The pH of a solution of `H_(2)SO_(4)` is 1. Assuming complete ionisation, find the molarity of `H_(2)SO_(4)` solution :

Options

A. `0.1`
B. `0.2`
C. `0.05`
D. `2.0`

Answer

To find the molarity of the `H₂SO₄` solution given that the pH is 1, we can follow these steps:

### Step 1: Understand the relationship between pH and H⁺ concentration
The pH of a solution is related to the concentration of hydrogen ions (H⁺) by the formula:
\[ \text{pH} = -\log[H^+] \]

### Step 2: Calculate the H⁺ concentration from the pH
Given that the pH is 1, we can find the concentration of H⁺ ions:
\[ [H^+] = 10^{-\text{pH}} = 10^{-1} = 0.1 \, \text{M} \]

### Step 3: Consider the complete ionization of `H₂SO₄`
Sulfuric acid (`H₂SO₄`) completely ionizes in solution as follows:
\[ H₂SO₄ \rightarrow 2H^+ + SO₄^{2-} \]
This means that for every 1 mole of `H₂SO₄`, 2 moles of H⁺ are produced.

### Step 4: Set up the relationship between H⁺ concentration and `H₂SO₄` molarity
Let the molarity of `H₂SO₄` be \( M \). Since `H₂SO₄` produces 2 moles of H⁺ for every mole of `H₂SO₄`, the concentration of H⁺ ions can be expressed as:
\[ [H^+] = 2M \]

### Step 5: Substitute the known H⁺ concentration
From Step 2, we know that:
\[ [H^+] = 0.1 \, \text{M} \]
Thus, we can set up the equation:
\[ 2M = 0.1 \]

### Step 6: Solve for M
To find the molarity \( M \) of `H₂SO₄`, we rearrange the equation:
\[ M = \frac{0.1}{2} = 0.05 \, \text{M} \]

### Final Answer
The molarity of the `H₂SO₄` solution is:
\[ \text{Molarity of } H₂SO₄ = 0.05 \, \text{M} \]

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