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TABLE 16.1 Mass and volume data for titration of primary standard acid and unkno

ID: 1042915 • Letter: T

Question

TABLE 16.1 Mass and volume data for titration of primary standard acid and unknown acid with sodium hydroxide. Run 1 Run 2 Run 3 A Mass of weighing paper (g) 38093370 3220 Mass of weighing paper and ??:04.2 H20 (g) Initial reading of buret (mL) .1 2.llo 1,05 26.7 24.8l 34.99 Run 1 Run 2 Run 3 , 2-.25 n,2-,252-25 Final reading of buret (mL) B Unknown number2 Mass of unknown acid suggested for titration (g) Number of reactive protons per molecule of unknown acid Mass of weighing paper (g) Mass of weighing paper and unknown 3813 .383%2 Initial reading of buret (mL) 2.89 Final reading of buret (mL)

Explanation / Answer

A)

Molar mass of H2C2O4.2H2O = 126.0654 g/mol

For Run 1,

Mass of H2C2O4.2H2O used (g) = 0.1734

Moles of H2C2O4.2H2O (mmol) = 0.1734 * 1000 / 126.0654

= 1.3755

Moles of protons reacted(mmol) = Moles of H2C2O4.2H2O (mmol) * No. of protons available for reaction

= 2.7510

Moles of OH- which reacted (mmol) = 2.7510 (Acid base neutralization)

Volume of NaOH solution (mL) = 22.9

Molarity of NaOH solution (M) = 2.7510 / 22.9

= 0.12 M

Similarly for run 2 & 3 we get,

A

Run 1

Run 2

Run 3

Mass of H2C2O4.2H2O used (g)

0.1734

0.2542

0.2203

Moles of H2C2O4.2H2O (mmol)

1.3755

2.0164

1.7475

No. of protons available for reaction

2

2

2

Moles of OH- which reacted (mmol)

2.7510

4.0328

3.4950

Volume of NaOH solution (mL)

22.9

22.65

33.94

Molarity of NaOH solution (M)

0.120

0.178

0.103

Average molarity of NaOH solution (M)

0.134

B)

For run 1,

Mass of unknown acid used (g) = 0.1838

Volume of NaOH solution (mL) = 17.79

Moles of NaOH which reacted (mmol) = 17.79 * Average molarity of NaOH solution (M) in part A

= 17.79 * 0.134 = 2.38

Moles of unknown acid (mmol) = Moles of NaOH which reacted (mmol) / No. of protons available for reaction

= 2.38 / 2 = 1.19

Molecular weight of unknown acid (g/mol) = Mass of unknown acid used (g) * 1000 / Moles of unknown acid (mmol)

= 0.1838 * 1000 / 1.19

= 155.39

Similarly for run 2 & 3 we get,

B

Run 1

Run 2

Run 3

Mass of unknown acid used (g)

0.1838

0.1942

0.1829

Volume of NaOH solution (mL)

17.79

18.70

17.50

Moles of NaOH which reacted (mmol)

2.38

2.50

2.34

No. of protons available for reaction

2

2

2

Moles of unknown acid (mmol)

1.19

1.25

1.17

Molecular weight of unknown acid (g/mol)

154.53

155.33

156.32

Average MW of unknown acid (g/mol)

155.39

A

Run 1

Run 2

Run 3

Mass of H2C2O4.2H2O used (g)

0.1734

0.2542

0.2203

Moles of H2C2O4.2H2O (mmol)

1.3755

2.0164

1.7475

No. of protons available for reaction

2

2

2

Moles of OH- which reacted (mmol)

2.7510

4.0328

3.4950

Volume of NaOH solution (mL)

22.9

22.65

33.94

Molarity of NaOH solution (M)

0.120

0.178

0.103

Average molarity of NaOH solution (M)

0.134