Which is cheaper : `40%` hydrochloric acid at the rate of 50 paise per kilogram or `80%H_(2)SO_(4)` at the rate of 25 paise per kilogram to completely neutralize 7 kg of caustic potash?

Explanation

To determine which acid is cheaper to completely neutralize 7 kg of caustic potash (KOH), we will compare the costs of using 40% hydrochloric acid (HCl) and 80% sulfuric acid (H₂SO₄).

### Step-by-Step Solution:

1. **Identify the Neutralization Reactions**:
- For HCl:
\[ \text{HCl} + \text{KOH} \rightarrow \text{KCl} + \text{H}_2\text{O} \]
- For H₂SO₄:
\[ \text{H}_2\text{SO}_4 + 2\text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O} \]

2. **Calculate the Amount of KOH**:
- Given that we need to neutralize 7 kg of KOH, which is equivalent to 7000 grams.

3. **Calculate the Amount of HCl Required**:
- The molar mass of KOH = 56 g/mol.
- From the reaction, 1 mole of HCl neutralizes 1 mole of KOH.
- Therefore, 7000 g of KOH will require:
\[ \text{Moles of KOH} = \frac{7000 \text{ g}}{56 \text{ g/mol}} = 125 \text{ moles} \]
- Thus, 125 moles of HCl are needed, which is:
\[ \text{Mass of HCl} = 125 \text{ moles} \times 36.5 \text{ g/mol} = 4562.5 \text{ g} = 4.5625 \text{ kg} \]

4. **Calculate the Cost of HCl**:
- The concentration of HCl is 40%, so the effective mass of HCl in 4.5625 kg is:
\[ \text{Mass of HCl} = 0.40 \times 4.5625 \text{ kg} = 1.825 \text{ kg} \]
- The cost of HCl is 50 paise per kg, so:
\[ \text{Cost} = 1.825 \text{ kg} \times 0.50 \text{ Rs/kg} = 0.9125 \text{ Rs} \approx 91.25 \text{ paise} \]

5. **Calculate the Amount of H₂SO₄ Required**:
- The molar mass of H₂SO₄ = 98 g/mol.
- From the reaction, 1 mole of H₂SO₄ neutralizes 2 moles of KOH.
- Therefore, for 7000 g of KOH, we need:
\[ \text{Moles of H₂SO₄} = \frac{125 \text{ moles}}{2} = 62.5 \text{ moles} \]
- Thus, the mass of H₂SO₄ required is:
\[ \text{Mass of H₂SO₄} = 62.5 \text{ moles} \times 98 \text{ g/mol} = 6125 \text{ g} = 6.125 \text{ kg} \]

6. **Calculate the Cost of H₂SO₄**:
- The concentration of H₂SO₄ is 80%, so the effective mass of H₂SO₄ in 6.125 kg is:
\[ \text{Mass of H₂SO₄} = 0.80 \times 6.125 \text{ kg} = 4.9 \text{ kg} \]
- The cost of H₂SO₄ is 25 paise per kg, so:
\[ \text{Cost} = 4.9 \text{ kg} \times 0.25 \text{ Rs/kg} = 1.225 \text{ Rs} \approx 122.5 \text{ paise} \]

7. **Compare the Costs**:
- Cost of HCl: 91.25 paise
- Cost of H₂SO₄: 122.5 paise
- **Conclusion**: HCl is cheaper.

### Final Answer:
**40% hydrochloric acid is cheaper to completely neutralize 7 kg of caustic potash.**