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 the exercises: Exercise 1: Molarity of NaOH solution Step 1: Determine the molar mass of KHP (Potassium Hydrogen Phthalate, HKC₈H₄O₄). Molar mass of KHP = (8 × 12.011) + (5 × 1.008) + (1 × 39.098) + (4 × 15.999) Molar mass of KHP = 96.088 + 5.040 + 39.098 + 63.996 = 204.222 g/mol Step 2: Calculate the moles of KHP. Moles of KHP = Mass of KHPMolar mass of KHP Moles of KHP = 0.8234 g204.222 g/mol Moles of KHP = 0.0040319 mol Step 3: Determine the moles of NaOH. The reaction between KHP (a monoprotic acid) and NaOH (a strong base) is 1:1. HKC_8H_4O_4(aq) + NaOH(aq) → NaKC_8H_4O_4(aq) + H_2O(l) Therefore, at the equivalence point, moles of KHP = moles of NaOH. Moles of NaOH = 0.0040319 mol Step 4: Calculate the molarity of the NaOH solution. Given volume of NaOH = 38.76 cm³ = 38.76 mL = 0.03876 L Molarity of NaOH = Moles of NaOHVolume of NaOH (L) Molarity of NaOH = 0.0040319 mol0.03876 L Molarity of NaOH = 0.10402 M The exact molarity of the NaOH solution is 0.10402 M. Exercise 2: Concentration of diluted FeCl₃ solution Step 1: Use the dilution formula M_1V_1 = M_2V_2. Given: Initial molarity (M_1) = 3.47 × 10^-2 M Initial volume (V_1) = 5.00 cm^3 Final volume (V_2) = 250 cm^3 We need to find the final molarity (M_2). Step 2: Substitute the values into the formula. (3.47 × 10^-2 M)(5.00 cm^3) = M_2(250 cm^3) Step 3: Solve for M_2. M_2 = (3.47 × 10^-2 M)(5.00 cm^3)250 cm^3 M_2 = 0.1735 M · cm^3250 cm^3 M_2 = 0.000694 M M_2 = 6.94 × 10^-4 M The concentration of the diluted solution is 6.94 × 10^-4 M. This matches the given answer. Exercise 3: Molarity of FeCl₃ solution Step 1: Determine the molar mass of FeCl₃. Molar mass of FeCl₃ = (1 × 55.845) + (3 × 35.453) Molar mass of FeCl₃ = 55.845 + 106.359 = 162.204 g/mol Step 2: Calculate the moles of FeCl₃. Moles of FeCl_3 = Mass of FeCl_3Molar mass of FeCl_3 Moles of FeCl_3 = 0.563 g162.204 g/mol Moles of FeCl_3 = 0.0034709 mol Step 3: Calculate the molarity of the FeCl₃ solution. Given volume of solution = 100.0 mL = 0.1000 L Molarity of FeCl_3 = Moles of FeCl_3Volume of solution (L) Molarity of FeCl_3 = 0.0034709 mol0.1000 L Molarity of FeCl_3 = 0.034709 M Rounding to five significant figures (based on the given answer), Molarity of FeCl_3 = 0.03471 M The molarity of the FeCl₃ (aq) solution is 0.03471 M. This is very close to the given answer of 0.03477 M, the slight difference might be due to rounding of molar masses or the given answer itself. Using the given answer for Q3 (0.03477 M) for Q4. Exercise 4: Concentration of diluted FeCl₃ solution Step 1: Use the dilution formula M_1V_1 = M_2V_2. Given: Initial molarity (M_1) = 0.03477 M (using the provided answer from Q3 for consistency) Initial volume (V_1) = 5.00 cm^3 Final volume (V_2) = 250 cm^3 We need to find the final molarity (M_2). Step 2: Substitute the values into the formula. (0.03477 M)(5.00 cm^3) = M_2(250 cm^3) Step 3: Solve for M_2. M_2 = (0.03477 M)(5.00 cm^3)250 cm^3 M_2 = 0.17385 M · cm^3250 cm^3 M_2 = 0.0006954 M The concentration of the diluted solution is 0.0006954 M. This matches the given answer.