One litre of a solution contains 18.9 gm of `HNO_(3)` and one lire of another solution contains 3.2 gm of NaOH. In what volume ratio must these solutions be mixed to obtain a neutral solution?

Answer

To solve the problem of mixing two solutions to obtain a neutral solution, we need to calculate the equivalent weights of the acids and bases involved and then find the volume ratio in which they should be mixed.

### Step-by-Step Solution:

1. **Determine the molecular weights:**
- The molecular weight of Nitric Acid (HNO₃) is approximately 63 g/mol.
- The molecular weight of Sodium Hydroxide (NaOH) is approximately 40 g/mol.

2. **Calculate the number of equivalents:**
- For HNO₃:
\[ \text{Equivalents of HNO}_3 = \frac{\text{mass}}{\text{molecular weight}} = \frac{18.9 \, \text{g}}{63 \, \text{g/mol}} = 0.3 \, \text{equivalents} \]
- For NaOH:
\[ \text{Equivalents of NaOH} = \frac{\text{mass}}{\text{molecular weight}} = \frac{3.2 \, \text{g}}{40 \, \text{g/mol}} = 0.08 \, \text{equivalents} \]

3. **Set up the equation for neutralization:**
- For a neutral solution, the number of equivalents of acid must equal the number of equivalents of base:
\[ \text{Equivalents of HNO}_3 = \text{Equivalents of NaOH} \]

4. **Calculate the volume ratio:**
- Let \( V_1 \) be the volume of HNO₃ solution and \( V_2 \) be the volume of NaOH solution.
- The equivalents can also be expressed in terms of volume:
\[ \frac{0.3 \, \text{equivalents}}{V_1} = \frac{0.08 \, \text{equivalents}}{V_2} \]
- Rearranging gives:
\[ \frac{V_1}{V_2} = \frac{0.3}{0.08} = \frac{30}{8} = \frac{15}{4} \]

5. **Conclusion:**
- The volume ratio in which the two solutions should be mixed to obtain a neutral solution is \( 15:4 \).