Mass of weighing paper 0.4452
Mass of weighing paper and H2C2O4*2H2O 0.6812
Initial reading of buret (mL) 0
Final reading of buret (mL) 32.00
Part A questions
Mass of H2C2O4*2H2O used (g)
0.6812 – 0.4452 = 0.2360 grams
Moles of H2C2O4*2H2O (mol)
0.2360 grams @ 126.07 g/mol = 0.001872 moles acid
Number of protons available for reaction OH-
2 H+ / molecule acid
Moles of OH- which reacted *Mol) @ 2 NaOH & 1 H2C2O4 –>
0.001872 moles acid => twice => 0.003744 moles NaOH
Volume of NaOH solution used (mL)
Molarity of NaOH solution (M)
0.003744 moles NaOH / 0.03200 litres = 0.1170 Molar NaOH
Average molarity of NaOH (M)
you will need to plug the other two trialls through & use the average molarity in the next procedure’s questions
Part B Run 1
Mass of unknown acid suggested for titration (0.25 g)
3 reactive protons per molecule of unknown acid
Mass of weighing pape(0.4452g)
Mass of weighing paper and unknown (0.6952 g)
Initial reading of buret(0mL)
Final reading of buret
Part B questions Run1
Mass of unknown acid used for titration (g)
(0.6952 g) – (0.4452g) = 0.2500 g
Volume of NaOH solution used
Moles of NaOH which reacted (mol)
0.03890 litres @ 0.1170 Molar NaOH = 0.0045513 moles NaOH
Moles of unknown acid 1 H3X & 3 NaOH –> ….
0.0045513 moles NaOH @ 1 mole acidH3X / 3 moles NaOH = 0.001517 moles of acid
Molecular weight of unknown acid (g/mol)
0.2500 g acid / 0.0045513 moles NaOH = 54.93 g/mol
Average molecular weight of unknown acid (g/mol)
remember to do part A for each trial, average the molarity of NaOH,…
then use that average NaOH molarity for part B
This Site Might Help You.
Mass and volume data for titration of primary standard acid and unknown acid with sodium hydroxide?
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