If `2` liter of `9.8% w//w H_(2)SO_(4) (d = 1.5 g//mL)` solution is mixed with `3` litre of `1 M KOH` solution then the concentration of `H^(+)` if solution is acidic or concentration of `OH^(-)` if solution is basic in the final solution is:

Options

A. `0`
B. `(3)/(10) N`
C. `(3)/(5) M`
D. `(2)/(5) M`

Explanation

To solve the question, we need to determine the concentration of H⁺ ions or OH⁻ ions in the final solution after mixing the given solutions. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the mass of H₂SO₄ in the solution
Given:
- Volume of H₂SO₄ solution = 2 liters = 2000 mL
- Density of H₂SO₄ solution = 1.5 g/mL

Using the formula:
\[ \text{Mass} = \text{Density} \times \text{Volume} \]
\[ \text{Mass of H₂SO₄ solution} = 1.5 \, \text{g/mL} \times 2000 \, \text{mL} = 3000 \, \text{g} \]

**Hint:** Remember that density is mass per unit volume.

### Step 2: Calculate the mass of H₂SO₄ in the solution using weight percent
Given:
- Weight percent of H₂SO₄ = 9.8%

Using the formula:
\[ \text{Mass of H₂SO₄} = \left(\frac{\text{Weight percent}}{100}\right) \times \text{Total mass of solution} \]
\[ \text{Mass of H₂SO₄} = \left(\frac{9.8}{100}\right) \times 3000 \, \text{g} = 294 \, \text{g} \]

**Hint:** Weight percent indicates how many grams of solute are present in 100 grams of solution.

### Step 3: Calculate the number of moles of H₂SO₄
Given:
- Molar mass of H₂SO₄ = 98 g/mol

Using the formula:
\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} \]
\[ \text{Number of moles of H₂SO₄} = \frac{294 \, \text{g}}{98 \, \text{g/mol}} = 3 \, \text{moles} \]

**Hint:** Moles can be calculated by dividing the mass of the substance by its molar mass.

### Step 4: Calculate the number of moles of KOH
Given:
- Volume of KOH solution = 3 liters
- Molarity of KOH = 1 M

Using the formula:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume} \]
\[ \text{Number of moles of KOH} = 1 \, \text{mol/L} \times 3 \, \text{L} = 3 \, \text{moles} \]

**Hint:** Molarity is defined as moles of solute per liter of solution.

### Step 5: Determine the reaction between H₂SO₄ and KOH
The balanced reaction is:
\[ \text{H₂SO₄} + 2 \text{KOH} \rightarrow \text{K₂SO₄} + 2 \text{H₂O} \]

From the reaction:
- 1 mole of H₂SO₄ reacts with 2 moles of KOH.

### Step 6: Calculate how much H₂SO₄ reacts with KOH
Since we have 3 moles of KOH, we can determine how many moles of H₂SO₄ it can neutralize:
\[ \text{Moles of H₂SO₄ that can react} = \frac{3 \, \text{moles KOH}}{2} = 1.5 \, \text{moles H₂SO₄} \]

### Step 7: Calculate the remaining moles of H₂SO₄
Total moles of H₂SO₄ = 3 moles
Moles of H₂SO₄ that reacted = 1.5 moles
Remaining moles of H₂SO₄:
\[ \text{Remaining moles of H₂SO₄} = 3 - 1.5 = 1.5 \, \text{moles} \]

### Step 8: Calculate the concentration of H⁺ ions
Since each mole of H₂SO₄ produces 2 moles of H⁺ ions:
\[ \text{Moles of H⁺} = 2 \times 1.5 = 3 \, \text{moles} \]

### Step 9: Calculate the total volume of the final solution
Total volume = Volume of H₂SO₄ + Volume of KOH = 2 L + 3 L = 5 L

### Step 10: Calculate the concentration of H⁺ ions in the final solution
Using the formula:
\[ \text{Concentration of H⁺} = \frac{\text{Number of moles of H⁺}}{\text{Total volume in L}} \]
\[ \text{Concentration of H⁺} = \frac{3 \, \text{moles}}{5 \, \text{L}} = 0.6 \, \text{M} \]

**Final Answer:** The concentration of H⁺ ions in the final solution is **0.6 M**.