10ml of `((N)/(2))` HCl, 30 ml of `((N)/(10)) HNO_(3)` and 75 ml of `((N)/(5)) HNO_(3)` are mixed, the normality of `H^(+)` in the resulting solution is -

Options

A. 0.2
B. 0.1
C. 0.5
D. 0.25

Explanation

To find the normality of \( H^+ \) ions in the resulting solution after mixing the given acids, we will follow these steps:

### Step 1: Calculate the equivalence of \( HCl \)
Given:
- Volume of \( HCl = 10 \, \text{ml} \)
- Normality of \( HCl = \frac{N}{2} \)

First, convert the volume from ml to liters:
\[ \text{Volume in liters} = \frac{10 \, \text{ml}}{1000} = 0.01 \, \text{L} \]

Now, calculate the equivalence of \( HCl \):
\[ \text{Equivalence of } HCl = \text{Normality} \times \text{Volume in L} = \frac{N}{2} \times 0.01 \]

### Step 2: Calculate the equivalence of \( HNO_3 \) (30 ml)
Given:
- Volume of \( HNO_3 = 30 \, \text{ml} \)
- Normality of \( HNO_3 = \frac{N}{10} \)

Convert the volume from ml to liters:
\[ \text{Volume in liters} = \frac{30 \, \text{ml}}{1000} = 0.03 \, \text{L} \]

Now, calculate the equivalence of this \( HNO_3 \):
\[ \text{Equivalence of } HNO_3 = \text{Normality} \times \text{Volume in L} = \frac{N}{10} \times 0.03 \]

### Step 3: Calculate the equivalence of \( HNO_3 \) (75 ml)
Given:
- Volume of \( HNO_3 = 75 \, \text{ml} \)
- Normality of \( HNO_3 = \frac{N}{5} \)

Convert the volume from ml to liters:
\[ \text{Volume in liters} = \frac{75 \, \text{ml}}{1000} = 0.075 \, \text{L} \]

Now, calculate the equivalence of this \( HNO_3 \):
\[ \text{Equivalence of } HNO_3 = \text{Normality} \times \text{Volume in L} = \frac{N}{5} \times 0.075 \]

### Step 4: Sum the equivalences
Now, we will add all the equivalences calculated:
\[ \text{Total Equivalence} = \left( \frac{N}{2} \times 0.01 \right) + \left( \frac{N}{10} \times 0.03 \right) + \left( \frac{N}{5} \times 0.075 \right) \]

### Step 5: Calculate the total volume
The total volume of the mixed solution:
\[ \text{Total Volume} = 10 \, \text{ml} + 30 \, \text{ml} + 75 \, \text{ml} = 115 \, \text{ml} = 0.115 \, \text{L} \]

### Step 6: Calculate the normality of \( H^+ \) ions
The normality of \( H^+ \) ions in the resulting solution is given by:
\[ \text{Normality of } H^+ = \frac{\text{Total Equivalence}}{\text{Total Volume in L}} \]

Substituting the values we calculated:
\[ \text{Normality of } H^+ = \frac{\left( \frac{N}{2} \times 0.01 + \frac{N}{10} \times 0.03 + \frac{N}{5} \times 0.075 \right)}{0.115} \]

### Final Calculation
Now, we can simplify and calculate the normality:
1. Calculate each term:
- \( \frac{N}{2} \times 0.01 = \frac{N \times 0.01}{2} = \frac{0.01N}{2} = 0.005N \)
- \( \frac{N}{10} \times 0.03 = \frac{N \times 0.03}{10} = \frac{0.03N}{10} = 0.003N \)
- \( \frac{N}{5} \times 0.075 = \frac{N \times 0.075}{5} = \frac{0.075N}{5} = 0.015N \)

2. Add these values:
\[ 0.005N + 0.003N + 0.015N = 0.023N \]

3. Now, substitute back into the normality equation:
\[ \text{Normality of } H^+ = \frac{0.023N}{0.115} = 0.2N \]

### Conclusion
The normality of \( H^+ \) in the resulting solution is \( 0.2N \).

---