`600mL` of `0.01MHCl` is mixed with `400mL` of `0.01MH_(2)SO_(4)`.The `pH` of the mixture is......`10^(-2)` (Nearest integer) [Given `log2=0.30],[log3=0.48],[log5=0.69],[log7=0.84],[log11=1.04]`

Answer

To solve the problem, we need to calculate the pH of the mixture formed by mixing 600 mL of 0.01 M HCl with 400 mL of 0.01 M H₂SO₄.

### Step-by-Step Solution:

1. **Calculate the millimoles of HCl:**
\[ \text{Millimoles of HCl} = \text{Molarity} \times \text{Volume} = 0.01 \, \text{M} \times 600 \, \text{mL} = 6 \, \text{mmol} \]

2. **Calculate the millimoles of H₂SO₄:**
\[ \text{Millimoles of H₂SO₄} = \text{Molarity} \times \text{Volume} = 0.01 \, \text{M} \times 400 \, \text{mL} = 4 \, \text{mmol} \]
Since each molecule of H₂SO₄ produces 2 H⁺ ions, the total millimoles of H⁺ from H₂SO₄ is:
\[ \text{H⁺ from H₂SO₄} = 4 \, \text{mmol} \times 2 = 8 \, \text{mmol} \]

3. **Calculate the total millimoles of H⁺ ions:**
\[ \text{Total H⁺} = \text{H⁺ from HCl} + \text{H⁺ from H₂SO₄} = 6 \, \text{mmol} + 8 \, \text{mmol} = 14 \, \text{mmol} \]

4. **Calculate the total volume of the mixture:**
\[ \text{Total Volume} = 600 \, \text{mL} + 400 \, \text{mL} = 1000 \, \text{mL} \]

5. **Calculate the concentration of H⁺ ions:**
\[ \text{Concentration of H⁺} = \frac{\text{Total H⁺}}{\text{Total Volume}} = \frac{14 \, \text{mmol}}{1000 \, \text{mL}} = 0.014 \, \text{M} = 14 \times 10^{-3} \, \text{M} \]

6. **Calculate the pH:**
\[ \text{pH} = -\log[\text{H⁺}] = -\log(14 \times 10^{-3}) \]
This can be rewritten using logarithmic properties:
\[ \text{pH} = -\log(14) - \log(10^{-3}) = -\log(14) + 3 \]

7. **Estimate \(-\log(14)\):**
Using logarithmic approximations, we can estimate:
\[ \log(14) \approx 1.146 \quad (\text{since } 10^{1.146} \approx 14) \]
Therefore:
\[ -\log(14) \approx -1.146 \]

8. **Final calculation of pH:**
\[ \text{pH} = -(-1.146) + 3 = 1.146 + 3 = 4.146 \]

9. **Round to the nearest integer:**
The nearest integer to 4.146 is **4**.

### Final Answer:
The pH of the mixture is **4**.