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Question

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VIPSolutions · Accepted Answer

Let's solve each question step by step:

---

## Q1a. Classify as Non-electrolyte (N), Weak Electrolyte (W), or Strong Electrolyte (S)

i. **HNO₂**  
- Weak acid (partially ionizes in water)  
- **Answer:** W

ii. **C₆H₆**  
- Benzene, a covalent compound (does not ionize)  
- **Answer:** N

iii. **NaH**  
- Ionic compound, strong base (fully dissociates)  
- **Answer:** S

---

## Q1b. Soluble (S) or Insoluble (IS) in Water

i. **PbSO₄**  
- Most sulfates are soluble, but *lead(II) sulfate* is insoluble  
- **Answer:** IS

ii. **NH₄Cl**  
- Ammonium salts are soluble  
- **Answer:** S

iii. **K₃PO₄**  
- Potassium salts are soluble  
- **Answer:** S

---

## Q1c. Neutralization: Ba(OH)₂(aq) and HF(aq)

### i. **Molecular Equation:**
$$
Ba(OH)_2(aq) + 2HF(aq) \rightarrow BaF_2(s) + 2H_2O(l)
$$

### ii. **Ionic Equation:**
$$
Ba^{2+}(aq) + 2OH^-(aq) + 2H^+(aq) + 2F^-(aq) \rightarrow BaF_2(s) + 2H_2O(l)
$$

### iii. **Net Ionic Equation:**
$$
Ba^{2+}(aq) + 2F^-(aq) \rightarrow BaF_2(s)
$$

---

## Q1d. Molarity and Dilution Calculations

#### 1. **Calculate Molarity:**
- Mass of NaOH = 60 g  
- Molar mass = 40 g/mol  
- Moles NaOH = $ \frac{60}{40} = 1.5 $ mol  
- Volume = 3.00 L  
- Molarity = $ \frac{1.5}{3.00} = 0.50 $ M

#### 2. **Volume Required for Dilution:**
- Use the formula $ M_1V_1 = M_2V_2 $
- $ M_1 = 0.50 $ M, $ M_2 = 0.125 $ M, $ V_2 = 250.0 $ mL = 0.250 L  
$$
V_1 = \frac{M_2V_2}{M_1} = \frac{0.125 \times 0.250}{0.50} = 0.0625 \text{ L} = 62.5 \text{ mL}
$$

---

## Q1e. Oxidation Numbers in $(VO_4)^{3-}$

Let V = x  
Oxygen always = -2  
$$
x + 4(-2) = -3 \implies x - 8 = -3 \implies x = +5
$$

- **V:** +5  
- **O:** -2

---

## Q1f. Predicting Redox Reactions (Cd + AgNO₃)

### i. Will the reaction occur?
- Check activity series: Cd is above Ag → **Yes**

### ii. Element oxidized
- **Cd** (Cd → Cd²⁺)

### iii. Oxidizing agent
- **AgNO₃** (Ag⁺ is reduced)

---

# **Summary Table**

| Q | Answer |
|---|--------|
| 1a(i) | W |
| 1a(ii) | N |
| 1a(iii) | S |
| 1b(i) | IS |
| 1b(ii) | S |
| 1b(iii) | S |
| 1c(i) | Ba(OH)₂(aq) + 2HF(aq) → BaF₂(s) + 2H₂O(l) |
| 1c(ii) | Ba²⁺ + 2OH⁻ + 2H⁺ + 2F⁻ → BaF₂(s) + 2H₂O |
| 1c(iii) | Ba²⁺ + 2F⁻ → BaF₂(s) |
| 1d | Molarity = 0.50 M; Volume = 62.5 mL |
| 1e(i) | +5 |
| 1e(ii) | -2 |
| 1f(i) | Yes |
| 1f(ii) | Cd |
| 1f(iii) | AgNO₃ (Ag⁺) |

If you have any step you want further explained, just ask!

VIPSolutions · Answer

# Detailed Explanation of the Problems

Let's break down each question for clarity.

---

## Q1a. Classify Compounds

### Classification Types:
- **Non-electrolyte (N):** Does not dissociate in water.
- **Weak Electrolyte (W):** Partially dissociates in water.
- **Strong Electrolyte (S):** Fully dissociates in water.

### Compounds:
1. **HNO₂ (Nitrous Acid)**
   - **Classification:** Weak Electrolyte (W)
   - **Explanation:** It partially ionizes in solution, meaning it doesn't fully dissociate into ions.

2. **C₆H₆ (Benzene)**
   - **Classification:** Non-electrolyte (N)
   - **Explanation:** As a covalent compound, it does not ionize in water.

3. **NaH (Sodium Hydride)**
   - **Classification:** Strong Electrolyte (S)
   - **Explanation:** It is an ionic compound and dissociates completely in water, acting as a strong base.

---

## Q1b. Solubility in Water

### Solubility Types:
- **Soluble (S):** Dissolves in water.
- **Insoluble (IS):** Does not dissolve significantly in water.

