand the Given Information
- The normality of the acetic acid solution is given as 0.1 N (decinormal).
- The extent of ionization (α) is given as 1.3%.

### Step 2: Convert Percent Ionization to a Decimal
The percent ionization can be converted to a decimal by dividing by 100:
\[ \alpha = \frac{1.3}{100} = 0.013 \]

### Step 3: Write the Dissociation Reaction
The dissociation of acetic acid in water can be represented as:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

### Step 4: Determine the Initial Concentration
Since the normality (0.1 N) is equivalent to molarity for monoprotic acids, we can say:
\[ C = 0.1 \, \text{M} \]

### Step 5: Calculate the Concentration of H⁺ Ions
The concentration of H⁺ ions produced at equilibrium can be calculated using:
\[ [\text{H}^+] = C \cdot \alpha \]
Substituting the values we have:
\[ [\text{H}^+] = 0.1 \, \text{M} \cdot 0.013 = 0.0013 \, \text{M} \]

### Step 6: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log [\text{H}^+] \]
Substituting the value of [H⁺]:
\[ \text{pH} = -\log(0.0013) \]

### Step 7: Use Logarithmic Properties
Using the property of logarithms:
\[ \log(0.0013) = \log(1.3 \times 10^{-3}) = \log(1.3) + \log(10^{-3}) \]
\[ \log(10^{-3}) = -3 \]
Thus:
\[ \text{pH} = -(\log(1.3) - 3) \]
Given that \(\log(1.3) = 0.11\):
\[ \text{pH} = - (0.11 - 3) = 3 - 0.11 = 2.89 \]

### Final Answer
The pH of the solution is:
\[ \text{pH} = 2.89 \]

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