Let's solve each part step by step. --- ### **7.1.1 Define the term *hydrolysis* of a salt. (2)** **Answer:** *Hydrolysis of a salt* is the reaction of the ions of a dissolved salt with water to produce either acidic or basic solutions. It occurs when the salt is formed from a weak acid or a weak base (or both). --- ### **7.1.2 Will the pH of the solution be GREATER THAN, SMALLER THAN or EQUAL TO 7? Write a relevant equation to support your answer. (3)** **Explanation:** - Ammonium chloride (**NH₄Cl**) is a salt of a *weak base* (NH₃) and a *strong acid* (HCl). - When dissolved in water, it dissociates into NH₄⁺ and Cl⁻. - **NH₄⁺** (ammonium ion) acts as a weak acid and hydrolyzes in water: \[ \mathrm{NH_4^+ (aq) + H_2O(l) \rightarrow NH_3(aq) + H_3O^+(aq)} \] - This produces hydronium ions (H₃O⁺), making the solution acidic. **Answer:** The pH will be **smaller than 7** because the solution is acidic. --- ### **7.2.1 Calculate the number of moles of H₂SO₄ present in this solution. (2)** Given: - Mass of H₂SO₄ = 7.35 g - Volume of solution = 500 cm³ = 0.5 L - Molar mass of H₂SO₄ = 2(1) + 32 + 4(16) = 2 + 32 + 64 = **98 g/mol** \[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} \] \[ \text{Number of moles of H}_2\text{SO}_4 = \frac{7.35}{98} = 0.075 \text{ moles} \] --- ### **7.2.2 Calculate the mass of NaOH added to the H₂SO₄ solution. Assume that the volume of the solution does not change. (9)** #### **Step 1: Calculate [H₃O⁺] after reaction** Given pH = 1.3 \[ \text{pH} = -\log[\text{H}_3\text{O}^+] \] \[ [\text{H}_3\text{O}^+] = 10^{-\text{pH}} = 10^{-1.3} \approx 0.0501\ \text{mol/L} \] Total volume = 0.5 L \[ \text{Moles of H}_3\text{O}^+ = 0.0501 \times 0.5 = 0.02505\ \text{mol} \] #### **Step 2: Initial moles of H₂SO₄** Already calculated: 0.075 mol #### **Step 3: Reaction between H₂SO₄ and NaOH** \[ \text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \] Each mole of H₂SO₄ reacts with 2 moles of NaOH. #### **Step 4: Moles of H₂SO₄ left after reaction** After addition of NaOH, some H₂SO₄ remains (because the final solution is still acidic). Let \( x \) = moles of H₂SO₄ left after reaction = moles of H₃O⁺ = 0.02505 mol So, moles of H₂SO₄ that reacted = initial - remaining = 0.075 - 0.02505 = **0.04995 mol** #### **Step 5: Moles of NaOH that reacted** From the balanced equation, 1 mol H₂SO₄ reacts with 2 mol NaOH: \[ \text{Moles of NaOH} = 2 \times \text{Moles of H}_2\text{SO}_4 \text{ that reacted} \] \[ = 2 \times 0.04995 = 0.0999 \text{ mol} \] #### **Step 6: Mass of NaOH added** Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol \[ \text{Mass of NaOH} = 0.0999 \times 40 = 3.996 \text{ g} \approx 4.00 \text{ g} \] --- ## **Final Answers** ### **7.1.1** Hydrolysis of a salt is the reaction of the ions of a dissolved salt with water to produce acidic or basic solutions. ### **7.1.2** The pH will be **smaller than 7**. Relevant equation: \[ \mathrm{NH_4^+ (aq) + H_2O(l) \rightarrow NH_3(aq) + H_3O^+(aq)} \] ### **7.2.1** Number of moles of H₂SO₄ in solution = **0.075 mol** ### **7.2.2** Mass of NaOH added = **4.00 g**