100 mL of `0.5` N NaOH solution is added to 10 mL of 3 N `H_(2)SO_(4)` solution and 20 mL of 1 N HCl solution . The mixture is

Options

A. acidic
B. alkaline
C. neutral
D. none of these

Answer

To solve the problem, we need to determine the equivalents of the acids and bases involved in the reaction and see if they neutralize each other.

### Step-by-Step Solution:

1. **Calculate the equivalents of NaOH (base)**:
- The formula for calculating equivalents is:
\[ \text{Equivalents} = \text{Volume (L)} \times \text{Normality (N)} \]
- For NaOH:
- Volume = 100 mL = 0.1 L
- Normality = 0.5 N
- Therefore, the equivalents of NaOH:
\[ \text{Equivalents of NaOH} = 0.1 \, \text{L} \times 0.5 \, \text{N} = 0.05 \, \text{equivalents} = 50 \, \text{meq} \]

2. **Calculate the equivalents of H₂SO₄ (acid)**:
- For H₂SO₄:
- Volume = 10 mL = 0.01 L
- Normality = 3 N
- Therefore, the equivalents of H₂SO₄:
\[ \text{Equivalents of H₂SO₄} = 0.01 \, \text{L} \times 3 \, \text{N} = 0.03 \, \text{equivalents} = 30 \, \text{meq} \]

3. **Calculate the equivalents of HCl (acid)**:
- For HCl:
- Volume = 20 mL = 0.02 L
- Normality = 1 N
- Therefore, the equivalents of HCl:
\[ \text{Equivalents of HCl} = 0.02 \, \text{L} \times 1 \, \text{N} = 0.02 \, \text{equivalents} = 20 \, \text{meq} \]

4. **Calculate the total equivalents of acids**:
- Total equivalents of acids (H₂SO₄ + HCl):
\[ \text{Total equivalents of acids} = 30 \, \text{meq} + 20 \, \text{meq} = 50 \, \text{meq} \]

5. **Compare the equivalents of acids and bases**:
- Equivalents of NaOH (base) = 50 meq
- Total equivalents of acids = 50 meq
- Since the equivalents of acids and bases are equal, they will neutralize each other.

6. **Conclusion**:
- The mixture is neutral because the equivalents of acid and base are equal.

### Final Answer:
The mixture is neutral.