# 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)

(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

• 1

RE:

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

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