100 c.c of N/10 NaOH solution is mixed with 100 c.c of N/5 HCI solution and the whole volume is made to 1 litre . The pH of the resulting solution will be

Options

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

Answer

To solve the problem, we need to determine the pH of the resulting solution after mixing the given volumes of NaOH and HCl solutions. Here’s a step-by-step solution:

### Step 1: Calculate the number of equivalents of NaOH and HCl

1. **Calculate the equivalents of NaOH:**
- Normality (N) of NaOH = 1/10 N
- Volume of NaOH = 100 c.c = 0.1 L
- Equivalents of NaOH = Normality × Volume (in L)
\[ \text{Equivalents of NaOH} = \frac{1}{10} \times 0.1 = 0.01 \text{ equivalents} \]

2. **Calculate the equivalents of HCl:**
- Normality (N) of HCl = 1/5 N
- Volume of HCl = 100 c.c = 0.1 L
- Equivalents of HCl = Normality × Volume (in L)
\[ \text{Equivalents of HCl} = \frac{1}{5} \times 0.1 = 0.02 \text{ equivalents} \]

### Step 2: Determine the neutralization reaction

- The reaction between NaOH and HCl is:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

### Step 3: Calculate the remaining equivalents after neutralization

- Since HCl is in excess:
\[ \text{Remaining HCl} = \text{Equivalents of HCl} - \text{Equivalents of NaOH} = 0.02 - 0.01 = 0.01 \text{ equivalents} \]
- All of the NaOH will react, leaving 0.01 equivalents of HCl unreacted.

### Step 4: Calculate the concentration of HCl in the final solution

- The total volume of the solution after mixing is 1 L.
- Concentration of HCl in the final solution:
\[ \text{Concentration of HCl} = \frac{\text{Remaining equivalents of HCl}}{\text{Total volume}} = \frac{0.01}{1} = 0.01 \text{ N} \]

### Step 5: Calculate the pH of the solution

- The pH of a solution is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
- Since HCl is a strong acid, it completely dissociates in solution:
\[ [\text{H}^+] = 0.01 \text{ N} \]
- Therefore,
\[ \text{pH} = -\log(0.01) = 2 \]

### Final Answer:
The pH of the resulting solution will be **2**.

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