PROCEDURE : place 100cm3 of HCL into a 250cm3 conical flask. Add 5cm3 of methyl ethanoate into the flask.Start the stopwatch.Pipette immediately 5cm3 of solution from the flask and transfer the sample to a 250cm3 titration flask containing 100cm3 of ice water.note the time.when all the 5.0cm3 is transferred, add 1 or 2 drops of phenolpthalein and titrate this solution immediately with NaOH.Repeat the procedure for sample 2,3 and 4 withdrwan from the 250cm3 conical flask in the specific time intervals 10,20 and30 minutes. Repeat procedure above but substitute HCL with distilled water to prepare sample 5. Repeat the procedure for sample 5 after 30 minutes have elapsed.
(A) write a balaced equation for the hydrolysis if methyl ethanoate
(B)What is the purpose of titrating the sample of reaction mixture (HCL + methyl ethanoate) with NaOH.
(C)Why was sample of reaction mixture added to 100cm3 of ice water before titration
(D)Sketch an expected graph for the hydrolysis of the ester conducted at the same temperature but with 0.1 mol dm3 ethanoic acid.Explain your answer.
(E)Based on the results of the experiments for samples 4 and 5, state role of hydrochloric acid in the experiments
My Answers:
(A)CH3COOCH3(l) + H20(l)-->CH3COOH(l) + CH3OH(l)
(B) To provide the OH- ions, so it will react with H+ from HCl to form H20 molecules.
(C) So the methly ethanoate can hydrolysis the water molecules to form ethanoic acid and methanol
(D) curve graph. this is because ethanoic acid is a weaker acid compare to hydrochloric acid. less sodium hydroxide is needed due to the partial dissociation of H+ ions to react.