On adding which of the following the pH of 20mL of 0.1N HCl will not alter

Options

A. 1 mL of 1 HCl
B. 20 mL of distilled water
C. 1 mL of 0.1 N NaOH
D. 500 mL of HCl of pH =1

Explanation

To determine which addition will not alter the pH of 20 mL of 0.1 N HCl, we can follow these steps:

### Step 1: Calculate the initial pH of 0.1 N HCl
- The concentration of H⁺ ions in 0.1 N HCl is 0.1 M (since HCl is a strong acid and fully dissociates).
- The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
- Substituting the concentration:
\[ \text{pH} = -\log(0.1) = 1 \]

### Step 2: Calculate the number of moles of HCl in 20 mL
- The number of moles can be calculated using the formula:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in L)} \]
- Convert 20 mL to liters:
\[ 20 \, \text{mL} = 0.020 \, \text{L} \]
- Now calculate the moles:
\[ \text{Number of moles} = 0.1 \, \text{mol/L} \times 0.020 \, \text{L} = 0.002 \, \text{mol} = 2 \times 10^{-3} \, \text{mol} \]

### Step 3: Consider the addition of different substances
- We need to analyze the effect of adding different substances on the pH of the solution. The addition of a strong acid (like more HCl) will not change the pH significantly if the concentration remains the same.
- If we add a neutral substance (like water or a salt that does not affect acidity), the pH will also remain unchanged.

### Step 4: Calculate the new concentration after adding 500 mL of HCl
- If we add 500 mL of 0.1 N HCl:
- Calculate the moles in 500 mL:
\[ \text{Number of moles in 500 mL} = 0.1 \, \text{mol/L} \times 0.500 \, \text{L} = 0.050 \, \text{mol} = 50 \times 10^{-3} \, \text{mol} \]
- Total moles after addition:
\[ \text{Total moles} = 2 \times 10^{-3} + 50 \times 10^{-3} = 52 \times 10^{-3} \, \text{mol} \]
- Total volume after addition:
\[ \text{Total volume} = 20 \, \text{mL} + 500 \, \text{mL} = 520 \, \text{mL} = 0.520 \, \text{L} \]
- New concentration of H⁺ ions:
\[ \text{New concentration} = \frac{52 \times 10^{-3} \, \text{mol}}{0.520 \, \text{L}} = 0.1 \, \text{mol/L} \]
- New pH:
\[ \text{pH} = -\log(0.1) = 1 \]

### Conclusion
- The pH remains unchanged at 1 when 500 mL of 0.1 N HCl is added to the solution. Therefore, the correct answer is that adding 500 mL of 0.1 N HCl will not alter the pH.