1 ml of `0.1` M HCl is added into 99 ml of water . Assume volumes are additive, what is pH of resulting solution .

Options

A. 3
B. 1
C. 2
D. 4

Explanation

To find the pH of the resulting solution when 1 ml of 0.1 M HCl is added to 99 ml of water, we can follow these steps:

### Step 1: Calculate the number of moles of HCl
The number of moles of HCl can be calculated using the formula:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume} \]
Given that the molarity of HCl is 0.1 M and the volume is 1 ml (which is 0.001 L), we can calculate:
\[ \text{Number of moles of HCl} = 0.1 \, \text{mol/L} \times 0.001 \, \text{L} = 0.0001 \, \text{mol} \]

### Step 2: Calculate the total volume of the solution
The total volume of the solution after adding 1 ml of HCl to 99 ml of water is:
\[ \text{Total volume} = 1 \, \text{ml} + 99 \, \text{ml} = 100 \, \text{ml} = 0.1 \, \text{L} \]

### Step 3: Calculate the concentration of HCl in the resulting solution
The concentration of HCl in the resulting solution can be calculated using the formula:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Total volume}} \]
Substituting the values we have:
\[ \text{Concentration of HCl} = \frac{0.0001 \, \text{mol}}{0.1 \, \text{L}} = 0.001 \, \text{mol/L} = 10^{-3} \, \text{M} \]

### Step 4: Calculate the pH of the solution
For strong acids like HCl, the concentration of hydrogen ions \([H^+]\) is equal to the concentration of the acid. Therefore:
\[ [H^+] = 10^{-3} \, \text{M} \]
The pH can be calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the value:
\[ \text{pH} = -\log(10^{-3}) = 3 \]

### Final Answer
The pH of the resulting solution is **3**.

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