10mL of N-HCl, 20mL of `N//2 H_(2)SO_(4) and 30mL N//3 HNO_(3)` are mixed togeher and volume made to one litre. The normally of `H^(+)` in the resulting solution is:

Options

A. 3N/100
B. N/10
C. N/20
D. N/40

Answer

To solve the problem of finding the normality of \( H^+ \) ions in the resulting solution when mixing 10 mL of 1 N HCl, 20 mL of \( \frac{1}{2} \) N \( H_2SO_4 \), and 30 mL of \( \frac{1}{3} \) N \( HNO_3 \), follow these steps:

### Step 1: Identify the normalities and volumes of each acid
- For HCl:
- Normality (\( N_1 \)) = 1 N
- Volume (\( V_1 \)) = 10 mL

- For \( H_2SO_4 \):
- Normality (\( N_2 \)) = \( \frac{1}{2} \) N
- Volume (\( V_2 \)) = 20 mL

- For \( HNO_3 \):
- Normality (\( N_3 \)) = \( \frac{1}{3} \) N
- Volume (\( V_3 \)) = 30 mL

### Step 2: Convert volumes to liters
- Convert the volumes from mL to L:
- \( V_1 = 10 \, \text{mL} = 0.01 \, \text{L} \)
- \( V_2 = 20 \, \text{mL} = 0.02 \, \text{L} \)
- \( V_3 = 30 \, \text{mL} = 0.03 \, \text{L} \)

### Step 3: Calculate the equivalents of \( H^+ \) ions from each acid
- For HCl:
- Equivalents of \( H^+ \) = \( N_1 \times V_1 = 1 \, \text{N} \times 0.01 \, \text{L} = 0.01 \, \text{equivalents} \)

- For \( H_2SO_4 \):
- Each molecule of \( H_2SO_4 \) provides 2 \( H^+ \) ions.
- Equivalents of \( H^+ \) = \( N_2 \times V_2 \times 2 = \frac{1}{2} \, \text{N} \times 0.02 \, \text{L} \times 2 = 0.02 \, \text{equivalents} \)

- For \( HNO_3 \):
- Each molecule of \( HNO_3 \) provides 1 \( H^+ \) ion.
- Equivalents of \( H^+ \) = \( N_3 \times V_3 = \frac{1}{3} \, \text{N} \times 0.03 \, \text{L} = 0.01 \, \text{equivalents} \)

### Step 4: Calculate the total equivalents of \( H^+ \) ions
- Total equivalents of \( H^+ \):
\[ \text{Total} = 0.01 + 0.02 + 0.01 = 0.04 \, \text{equivalents} \]

### Step 5: Calculate the final normality of \( H^+ \) ions in the 1 L solution
- The total volume of the solution is 1 L (1000 mL).
- Normality of \( H^+ \) ions (\( N_f \)):
\[ N_f = \frac{\text{Total equivalents}}{\text{Total volume in L}} = \frac{0.04 \, \text{equivalents}}{1 \, \text{L}} = 0.04 \, \text{N} \]

### Final Answer
The normality of \( H^+ \) ions in the resulting solution is \( 0.04 \, \text{N} \).

---