The chemiluminescence part of the experiment, we had to make four solutions labeled as ‘stock solution A, solution A, stock solution B, and solution B’. For the ‘stock solution A’ we put the luminol product, (0.242 g) in a 25 mL Erlenmeyer flask and dissolve it with 2 mL of 3M NaOH. Then we took 1 mL of the ‘stock solution A’ and diluted in 9 mL of water using a 50 mL beaker. Solution A. For the ‘stock solution B’ we mixed 4 mL of potassium ferricyanide solution and 4 mL of hydrogen peroxide solution using a 25 mL Erlenmeyer flask. Then for Solution B we diluted 4 mL of ‘stock solution B’ with 16 mL of water using a 50 mL beaker. The luminesce solution of the experiment, was obtain by diluting 3 mL of ‘solution A’ with 16 mL of water using a 150 mL beaker. Then in a dark room, we poured ‘solution …show more content…
A: The compound that is being oxidized in the luminol synthesis reaction is Na2S2O4, (Sodium Hydrosulfite. Since sodium hydrosulfite is acting as the reducing agent, in this experiment.
2. What is the purpose of acetic acid in the luminol synthesis reaction? How would the amount of the luminol product obtained at the end of the reaction be affected if the number of moles of NaOH added at the beginning of the reaction and the acetic acid added at the end of the reaction added were reversed? A: We know that with the addition of acetic acid in the luminol synthesis since it will protonate the dianion while reaction in the luminol synthesis so it will allow the product to separate completely from water. If we were to add NaOH at the beginning of the reaction, while adding acetic acid towards the end of the reaction, it wouldn’t affect it since we would still get a luminol produced.
3. In the luminol synthesis reaction, why was cold water used for rinsing the filtration step? How could the efficiency of the filtration be affected if hot water was used