This biology question covers important biological concepts and processes. The step-by-step explanation below helps you understand the underlying mechanisms and reasoning.
ScanSolve AI Answer
Step 1: Calculate the average volume of NaOH used in the two titrations. V_médio\ NaOH = (V_1 + V_2)/(2) V_médio\ NaOH = 24.78 \, mL + 25.80 \, mL2 = 50.58 \, mL2 = 25.29 \, mL Step 2: Convert volumes to liters. V_NaOH = 25.29 \, mL = 25.29 × 10^-3 \, L = 0.02529 \, L V_HCl = 25 \, mL = 25 × 10^-3 \, L = 0.025 \, L Step 3: Calculate the number of moles of NaOH used at the equivalence point (1:1 stoichiometry with HCl). n_NaOH = c_NaOH × V_NaOH n_NaOH = 0.1073 \, mol/L × 0.02529 \, L First, 0.1073 × 0.02529 = 0.1073 × (0.025 + 0.00029) = (0.1073 × 0.025) + (0.1073 × 0.00029) 0.1073 × 0.025 = 0.0026825 0.1073 × 0.00029 = 0.000031117 n_NaOH = 0.0026825 + 0.000031117 = 0.002713617 \, mol Step 4: Since n_HCl = n_NaOH, c_HCl = n_HClV_HCl = n_NaOHV_HCl c_HCl = 0.002713617 \, mol0.025 \, L c_HCl = (0.002713617)/(0.025) = 0.10854468 \, mol/L Considering significant figures (4 significant figures), c_HCl = 0.1085 \, mol/L 0.1085 mol/L