Calculate the pH of a solution which is `1xx10^(-3)` M with respect to sulphuric acid.

Answer

To calculate the pH of a `1 x 10^(-3)` M solution of sulfuric acid (H₂SO₄), follow these steps:

### Step 1: Understand the dissociation of sulfuric acid
Sulfuric acid is a strong dibasic acid, meaning it dissociates completely in water. The dissociation can be represented as:
\[ \text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-} \]

### Step 2: Calculate the concentration of hydrogen ions
Since sulfuric acid dissociates to produce two hydrogen ions for every molecule of H₂SO₄, the concentration of hydrogen ions \([\text{H}^+]\) will be twice the concentration of the sulfuric acid solution.
\[ [\text{H}^+] = 2 \times [\text{H}_2\text{SO}_4] = 2 \times (1 \times 10^{-3}) = 2 \times 10^{-3} \, \text{M} \]

### Step 3: Use the formula for pH
The pH of a solution is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]

### Step 4: Substitute the hydrogen ion concentration into the pH formula
Now, substitute the concentration of hydrogen ions into the pH formula:
\[ \text{pH} = -\log(2 \times 10^{-3}) \]

### Step 5: Break down the logarithm
Using the properties of logarithms, we can separate the terms:
\[ \text{pH} = -\log(2) - \log(10^{-3}) \]
\[ \text{pH} = -\log(2) + 3 \]

### Step 6: Calculate \(-\log(2)\)
The value of \(\log(2)\) is approximately \(0.301\):
\[ \text{pH} = -0.301 + 3 \]
\[ \text{pH} = 3 - 0.301 \]
\[ \text{pH} \approx 2.699 \]

### Conclusion
The pH of the `1 x 10^(-3)` M sulfuric acid solution is approximately **2.699**.

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