What will be the pH of `1xx10^(-4)` M `H_(2)SO_(4)` solution ?

Options

A. 10.4
B. 3.7
C. 3
D. 13

Explanation

To find the pH of a \(1 \times 10^{-4}\) M \(H_2SO_4\) solution, we can follow these steps:

### Step 1: Determine the dissociation of \(H_2SO_4\)
Sulfuric acid (\(H_2SO_4\)) 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 \(H_2SO_4\), we get 2 moles of \(H^+\) ions.

### Step 2: Calculate the concentration of \(H^+\) ions
Given the initial concentration of \(H_2SO_4\) is \(1 \times 10^{-4}\) M, the concentration of \(H^+\) ions will be:
\[ [H^+] = 2 \times (1 \times 10^{-4}) = 2 \times 10^{-4} \text{ M} \]

### Step 3: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the value of \([H^+]\):
\[ \text{pH} = -\log(2 \times 10^{-4}) \]

### Step 4: Simplify the logarithm
Using the properties of logarithms:
\[ \text{pH} = -(\log(2) + \log(10^{-4})) = -(\log(2) - 4) \]
Since \(\log(10^{-4}) = -4\).

### Step 5: Calculate \(\log(2)\)
The approximate value of \(\log(2)\) is \(0.301\). Therefore:
\[ \text{pH} = -0.301 + 4 = 3.699 \]

### Step 6: Round the pH value
Rounding \(3.699\) gives us:
\[ \text{pH} \approx 3.7 \]

### Final Answer
Thus, the pH of the \(1 \times 10^{-4}\) M \(H_2SO_4\) solution is approximately **3.7**.

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