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 x 10^(-4)` M solution of `H2SO4`, we can follow these steps:

### Step 1: Understand the dissociation of `H2SO4`
`H2SO4` (sulfuric acid) is a strong acid that completely dissociates in water. The dissociation can be represented as:
\[ H2SO4 \rightarrow 2H^+ + SO_4^{2-} \]
This means that for every mole of `H2SO4`, two moles of `H^+` ions are produced.

### Step 2: Calculate the concentration of `H^+` ions
Given the concentration of `H2SO4` is `1 x 10^(-4)` M, the concentration of `H^+` ions will be:
\[ [H^+] = 2 \times [H2SO4] = 2 \times (1 x 10^{-4}) = 2 x 10^{-4} \text{ M} \]

### Step 3: Calculate the pH
The pH of a solution is calculated using the formula:
\[ pH = -\log[H^+] \]
Substituting the value of `[H^+]` we found:
\[ pH = -\log(2 x 10^{-4}) \]

### Step 4: Break down the logarithm
Using the properties of logarithms:
\[ \log(ab) = \log(a) + \log(b) \]
We can express:
\[ \log(2 x 10^{-4}) = \log(2) + \log(10^{-4}) \]
Where:
- \(\log(10^{-4}) = -4\)
- \(\log(2) \approx 0.301\)

So,
\[ \log(2 x 10^{-4}) = 0.301 - 4 = -3.699 \]

### Step 5: Calculate the pH
Now substituting back into the pH formula:
\[ pH = -(-3.699) = 3.699 \]

### Step 6: Round the pH value
The pH can be rounded to:
\[ pH \approx 3.7 \]

### Final Answer
Thus, the pH of the `1 x 10^(-4)` M `H2SO4` solution is approximately **3.7**.

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