This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here are the solutions to Question 7: 7.1 7.1.1 Define the term hydrolysis. Hydrolysis is a chemical reaction in which water reacts with a substance, often a salt, to produce an acidic or basic solution by breaking a bond. 7.1.2 Will the final solution be ACIDIC or BASIC? The final solution will be ACIDIC. 7.1.3 Write down a balanced chemical equation that explains the answer to QUESTION 7.1.2. The salt formed is ammonium chloride (NH_4Cl). The ammonium ion (NH_4^+) is the conjugate acid of a weak base (NH_3) and will hydrolyze in water to produce hydronium ions. NH_4^+(aq) + H_2O(l) NH_3(aq) + H_3O^+(aq) 7.2 7.2.1 Is HCl a STRONG or a WEAK acid? HCl is a STRONG acid. 7.2.2 Explain the answer to QUESTION 7.2.1. A strong acid, like HCl, dissociates completely in water to produce a high concentration of hydrogen ions (H^+ or H_3O^+). 7.2.3 Write down the name of the indicator that is most suitable for this reaction. Methyl Orange (or Methyl Red). 7.2.4 Explain your choice of Indicator. The titration is between a strong acid (HCl) and a weak base (NH_3). The equivalence point for such a titration occurs at a pH less than 7 (acidic). Methyl Orange changes color in the acidic pH range (3.1-4.4), making it suitable for detecting the equivalence point. 7.2.5 Calculate the number of moles of NH_3 in 25 cm^3 of the ammonia. Step 1: Convert the volume to dm^3. V_NH_3 = 25 cm^3 = 25 × 10^-3 dm^3 = 0.025 dm^3 Step 2: Calculate the number of moles using the formula n = C × V. n_NH_3 = C_NH_3 × V_NH_3 n_NH_3 = 0.1 mol · dm^-3 × 0.025 dm^3 n_NH_3 = 0.0025 mol 7.2.6 Calculate the pH of the final solution if the concentration of the original HCl was 0.1 mol·dm^-3. Step 1: Calculate the initial moles of NH_3 and HCl. n_NH_3 = C_NH_3 × V_NH_3 = 0.1 mol · dm^-3 × 0.025 dm^3 = 0.0025 mol n_HCl = C_HCl × V_HCl = 0.1 mol · dm^-3 × 0.040 dm^3 = 0.0040 mol Step 2: Determine the moles of HCl remaining after the reaction. The reaction is HCl(aq) + NH_3(aq) NH_4Cl(aq). The molar ratio is 1:1. Since NH_3 is the limiting reactant, 0.0025 mol of HCl will react. n_HCl, remaining = n_HCl, initial - n_NH_3, reacted n_HCl, remaining = 0.0040 mol - 0.0025 mol = 0.0015 mol Step 3: Calculate the total volume of the solution. V_total = V_NH_3 + V_HCl = 0.025 dm^3 + 0.040 dm^3 = 0.065 dm^3 Step 4: Calculate the concentration of the remaining HCl (which is [H^+]). [H^+] = n_HCl, remainingV_total = 0.0015 mol0.065 dm^3 [H^+] ≈ 0.023077 mol · dm^-3 Step 5: Calculate the pH of the solution. pH = -_10[H^+] pH = -_10(0.023077) pH ≈ 1.64 That's 2 down. 3 left today — send the next one.