100 ml 0.2 M NaOH is exactly neutralised by a mixture of which of the following ?

Options

A. 100 ml of `0.1 M HCl + 100 ml` of `0.1 M H_(2)SO_(4)`
B. 100 ml of `0.1 M HCl + 50 ml` of `0.1 M H_(2)SO_(4)`
C. 50 ml `0.1 M HCl + 50 ml` of `0.1 M H_(2)SO_(4)`
D. 50 ml of `0.1 M HCl + 100 ml` of `0.1 M H_(2)SO_(4)`

Explanation

To solve the problem of determining which mixture of acids will exactly neutralize 100 ml of 0.2 M NaOH, we can follow these steps:

### Step-by-Step Solution:

1. **Calculate the moles of NaOH:**
- Molarity (M) = moles of solute / volume of solution in liters.
- Given: Volume of NaOH = 100 ml = 0.1 L, Molarity of NaOH = 0.2 M.
- Moles of NaOH = Molarity × Volume = 0.2 mol/L × 0.1 L = 0.02 moles.
- Since 1 mole = 1000 millimoles, we convert moles to millimoles:
- 0.02 moles = 20 millimoles of NaOH.

**Hint:** Remember that 1 mole = 1000 millimoles for easy conversion.

2. **Identify the neutralization reaction:**
- The neutralization reaction between NaOH and acids (HCl and H2SO4) can be represented as:
- NaOH + HCl → NaCl + H2O (1:1 ratio)
- 2 NaOH + H2SO4 → Na2SO4 + 2 H2O (2:1 ratio)
- Therefore, 1 mole of HCl neutralizes 1 mole of NaOH, and 1 mole of H2SO4 neutralizes 2 moles of NaOH.

3. **Determine the total moles of H+ required:**
- Since we have 20 millimoles of NaOH, we need 20 millimoles of H+ to completely neutralize it.

**Hint:** The total moles of H+ required for neutralization equals the moles of OH- from NaOH.

4. **Evaluate the options:**
- **Option A:** 100 ml of 0.1 M HCl + 100 ml of 0.1 M H2SO4
- Moles of HCl = 0.1 mol/L × 0.1 L = 0.01 moles = 10 millimoles.
- Moles of H2SO4 = 0.1 mol/L × 0.1 L × 2 = 0.02 moles = 20 millimoles.
- Total H+ = 10 + 20 = 30 millimoles (not suitable).

- **Option B:** 100 ml of 0.1 M HCl + 50 ml of 0.1 M H2SO4
- Moles of HCl = 0.1 mol/L × 0.1 L = 10 millimoles.
- Moles of H2SO4 = 0.1 mol/L × 0.05 L × 2 = 0.01 moles = 10 millimoles.
- Total H+ = 10 + 10 = 20 millimoles (suitable).

- **Option C:** 50 ml of 0.1 M HCl + 50 ml of 0.1 M H2SO4
- Moles of HCl = 0.1 mol/L × 0.05 L = 5 millimoles.
- Moles of H2SO4 = 0.1 mol/L × 0.05 L × 2 = 10 millimoles.
- Total H+ = 5 + 10 = 15 millimoles (not suitable).

- **Option D:** 50 ml of 0.1 M HCl + 100 ml of 0.1 M H2SO4
- Moles of HCl = 0.1 mol/L × 0.05 L = 5 millimoles.
- Moles of H2SO4 = 0.1 mol/L × 0.1 L × 2 = 20 millimoles.
- Total H+ = 5 + 20 = 25 millimoles (not suitable).

5. **Conclusion:**
- The only option that provides exactly 20 millimoles of H+ to neutralize the 20 millimoles of OH- from NaOH is **Option B**.

**Final Answer:** Option B (100 ml of 0.1 M HCl + 50 ml of 0.1 M H2SO4) is the correct mixture that will exactly neutralize 100 ml of 0.2 M NaOH.