REPORT SHEET EXPERIMENT Determining the Buffer Capacity of Antacids 25 Titration
ID: 493276 • Letter: R
Question
REPORT SHEET EXPERIMENT Determining the Buffer Capacity of Antacids 25 Titration of Antacids 1, Concentration of HCI 2. Concentration of NaoH 3. Mass of antacid #1 Run 1) Run 2) Run 3) .31 4. Original volume of HCI solution in buret 40 Run 2) Run 1) Run 3) 5. Final volume of HCI solution in buret Run 2) 18. Run 1) Run 3) mL 6. Volume utilized (final initial) Run 2) 21 S mu Run 1) Run 3) 7. Moles of HCl utilized Run 2) DL mol mol Run 1) Run 3) 8. Original volume of NaoH solution in buret Run 2) Run 1) 08 Run 3 9. Final volume of NaoH solution in buret Run 1) M m Run 2) mL Run 3) mLExplanation / Answer
Volume of NaOH used
Run 1) 6mL
Run 2) 6mL
Run 3) 6mL
11)
Moles of NaOH used
Molarity X volume used
Run 1) 1 X 0.006mL = 0.006 moles
Run 2) 1 X 0.006mL = 0.006 moles
Run 3) 1 X 0.006mL = 0.006 moles
12) Moles of H+
Run 1) Moles of HCl – Moles of NaOH = 0.022 – 0.006 = 0.016
Run 2) Moles of HCl – Moles of NaOH = 0.0215 – 0.006 =0.0155
Run 3) Moles of HCl – Moles of NaOH = 0.021 – 0.006 =0.015
13) Moles of H+ neutralized per gram antacid#1
Run 1) 0.016 / 0.32 = 0.05
Run 2) 0.0155 / 0.31 = 0.05
Run 3) 0.015 / 0.30 = 0.05
19)
Moles of HCl utilized = Molarity X volume
Run 1) 1 X 0.021= 0.021
Run 2) 1 X 0.021 = 0.021
Run 3) 1 X 0.022 = 0.022
23)
Moles of NaOH utilized = Molarity X volume
Run 1) 1 X 0.015= 0.015
Run 2) 1 X 0.010 = 0.010
Run 3) 1 X 0.013 = 0.013
24.
Moles of H+
Run 1) Moles of HCl – Moles of NaOH = 0.021 – 0.015 = 0.006
Run 2) Moles of HCl – Moles of NaOH = 0.021 – 0.010 =0.011
Run 3) Moles of HCl – Moles of NaOH = 0.022 – 0.013 =0.009
25)
Moles of H+ per gram antacid
Run 1) 0.006 / 0.291 = 0.0206
Run 2) 0.011 / 0.297 = 0.037
Run 3)0.009 / 0.306 =0.0294
26) average = 0.0206 + 0.037 + 0.0294 / 3 = 0.029
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