1 c.c. of 0.1N HCl is added to 99 CC solution of NaCl. The pH of the resulting solution will be

Options

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

Answer

To find the pH of the resulting solution when 1 c.c. of 0.1N HCl is added to 99 c.c. of NaCl solution, we can follow these steps:

### Step 1: Understand the given information
1. **Volume of HCl solution**: 1 c.c. (which is equivalent to 1 mL)
2. **Normality of HCl**: 0.1 N
3. **Volume of NaCl solution**: 99 c.c. (which is equivalent to 99 mL)

### Step 2: Calculate the number of millimoles of HCl
Since HCl is a strong acid, it completely dissociates in solution. The number of millimoles of HCl can be calculated using the formula:
\[ \text{Millimoles of HCl} = \text{Normality} \times \text{Volume (in mL)} \]
\[ \text{Millimoles of HCl} = 0.1 \, \text{N} \times 1 \, \text{mL} = 0.1 \, \text{mmol} \]

### Step 3: Calculate the total volume of the solution
The total volume of the resulting solution after mixing HCl and NaCl is:
\[ \text{Total Volume} = \text{Volume of HCl} + \text{Volume of NaCl} = 1 \, \text{mL} + 99 \, \text{mL} = 100 \, \text{mL} \]

### Step 4: Calculate the concentration of H⁺ ions
The concentration of H⁺ ions in the solution can be calculated using the formula:
\[ \text{Concentration of H⁺} = \frac{\text{Number of moles of H⁺}}{\text{Total Volume (in L)}} \]
First, convert the total volume from mL to L:
\[ 100 \, \text{mL} = 0.1 \, \text{L} \]
Now, calculate the concentration:
\[ \text{Concentration of H⁺} = \frac{0.1 \, \text{mmol}}{100 \, \text{mL}} = \frac{0.1 \, \text{mmol}}{100 \, \text{mL}} = 0.001 \, \text{mol/L} = 10^{-3} \, \text{mol/L} \]

### Step 5: Calculate the pH of the solution
The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H⁺}] \]
Substituting the concentration of H⁺:
\[ \text{pH} = -\log(10^{-3}) = 3 \]

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

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