front 1 Would adding extra distilled water to the Erlenmeyer flask during the titration affect the calculated molarity of the N a O H solution? Explain.
| back 1 No, the added water does not affect the number of moles of N a O H or H C l added and the added water does not affect the volume of H C l or N a O H. Ex. From Brainly, https://brainly.com/question/37306660 amitkumar12
Final answer: Adding deionized water to an Erlenmeyer flask during a titration does not alter the number of moles of NaOH or HCI added and does not affect the volume of HCl or NaOH. Extra water dilutes the solution, but the amount of NaOH remains the same, keeping the titration process intact. Explanation: The answer to your question is C. No, the added deionized water does not affect the number of moles of NaOH or HCI added and the added water does not affect the volume of HCl or NaOH. This is because in a titration experiment, the number of moles of reactants is what's important, not the volume of the solution. When you add more deionized water to the Erlenmeyer flask, it would dilute the NaOH solution, but it wouldn't alter the amount of NaOH (the number of moles) in the solution. Thus, the titration's measurement is focused on the volume of solution needed to react completely with the substance in the flask, which is independent of the amount of water present. However, keep in mind that adding extra water may dilute the solution, which could make it harder to observe the colour change at the endpoint of the titration, especially for a weak acid, weak base titration. Learn more about Titration here: brainly.com/question/31271061 #SPJ11 |
front 2 Why is it important that air bubbles from the buret or valve be removed prior to starting the titration?
| back 2 The air bubbles may escape during the titration causing the final buret reading to be incorrectly high. |
front 3 Calculate the molarity of a NaOH solution if 24.70 mL of NaOH is titrated with 0.500 grams of KHP and 2.05 mL of 0.135 M HCl is required. | back 3 0.500g KHP x (1 mol KHP / 204.22g/mL KHP) = 0.00245 moles KHP = mol H+ 2.05 mL HCl x (1 L / 1000 mL) x (0.135 mol HCl / 1 L) = 0.000277 mol HCl = mol H+ added 0.00245 moles H+ from KHP + 0.000277 mol H+ from HCl = 0.00273 mol H+ = mol NaOH 0.00273 mol NaOH x (1 L / 24.70 ML NaOH) x (1000 mL / 1 L) = 0.111 M NaOH |