`100mL` of `" 0.1 M NaOH"` solution is titrated with `100mL` of `"0.5 M "H_(2)SO_(4)` solution. The `pH` of the resulting solution is `: (` For `H_(2)SO_(4), K_(a1)=10^(-2))`

Options

A. 7
B. 7.2
C. 7.4
D. 6.8

Explanation

To solve the problem of titrating 100 mL of 0.1 M NaOH with 100 mL of 0.5 M H₂SO₄ and finding the pH of the resulting solution, we can follow these steps:

### Step 1: Calculate the initial millimoles of NaOH and H₂SO₄
- **For NaOH:**
\[ \text{Millimoles of NaOH} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{M} \times 100 \, \text{mL} = 10 \, \text{mmol} \]

- **For H₂SO₄:**
\[ \text{Millimoles of H₂SO₄} = \text{Concentration} \times \text{Volume} = 0.5 \, \text{M} \times 100 \, \text{mL} = 50 \, \text{mmol} \]

### Step 2: Determine the limiting reagent
The reaction between NaOH and H₂SO₄ can be represented as:
\[ \text{NaOH} + \text{H₂SO₄} \rightarrow \text{NaHSO₄} + \text{H₂O} \]
From the stoichiometry of the reaction, 1 mole of NaOH reacts with 1 mole of H₂SO₄.

- Since we have 10 mmol of NaOH and 50 mmol of H₂SO₄, NaOH is the limiting reagent.

### Step 3: Calculate the amounts after the reaction
- **NaOH consumed:** 10 mmol
- **H₂SO₄ consumed:** 10 mmol (since it reacts in a 1:1 ratio with NaOH)
- **Remaining H₂SO₄:** 50 mmol - 10 mmol = 40 mmol
- **Products formed:** 10 mmol of NaHSO₄

### Step 4: Calculate the concentration of the resulting solution
After the reaction, we have:
- Total volume = 100 mL (NaOH) + 100 mL (H₂SO₄) = 200 mL
- Concentration of NaHSO₄ (formed):
\[ \text{Concentration of NaHSO₄} = \frac{10 \, \text{mmol}}{200 \, \text{mL}} = 0.05 \, \text{M} \]
- Remaining concentration of H₂SO₄:
\[ \text{Concentration of H₂SO₄} = \frac{40 \, \text{mmol}}{200 \, \text{mL}} = 0.2 \, \text{M} \]

### Step 5: Determine the pH of the resulting solution
Since we have a weak acid (NaHSO₄) and a strong base (NaOH), we can use the formula:
\[ \text{pH} = 7 + \frac{1}{2} pK_a + \log C \]
Where \( C \) is the concentration of the salt formed (NaHSO₄) and \( K_{a1} \) for H₂SO₄ is given as \( 10^{-2} \).

- Calculate \( pK_a \):
\[ pK_a = -\log(10^{-2}) = 2 \]
- Substitute values into the pH formula:
\[ \text{pH} = 7 + \frac{1}{2} \times 2 + \log(0.05) \]
\[ \text{pH} = 7 + 1 - 1.3 \quad (\text{since } \log(0.05) \approx -1.3) \]
\[ \text{pH} = 7.7 \]

### Final Answer
The pH of the resulting solution is approximately **7.7**.

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