Mass and volume data for titration of primary standard acid and unknown acid with sodium hydroxide?

part A

Run 1

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

Questions

Part A questions

Run1

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)

32.00 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

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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

(38.90 mL)

Questions

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

(38.90mL)

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

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Mass and volume data for titration of primary standard acid and unknown acid with sodium hydroxide?

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