Question

Question 4 3 pts A student titrated 5.00 mL of 50.00 mL reference solution with 0.692 M NaOH(aq) The information of the reference solution is given as Initial Chemical Mols ethyl 0.101 acetate ethanol 0.258 acetic 0.052 acid H20 1.034 HCI 0.075 The titration data for 4 trials is given by A student prepared 4 samples for titration and titrated the samples with NaOH. Here are the results: Volume Trial NaOH 21.48 21.43 22.53 4 21.39 Calculate the mols of acetic acid at equilibrium for the given data and report it in millimol (no units need to be entered). Use the average volume of NaOH rather than doing the calculation

Calculate the mols of acetic acid at equilibrium for the given data and report it in millimol (no units need to be entered). Use the average volume of NaOH rather than doing the calculation for each trial. (please show work)

Answer 1

The average volume of NaOH solution = (21.48 + 21.43 + 22.53 + 21.39) mL/4 = 21.7075 mL

Now, the no. of millimoles of NaOH = 21.7075 mL * 0.692 mmol/mL = 15.02159 mmol

Concept: At equilibrium, the no. of millimoles of acetic acid = the no. of millimoles of NaOH

Therefore, the no. of millimoles of acetic acid at equilibrium = 15.02159