Q Problem 17.41 6 of 27 Constants | Periodic Table How many milliliters of 8.50×102 MNaOH are required to titrate each of the following solutions to the equivalence point? Part A 45.0 mL of 9.50×102 MHNO3 V = mL Part B 35.0 mL of 8.50×102 MCH3COOH V = mL Request Answer Solution Ans. Part A: Moles of HNO3 in sample = Molarity x Volume of solution in liters = 9.50 x 10-2 M x 0.045 L = 0.004275 mol # Balanced reaction: HNO3(aq) + NaOH(aq) -----> NaCl(aq) + H2O(l) According to the stoichiometry of balanced reaction, 1 mol HNO3 is neutralized by 1 mol NaOH. So, Required moles of NaOH to reach endpoint = 0.004275 mol Now, Required volume of NaOH solution = Required moles of NaOH / Molarity = 0.004275 mol / (8.50 x 10-2 M) = 0.004275 mol / (8.50 x 10-2 mol / L) = 0.05029 L = 50.29 mL Part B: Moles of CH3COOH in sample = Molarity x Volume of solution in liters = 8.50 x 10-2 M x 0.035 L = 0.002975 mol # Balanced reaction: CH3COOH(aq) + NaOH(aq) -----> CH3COONa(aq) + H2O(l) According to the stoichiometry of balanced reaction, 1 mol CH3COOH is neutralized by 1 mol NaOH. So, Required moles of NaOH to reach endpoint = 0.002975 mol Now, Required volume of NaOH solution = Required moles of NaOH / Molarity = 0.002975 mol / (8.50 x 10-2 M) = 0.002975 mol / (8.50 x 10-2 mol / L) = 0.035 L = 35.0 mL.