`10mL` of a strong acid solution of `pH=2.000` are mixed with `990 mL` of another strong acid solution of `pH=4.000`. The `pH` of the resulting solution will be `:`

Options

A. `4.002`
B. `4.000`
C. `4.200`
D. `3.7`

Explanation

To solve the problem of finding the pH of the resulting solution after mixing two strong acid solutions, we will follow these steps:

### Step 1: Calculate the hydrogen ion concentration for both solutions

1. **For the first solution (pH = 2.000)**:
- The hydrogen ion concentration \([H^+]\) can be calculated using the formula:
\[ [H^+] = 10^{-\text{pH}} = 10^{-2} \, \text{mol/L} = 0.01 \, \text{mol/L} \]

2. **For the second solution (pH = 4.000)**:
- Similarly, calculate the hydrogen ion concentration:
\[ [H^+] = 10^{-4} \, \text{mol/L} = 0.0001 \, \text{mol/L} \]

### Step 2: Calculate the number of moles of hydrogen ions in each solution

1. **For the first solution (10 mL)**:
- Convert the volume to liters: \(10 \, \text{mL} = 0.010 \, \text{L}\)
- Calculate the number of moles:
\[ \text{Moles of } H^+ = [H^+] \times \text{Volume} = 0.01 \, \text{mol/L} \times 0.010 \, \text{L} = 0.0001 \, \text{mol} \]

2. **For the second solution (990 mL)**:
- Convert the volume to liters: \(990 \, \text{mL} = 0.990 \, \text{L}\)
- Calculate the number of moles:
\[ \text{Moles of } H^+ = [H^+] \times \text{Volume} = 0.0001 \, \text{mol/L} \times 0.990 \, \text{L} = 0.000099 \, \text{mol} \]

### Step 3: Calculate the total number of moles of hydrogen ions in the mixed solution

- Total moles of \(H^+\):
\[ \text{Total moles} = 0.0001 + 0.000099 = 0.000199 \, \text{mol} \]

### Step 4: Calculate the total volume of the mixed solution

- Total volume:
\[ \text{Total volume} = 10 \, \text{mL} + 990 \, \text{mL} = 1000 \, \text{mL} = 1 \, \text{L} \]

### Step 5: Calculate the final hydrogen ion concentration in the mixed solution

- Final concentration of \(H^+\):
\[ [H^+] = \frac{\text{Total moles}}{\text{Total volume}} = \frac{0.000199 \, \text{mol}}{1 \, \text{L}} = 0.000199 \, \text{mol/L} \]

### Step 6: Calculate the pH of the resulting solution

- Use the formula for pH:
\[ \text{pH} = -\log[H^+] = -\log(0.000199) \]

Using logarithmic properties:
- \(-\log(0.000199) \approx -\log(2 \times 10^{-4}) \approx 4 - 0.301 = 3.699\)

Thus, the pH of the resulting solution is approximately **3.70**.

### Final Answer:
The pH of the resulting solution will be approximately **3.70**.