What volume at NTP of gaseous ammonia will be required to be passed into `30 cm^(3)` of 1 N `H_(2)SO_(4)` solution to bring down the acid strength of the latter of 0.2 N ?

Answer

To solve the problem of determining the volume of gaseous ammonia (NH₃) required to reduce the acid strength of 30 cm³ of 1 N H₂SO₄ solution to 0.2 N, we can follow these steps:

### Step-by-Step Solution:

1. **Convert Volume to Liters**:
- The volume of H₂SO₄ solution is given as 30 cm³.
- Convert this to liters:
\[ 30 \, \text{cm}^3 = 30 \, \text{mL} = 0.030 \, \text{L} \]

2. **Calculate Initial Milliequivalents of H₂SO₄**:
- The normality (N) of the H₂SO₄ solution is 1 N.
- Milliequivalents (mEq) of H₂SO₄ can be calculated as:
\[ \text{mEq of H₂SO₄} = \text{Volume (L)} \times \text{Normality (N)} = 0.030 \, \text{L} \times 1 \, \text{N} = 0.030 \, \text{mEq} \]

3. **Calculate Final Milliequivalents of H₂SO₄**:
- The final normality is given as 0.2 N.
- Calculate the final mEq of H₂SO₄:
\[ \text{Final mEq of H₂SO₄} = 0.030 \, \text{L} \times 0.2 \, \text{N} = 0.006 \, \text{mEq} \]

4. **Determine the Change in Milliequivalents**:
- The change in milliequivalents of H₂SO₄ is:
\[ \text{Change in mEq} = \text{Initial mEq} - \text{Final mEq} = 0.030 \, \text{mEq} - 0.006 \, \text{mEq} = 0.024 \, \text{mEq} \]

5. **Relate Milliequivalents of NH₃ to H₂SO₄**:
- Since 1 mole of NH₃ reacts with 1 mole of H₂SO₄, the mEq of NH₃ required is equal to the change in mEq of H₂SO₄:
\[ \text{mEq of NH₃} = 0.024 \, \text{mEq} \]

6. **Calculate the Weight of NH₃ Required**:
- The equivalent weight of NH₃ is 17 g/mol (molar mass of NH₃).
- The weight of NH₃ can be calculated using the formula:
\[ \text{Weight of NH₃} = \text{mEq} \times \text{Equivalent weight} = 0.024 \, \text{mEq} \times 17 \, \text{g/mol} = 0.408 \, \text{g} \]

7. **Calculate the Volume of NH₃ at NTP**:
- At NTP, 1 mole of gas occupies 22.4 L.
- The volume of NH₃ can be calculated as:
\[ \text{Volume of NH₃} = \frac{\text{Weight of NH₃}}{\text{Molar mass of NH₃}} \times 22.4 \, \text{L} = \frac{0.408 \, \text{g}}{17 \, \text{g/mol}} \times 22.4 \, \text{L} \approx 0.5376 \, \text{L} \]
- Convert to mL:
\[ 0.5376 \, \text{L} = 537.6 \, \text{mL} \]

### Final Answer:
The volume of gaseous ammonia required at NTP is approximately **537.6 mL**.