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 solve the problem of finding 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
We start by calculating the number of moles of HCl in the solution. The formula for calculating moles from molarity is:

\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in liters)} \]

Given:
- Molarity of HCl = 0.1 M
- Volume of HCl = 1 ml = \(1 \times 10^{-3}\) L

Calculating the number of moles:

\[ \text{Number of moles of HCl} = 0.1 \, \text{mol/L} \times 1 \times 10^{-3} \, \text{L} = 1 \times 10^{-4} \, \text{moles} \]

### Step 2: Calculate the total volume of the solution
The total volume of the solution after adding HCl to water is:

\[ \text{Total Volume} = \text{Volume of HCl} + \text{Volume of Water} = 1 \, \text{ml} + 99 \, \text{ml} = 100 \, \text{ml} = 0.1 \, \text{L} \]

### Step 3: Calculate the concentration of HCl in the resulting solution
Now we need to find the concentration of HCl in the total volume:

\[ \text{Concentration of HCl} = \frac{\text{Number of moles of HCl}}{\text{Total Volume (in liters)}} \]

Substituting the values:

\[ \text{Concentration of HCl} = \frac{1 \times 10^{-4} \, \text{moles}}{0.1 \, \text{L}} = 1 \times 10^{-3} \, \text{M} \]

### Step 4: Calculate the pH of the solution
Since HCl is a strong acid, it completely dissociates in solution:

\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]

Thus, the concentration of \( \text{H}^+ \) ions is equal to the concentration of HCl:

\[ [\text{H}^+] = 1 \times 10^{-3} \, \text{M} \]

Now we can calculate the pH using the formula:

\[ \text{pH} = -\log[\text{H}^+] \]

Substituting the concentration:

\[ \text{pH} = -\log(1 \times 10^{-3}) = 3 \]

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

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