Objective question (single correct answer).
i. `H_(3) PO_(4)` is a tribasic acid and one of its salt is `NaH_(2) PO_(4)`. What volume of `1M NaOH` solution should be added to `12 g` of `NaH_(2) PO_(4)` to convert in into `Na_(3) PO_(4)` ?
a. `100 mL` b. 2 mol of `Ca (OH)_(2)` c. Both d. None
iii. The normality of a mixture obtained mixing `100 mL` of `0.2 m H_(2) SO_(4)` with `100 mL` of `0.2 M NaOH` is:
a. `0.05 N` b. `0.1 N` c. `0.15 N` d. `0.2 N`
iv `100 mL` solution of `0.1 N HCl` was titrated with `0.2 N` `NaOH` solutions. The titration was discontinued after adding `30 mL` of `NaOH` solution. The reamining titration was completed by adding `0.25 N KOH` solution. The volume of `KOH` required from completing the titration is:
a. `70 mL` b. `35 mL` c. `32 mL` d. `16 mL`

Explanation

### Step-by-Step Solution

#### Part i: Volume of 1M NaOH to Convert NaH₂PO₄ to Na₃PO₄

1. **Identify the Reaction**:
The reaction between sodium dihydrogen phosphate (NaH₂PO₄) and sodium hydroxide (NaOH) to form sodium phosphate (Na₃PO₄) and water can be represented as:
\[ \text{NaH}_2\text{PO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_3\text{PO}_4 + 2 \text{H}_2\text{O} \]

2. **Calculate Molar Mass of NaH₂PO₄**:
- Sodium (Na): 23 g/mol
- Hydrogen (H): 1 g/mol
- Phosphorus (P): 31 g/mol
- Oxygen (O): 16 g/mol
\[ \text{Molar Mass of NaH}_2\text{PO}_4 = 23 + (2 \times 1) + 31 + (4 \times 16) = 120 \text{ g/mol} \]

3. **Calculate Moles of NaH₂PO₄**:
\[ \text{Moles of NaH}_2\text{PO}_4 = \frac{\text{Given Mass}}{\text{Molar Mass}} = \frac{12 \text{ g}}{120 \text{ g/mol}} = 0.1 \text{ moles} \]

4. **Determine Moles of NaOH Required**:
From the balanced equation, 1 mole of NaH₂PO₄ reacts with 2 moles of NaOH:
\[ \text{Moles of NaOH} = 0.1 \text{ moles NaH}_2\text{PO}_4 \times 2 = 0.2 \text{ moles NaOH} \]

5. **Calculate Volume of 1M NaOH Required**:
Since the molarity (M) is defined as moles per liter:
\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.2 \text{ moles}}{1 \text{ M}} = 0.2 \text{ L} = 200 \text{ mL} \]

6. **Conclusion for Part i**:
The volume of 1M NaOH required to convert 12 g of NaH₂PO₄ to Na₃PO₄ is **200 mL**. Therefore, the correct answer is **D. None** (since 200 mL is not listed).

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#### Part ii: Normality of the Mixture of H₂SO₄ and NaOH

1. **Use the Formula for Normality**:
The normality of a mixture can be calculated using the formula:
\[ N_1V_1 = N_2V_2 \]
where \(N_1\) is the normality of H₂SO₄, \(V_1\) is the volume of H₂SO₄, \(N_2\) is the normality of NaOH, and \(V_2\) is the volume of NaOH.

2. **Given Values**:
- \(V_1 = 100 \text{ mL}\)
- \(N_1 = 0.2 \text{ N}\)
- \(V_2 = 100 \text{ mL}\)
- \(N_2 = 0.2 \text{ N}\)

3. **Calculate Normality of the Mixture**:
Since both solutions are equal in volume and normality, the resulting normality of the mixture is:
\[ N = \frac{N_1 + N_2}{2} = \frac{0.2 + 0.2}{2} = 0.2 \text{ N} \]

4. **Conclusion for Part ii**:
The normality of the mixture is **0.2 N**. Therefore, the correct answer is **D. 0.2 N**.

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#### Part iii: Volume of KOH Required to Complete the Titration

1. **Calculate Moles of H⁺ from HCl**:
\[ \text{Moles of H}^+ = \text{Normality} \times \text{Volume} = 0.1 \text{ N} \times 0.1 \text{ L} = 0.01 \text{ moles} \]

2. **Calculate Moles of NaOH Added**:
\[ \text{Moles of NaOH} = 0.2 \text{ N} \times 0.03 \text{ L} = 0.006 \text{ moles} \]

3. **Calculate Remaining Moles of H⁺**:
\[ \text{Remaining H}^+ = 0.01 - 0.006 = 0.004 \text{ moles} \]

4. **Calculate Volume of KOH Required**:
Using the normality of KOH (0.25 N):
\[ \text{Volume of KOH} = \frac{\text{Remaining H}^+}{\text{Normality}} = \frac{0.004}{0.25} = 0.016 \text{ L} = 16 \text{ mL} \]

5. **Conclusion for Part iii**:
The volume of KOH required to complete the titration is **16 mL**. Therefore, the correct answer is **D. 16 mL**.

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