pH of a 100 cc solution is 2. It will not change if

Options

A. 100 cc of water is added to it
B. 100 cc of 0.1 M HCl is added to it
C. 100 cc (N/100) HCl is added to it
D. 1 cc of 0.1 M HCl is added to it

Answer

To solve the problem, we need to determine under what conditions the pH of a solution remains unchanged when different amounts of substances are added to it. The initial pH of the solution is given as 2.

### Step-by-Step Solution:

1. **Understanding pH**:
The pH of a solution is defined as:
\[ \text{pH} = -\log[H^+] \]
Given that the pH of the solution is 2, we can calculate the concentration of hydrogen ions \([H^+]\):
\[ [H^+] = 10^{-2} \text{ M} = 0.01 \text{ M} \]

2. **Analyzing the Options**:
We will analyze each option to see whether the pH remains unchanged when the specified amount is added.

- **Option 1**: Adding 100 cc of water.
- When water is added, the total volume becomes 200 cc. The concentration of \([H^+]\) will decrease because the same amount of \(H^+\) is now in a larger volume.
- New concentration:
\[ [H^+] = \frac{0.01 \text{ M} \times 100 \text{ cc}}{200 \text{ cc}} = 0.005 \text{ M} \]
- New pH:
\[ \text{pH} = -\log(0.005) \approx 2.3 \]
- **Conclusion**: pH changes, so this option is incorrect.

- **Option 2**: Adding 100 cc of 0.1 M HCl.
- Adding 0.1 M HCl introduces additional \(H^+\) ions.
- Moles of \(H^+\) from 100 cc of 0.1 M HCl:
\[ \text{Moles} = 0.1 \text{ M} \times 0.1 \text{ L} = 0.01 \text{ moles} \]
- Total moles of \(H^+\) after addition:
\[ 0.01 \text{ moles (initial)} + 0.01 \text{ moles (added)} = 0.02 \text{ moles} \]
- New concentration in 200 cc:
\[ [H^+] = \frac{0.02 \text{ moles}}{0.2 \text{ L}} = 0.1 \text{ M} \]
- New pH:
\[ \text{pH} = -\log(0.1) = 1 \]
- **Conclusion**: pH changes, so this option is incorrect.

- **Option 3**: Adding 100 cc of 0.01 N (N/100) HCl.
- Normality (N) for HCl is equal to its molarity since it is a monoprotic acid.
- Thus, 0.01 N HCl is also 0.01 M.
- Moles of \(H^+\) from 100 cc of 0.01 M HCl:
\[ \text{Moles} = 0.01 \text{ M} \times 0.1 \text{ L} = 0.001 \text{ moles} \]
- Total moles of \(H^+\) after addition:
\[ 0.01 \text{ moles (initial)} + 0.001 \text{ moles (added)} = 0.011 \text{ moles} \]
- New concentration in 200 cc:
\[ [H^+] = \frac{0.011 \text{ moles}}{0.2 \text{ L}} = 0.055 \text{ M} \]
- New pH:
\[ \text{pH} = -\log(0.055) \approx 1.26 \]
- **Conclusion**: pH changes, so this option is incorrect.

- **Option 4**: Adding 1 cc of 0.1 M HCl.
- Moles of \(H^+\) from 1 cc of 0.1 M HCl:
\[ \text{Moles} = 0.1 \text{ M} \times 0.001 \text{ L} = 0.0001 \text{ moles} \]
- Total moles of \(H^+\) after addition:
\[ 0.01 \text{ moles (initial)} + 0.0001 \text{ moles (added)} = 0.0101 \text{ moles} \]
- New concentration in 101 cc:
\[ [H^+] = \frac{0.0101 \text{ moles}}{0.101 \text{ L}} \approx 0.1 \text{ M} \]
- New pH:
\[ \text{pH} = -\log(0.1) = 1 \]
- **Conclusion**: pH changes, so this option is incorrect.

3. **Final Conclusion**:
The only option that maintains the pH at 2 is when 100 cc of 0.01 N (N/100) HCl is added, as it does not significantly change the concentration of \(H^+\) ions in the solution.