`1 c.c` of `0.1 N HCl` is added to `1` litres of `0.1 N NaCl` solution. The pH of the resulting solution will be

Options

A. 7
B. 1
C. 3
D. 4

Explanation

To solve the problem of finding the pH of the resulting solution when `1 c.c` of `0.1 N HCl` is added to `1 litre` of `0.1 N NaCl`, we can follow these steps:

### Step 1: Convert the volume of HCl from cc to liters.
1 c.c = 0.001 liters.

### Step 2: Calculate the number of gram equivalents of HCl.
Using the formula:
\[ \text{Number of gram equivalents} = \text{Normality} \times \text{Volume (in liters)} \]
For HCl:
\[ \text{Normality} = 0.1 \, N, \quad \text{Volume} = 0.001 \, L \]
\[ \text{Number of gram equivalents} = 0.1 \times 0.001 = 0.0001 \, equivalents \]

### Step 3: Determine the total volume of the resulting solution.
The total volume of the solution after adding HCl to NaCl is:
\[ \text{Total Volume} = 1 \, L + 0.001 \, L = 1.001 \, L \]

### Step 4: Calculate the normality of HCl in the resulting solution.
The normality of HCl in the total solution can be calculated as:
\[ \text{Normality of HCl} = \frac{\text{Number of gram equivalents}}{\text{Total Volume}} \]
Substituting the values:
\[ \text{Normality of HCl} = \frac{0.0001}{1.001} \approx 0.000099 \, N \]

### Step 5: Calculate the pH of the resulting solution.
The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Since HCl is a strong acid, the concentration of H⁺ ions will be equal to the normality of HCl:
\[ \text{pH} = -\log(0.000099) \approx 4 \]

### Final Answer:
The pH of the resulting solution is approximately **4**.

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