CHEM.102 (part 2) The following link part(1) : https://www.chegg.com/homework-he

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CHEM.102 (part 2)

The following link part(1) :

https://www.chegg.com/homework-help/questions-and-answers/chem102-following-link-complate-lab-300-points-mpleted-please-q7276189

Determining the Buffer Capacity of Antacids Original volume of NaOH solution in buret Final volume of NaOH solution in buret Volume utilized (final-initial) Moles of NaOH utilized 24. Moles of H+ neutralized by antacid (moles HC1 -moles NaOH) 25. Moles of H+ neutralized per gram antacid #2 26. Average moles per gram antacid #2

Explanation / Answer

Calculate moles of HCl in part 1 :

run 1) 0.00206 moles          run 2) 0.0019 moles            run 3) 0.0019 moles

23) Moles of NaOH utilized: (Molarity x Volume in litres)

run 1) 0.1122 x 0.00307 L = 0.000344 moles

run 2) 0.1122 x 0.00207 L = 0.000302 moles

run 3) 0.1122 x 0.00240 L = 0.00024 moles

24) Moles of H+ neutralized ny antacid (moles of HCl - moles of NaOH)

run 1) 0.00206 - 0.000344 = 0.00171moles

run 2) 0.0019 - 0.000302 = 0.00159 moles

run 3) 0.0019 - 0.00026 = 0.00164 moles

25) Moles of H+ neutralized per gram antacid (moles of HCl/mass)

run 1) 0.00171 / 0.39 = 0.00438 mol/g

run 2) 0.00159 / 0.301 = 0.00528 mol/g

run 3) 0.00164/0.305 = 0.00537 mol/g

26) Average mol/g antacid : 0.015 mol/g

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