`H_(2)CO_(3)` ionises `1.5` percent in its decimolar solution pH of solution
will be `:-`

Options

A. `2.85`
B. `2.52`
C. `2.22`
D. `3.22`

Explanation

To find the pH of a decimolar solution of carbonic acid (H₂CO₃) that ionizes 1.5%, we can follow these steps:

### Step 1: Determine the initial concentration of H₂CO₃
The problem states that we have a decimolar solution, which means the initial concentration of H₂CO₃ is:
\[ [H₂CO₃] = 0.1 \, \text{M} \]

### Step 2: Calculate the amount of H₂CO₃ that ionizes
Since H₂CO₃ ionizes 1.5%, we can calculate the concentration of H₂CO₃ that ionizes:
\[ \text{Ionized concentration} = 0.1 \, \text{M} \times \frac{1.5}{100} = 0.0015 \, \text{M} \]

### Step 3: Determine the concentration of H⁺ ions produced
When H₂CO₃ ionizes, it produces 2 moles of H⁺ ions for every mole of H₂CO₃ that ionizes. Therefore, the concentration of H⁺ ions produced is:
\[ [H^+] = 2 \times 0.0015 \, \text{M} = 0.003 \, \text{M} \]

### Step 4: Calculate the pH of the solution
The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the concentration of H⁺ ions:
\[ \text{pH} = -\log(0.003) \]

### Step 5: Simplify the logarithm
Using the properties of logarithms:
\[ \text{pH} = -\log(3 \times 10^{-3}) = -\log(3) - \log(10^{-3}) \]
\[ \text{pH} = -\log(3) + 3 \]

### Step 6: Calculate the value of -log(3)
The value of \(-\log(3)\) is approximately \(0.477\). Thus:
\[ \text{pH} = 3 - 0.477 = 2.523 \]

### Final Answer
The pH of the solution is approximately:
\[ \text{pH} \approx 2.52 \]

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