0.049 g of `H_(2)SO_(4)` is dissolved per litre of the given solution . Calculate the pH of the solution.

Explanation

To calculate the pH of a solution containing 0.049 g of \( H_2SO_4 \) dissolved in 1 litre, follow these steps:

### Step 1: Calculate the number of moles of \( H_2SO_4 \)

To find the number of moles, use the formula:

\[ \text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

The molar mass of \( H_2SO_4 \) is approximately 98 g/mol.

\[ \text{Moles of } H_2SO_4 = \frac{0.049 \, \text{g}}{98 \, \text{g/mol}} \approx 0.0005 \, \text{mol} \]

### Step 2: Calculate the molarity of the solution

Since the solution is 1 litre, the molarity (M) is:

\[ \text{Molarity} = \frac{\text{moles}}{\text{volume (L)}} = \frac{0.0005 \, \text{mol}}{1 \, \text{L}} = 0.0005 \, \text{M} \]

### Step 3: Determine the concentration of \( H^+ \) ions

\( H_2SO_4 \) is a strong acid and dissociates completely in water. The dissociation can be represented as:

\[ H_2SO_4 \rightarrow 2H^+ + SO_4^{2-} \]

This means that for every mole of \( H_2SO_4 \), 2 moles of \( H^+ \) are produced. Therefore, the concentration of \( H^+ \) ions is:

\[ [H^+] = 2 \times \text{Molarity of } H_2SO_4 = 2 \times 0.0005 \, \text{M} = 0.001 \, \text{M} \]

### Step 4: Calculate the pH of the solution

The pH is calculated using the formula:

\[ \text{pH} = -\log[H^+] \]

Substituting the concentration of \( H^+ \):

\[ \text{pH} = -\log(0.001) = -\log(10^{-3}) = 3 \]

### Final Answer

The pH of the solution is **3**.

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