The pH of a mixture of 100 ml 1M `H_2SO_4` and 200 ml 1 N NaOH at `25^@C` is

Options

A. More than 7
B. Less than 7
C. Equal to 7
D. Can't predict

Answer

To find the pH of the mixture of 100 ml of 1 M H₂SO₄ and 200 ml of 1 N NaOH at 25°C, we will follow these steps:

### Step 1: Calculate the millimoles of H₂SO₄
- **Given**: Molarity of H₂SO₄ = 1 M, Volume = 100 ml
- **Calculation**:
\[ \text{Millimoles of H₂SO₄} = \text{Molarity} \times \text{Volume} = 1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.1 \, \text{mol} = 100 \, \text{mmol} \]

### Step 2: Calculate the millimoles of NaOH
- **Given**: Normality of NaOH = 1 N, Volume = 200 ml
- **Calculation**:
\[ \text{Millimoles of NaOH} = \text{Normality} \times \text{Volume} = 1 \, \text{N} \times 0.2 \, \text{L} = 0.2 \, \text{mol} = 200 \, \text{mmol} \]

### Step 3: Write the balanced chemical reaction
- The reaction between H₂SO₄ and NaOH is:
\[ \text{H₂SO₄} + 2 \text{NaOH} \rightarrow \text{Na₂SO₄} + 2 \text{H₂O} \]

### Step 4: Determine the limiting reactant
- From the balanced equation, 1 mole of H₂SO₄ reacts with 2 moles of NaOH.
- For 100 mmol of H₂SO₄, we need:
\[ 100 \, \text{mmol H₂SO₄} \times 2 = 200 \, \text{mmol NaOH} \]
- We have exactly 200 mmol of NaOH available, which means both reactants will completely neutralize each other.

### Step 5: Determine the pH of the resulting solution
- At complete neutralization, there are no excess acids or bases left in the solution. The resulting solution will be neutral.
- The pH of pure water at 25°C is 7.

### Final Answer:
The pH of the mixture is **7**.