10 ml of `M/(200) H_(2)SO_(4)` is mixed with 40 ml of `M/200 H_(2)SO_(4)`. The pH of the resulting solution is

Options

A. 1
B. 2
C. 2.3
D. None of these

Answer

To solve the problem step by step, we will calculate the pH of the resulting solution after mixing two solutions of sulfuric acid (H₂SO₄).

### Step 1: Determine the molarity of H₂SO₄ in each solution
We have two solutions of H₂SO₄:
- Solution 1: 10 mL of M/200 H₂SO₄
- Solution 2: 40 mL of M/200 H₂SO₄

The molarity of both solutions is M/200, which means:
\[ \text{Molarity} = \frac{M}{200} \]

### Step 2: Calculate the number of moles of H₂SO₄ in each solution
Using the formula for moles:
\[ \text{Moles} = \text{Molarity} \times \text{Volume (in L)} \]

For Solution 1:
\[ \text{Moles of H₂SO₄} = \frac{M}{200} \times \frac{10}{1000} = \frac{M \times 10}{200 \times 1000} = \frac{M \times 10}{200000} \]

For Solution 2:
\[ \text{Moles of H₂SO₄} = \frac{M}{200} \times \frac{40}{1000} = \frac{M \times 40}{200 \times 1000} = \frac{M \times 40}{200000} \]

### Step 3: Calculate the total moles of H₂SO₄ after mixing
Total moles of H₂SO₄:
\[ \text{Total Moles} = \frac{M \times 10}{200000} + \frac{M \times 40}{200000} = \frac{M \times (10 + 40)}{200000} = \frac{M \times 50}{200000} \]

### Step 4: Calculate the total volume of the resulting solution
Total volume after mixing:
\[ \text{Total Volume} = 10 \, \text{mL} + 40 \, \text{mL} = 50 \, \text{mL} = 0.050 \, \text{L} \]

### Step 5: Calculate the concentration of H⁺ ions
Since H₂SO₄ is a strong acid and dissociates completely in solution, it produces 2 moles of H⁺ ions for every mole of H₂SO₄. Therefore, the concentration of H⁺ ions will be:
\[ [H^+] = 2 \times \frac{\text{Total Moles of H₂SO₄}}{\text{Total Volume}} \]
\[ [H^+] = 2 \times \frac{\frac{M \times 50}{200000}}{0.050} \]
\[ [H^+] = 2 \times \frac{M \times 50}{200000 \times 0.050} = 2 \times \frac{M \times 50}{10000} = \frac{M \times 100}{10000} = \frac{M}{100} \]

### Step 6: Calculate the pH of the solution
Using the formula for pH:
\[ \text{pH} = -\log[H^+] \]
Substituting the concentration of H⁺:
\[ \text{pH} = -\log\left(\frac{M}{100}\right) \]

Assuming M = 1 (for simplicity):
\[ \text{pH} = -\log\left(\frac{1}{100}\right) = -\log(10^{-2}) = 2 \]

Thus, the pH of the resulting solution is **2**.