Calculate normality of the mixture obtained by mixing `100ml` of `0.1N HCl` and `50 ml` of `0.25N NaOH` solution

Options

A. ` 0.0467 N `
B. ` 0.0367 N `
C. ` 0.0267N `
D. ` 0.0167N `

Explanation

To calculate the normality of the mixture obtained by mixing 100 ml of 0.1N HCl and 50 ml of 0.25N NaOH, we will follow these steps:

### Step 1: Calculate the milliequivalents of HCl
The formula to calculate milliequivalents is:
\[ \text{Milliequivalents} = \text{Normality} \times \text{Volume (in L)} \times 1000 \]
For HCl:
- Normality (N) = 0.1N
- Volume (V) = 100 ml = 0.1 L

Calculating milliequivalents of HCl:
\[ \text{Milliequivalents of HCl} = 0.1 \times 0.1 \times 1000 = 10 \text{ milliequivalents} \]

### Step 2: Calculate the milliequivalents of NaOH
Using the same formula for NaOH:
- Normality (N) = 0.25N
- Volume (V) = 50 ml = 0.05 L

Calculating milliequivalents of NaOH:
\[ \text{Milliequivalents of NaOH} = 0.25 \times 0.05 \times 1000 = 12.5 \text{ milliequivalents} \]

### Step 3: Determine the limiting reactant
Now we compare the milliequivalents of HCl and NaOH:
- Milliequivalents of HCl = 10
- Milliequivalents of NaOH = 12.5

Since NaOH has more milliequivalents, it will completely neutralize HCl, and we need to find out how much NaOH is left after the reaction.

### Step 4: Calculate the remaining milliequivalents of NaOH
Remaining NaOH after neutralization:
\[ \text{Remaining NaOH} = \text{Initial NaOH} - \text{HCl} = 12.5 - 10 = 2.5 \text{ milliequivalents} \]

### Step 5: Calculate the total volume of the solution
Total volume of the solution after mixing:
\[ \text{Total Volume} = 100 \text{ ml} + 50 \text{ ml} = 150 \text{ ml} = 0.15 \text{ L} \]

### Step 6: Calculate the normality of the remaining NaOH
Normality (N) is calculated using the formula:
\[ \text{Normality} = \frac{\text{Milliequivalents}}{\text{Volume (in L)}} \]
For the remaining NaOH:
\[ \text{Normality of remaining NaOH} = \frac{2.5 \text{ milliequivalents}}{0.15 \text{ L}} = \frac{2.5}{0.15} \approx 0.0167 \text{ N} \]

### Final Answer
The normality of the mixture is approximately **0.0167 N**.

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