### Compounds:
1. **PbSO₄ (Lead(II) Sulfate)**
   - **Classification:** Insoluble (IS)
   - **Explanation:** Most sulfates are soluble, but lead sulfate is an exception.

2. **NH₄Cl (Ammonium Chloride)**
   - **Classification:** Soluble (S)
   - **Explanation:** Ammonium salts are generally soluble in water.

3. **K₃PO₄ (Potassium Phosphate)**
   - **Classification:** Soluble (S)
   - **Explanation:** Potassium salts are soluble in water.

---

## Q1c. Neutralization Reaction

### Molecular Equation:
- **Reactants:** $ \text{Ba(OH)}_2(aq) + 2 \text{HF}(aq) $
- **Products:** $ \text{BaF}_2(s) + 2 \text{H}_2\text{O}(l) $

### Ionic Equation:
- Includes all ions:
$$
\text{Ba}^{2+}(aq) + 2 \text{OH}^-(aq) + 2 \text{H}^+(aq) + 2 \text{F}^-(aq) \rightarrow \text{BaF}_2(s) + 2 \text{H}_2\text{O}(l)
$$

### Net Ionic Equation:
- Eliminates spectator ions:
$$
\text{Ba}^{2+}(aq) + 2 \text{F}^-(aq) \rightarrow \text{BaF}_2(s)
$$

---

## Q1d. Molarity and Dilution Calculations

### Step 1: Calculating Molarity of NaOH Solution
- **Given:**
  - Mass of NaOH = 60 g
  - Molar mass of NaOH = 40 g/mol
  - Volume of solution = 3.00 L
  
- **Moles of NaOH:**
$$
\text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{60 \text{ g}}{40 \text{ g/mol}} = 1.5 \text{ mol}
$$

- **Molarity (M):**
$$
M = \frac{\text{moles}}{\text{volume (L)}} = \frac{1.5 \text{ mol}}{3.00 \text{ L}} = 0.50 \text{ M}
$$

### Step 2: Volume Required for Dilution
- **Using dilution formula:** $ M_1V_1 = M_2V_2 $
- **Given:**
  - $ M_1 = 0.50 $ M
  - $ M_2 = 0.125 $ M
  - $ V_2 = 250.0 $ mL = 0.250 L

- **Calculating $ V_1 $:**
$$
V_1 = \frac{M_2V_2}{M_1} = \frac{0.125 \times 0.250}{0.50} = 0.0625 \text{ L} = 62.5 \text{ mL}
$$

---

## Q1e. Oxidation Numbers in $ (VO_4)^{3-} $

### Calculation of Oxidation Number:
1. Let $ V $ be $ x $.
2. Oxygen contributes -2.
3. Equation setup:
$$
x + 4(-2) = -3 \implies x - 8 = -3 \implies x = +5
$$

### Oxidation Numbers:
- **Vanadium (V):** +5
- **Oxygen (O):** -2

---

## Q1f. Predicting Redox Reactions

### Reaction:
$$
\text{Cd} + 2 \text{AgNO}_3 \rightarrow \text{Cd(NO}_3)_2 + 2 \text{Ag}
$$

### Step 1: Will the Reaction Occur?
- **Activity Series:** Cadmium (Cd) is more reactive than Silver (Ag).
- **Conclusion:** Yes, the reaction will occur.

### Step 2: Identify the Oxidizing Agent
- **Oxidation:** $ \text{Cd} \rightarrow \text{Cd}^{2+} $ (loses electrons)
- **Reduction:** $ \text{Ag}^{+} \rightarrow \text{Ag} $ (gains electrons)
- **Oxidizing Agent:** $ \text{Ag}^{+} $ (in AgNO₃)

---

# Summary of Answers

| Q | Answer |
|---|--------|
| 1a(i) | W |
| 1a(ii) | N |
| 1a(iii) | S |
| 1b(i) | IS |
| 1b(ii) | S |
| 1b(iii) | S |
| 1c(i) | $ \text{Ba(OH)}_2(aq) + 2\text{HF}(aq) \rightarrow \text{BaF}_2(s) + 2\text{H}_2\text{O}(l) $ |
| 1c(ii) | $ \text{Ba}^{2+} + 2\text{OH}^- + 2\text{H}^+ + 2\text{F}^- \rightarrow \text{BaF}_2(s) + 2\text{H}_2\text{O} $ |
| 1c(iii) | $ \text{Ba}^{2+} + 2\text{F}^- \rightarrow \text{BaF}_2(s) $ |
| 1d | Molarity = 0.50 M; Volume = 62.5 mL |
| 1e(i) | +5 |
| 1e(ii) | -2 |
| 1f(i) | Yes |
| 1f(ii) | Cd |
| 1f(iii) | AgNO₃ (Ag⁺) |

If you have further questions on any specific part, feel free to ask!

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