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Trial one Trial two Trial three Volume of oxalic acid (mL) 10.1 mL 10.2 mL 10.1...

Trial one

Trial two

Trial three

Volume of oxalic acid (mL)

10.1 mL

10.2 mL

10.1 mL

Moles of oxalic acid (moles)

Initial volume of NaOH (mL)

18.7 mL

26.5 mL

34.6 mL

Final volume of NaOH (mL)

26.5 mL

34.6 mL

42.6 mL

Delivered volume of NaOH (moles)

7.8 mL

8.1 mL

8 mL

Moles of NaOH (moles)

Molarity of NaOH (M)

Average Molarity NaOH (M)

Concentration of oxalic acid (M) = 0.2567

Trial one

Trial two

Trial three

Volume of acetic acid sol’n (mL)

10

10

10

Moles of acetic acid sol’n (moles)

Initial volume of NaOH (mL)

2.8

15.65

28.45

Final volume of NaOH (mL)

15.65

28.45

41.2

Delivered volume of NaOH (mL)

12.85

12.8

12.75

Average Molarity of NaOH (M)

*See table 1*

Molarity of acetic acid (M)

Average molarity of acetic acid (M)

H2C2O4 + 2 Na(OH) --> 2 Na (C2O4) + 2H2O

(Mol. Wt. oxalic acid = 126.1 g/mol)

Titration of vinegar lab

Using this information, how can the moles of oxalic acid and the molarity of NaOH?

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Homework Answers

Answer #1

The formula that we need to know before solving the question-

No.ofmolesin) Molarity (M)= Volumeinlitres(V)

A) Oxalic acid and NaOH

FOR TRIAL ONE

Volume of oxalic acid = 10.1 mL = (10.1/1000) L -----(Given)

= 0.0101 L

Concentration of oxalic acid (Molarity) = 0.2567 M -----(Given)

No. of moles of oxalic acid = (0.0101).(0.2567) -----using equation(a)

= 0.00259267

= 0.0026 moles

Balanced chemical equation for oxalic acid and NaOH:-

H2C2O4(aq) + 2 NaOH(aq) -----> Na2C2O4(aq) + 2 H2O

It is evident from the balanced chemical equation that for 1 mole of oxalic acid, 2 moles of NaOH are required for titration.

Hence, for 0.0026 moles of oxalic acid:-

No. of moles of NaOH required = 2(0.0026)

= 0.0052 moles

Initial volume of NaOH = 18.7 mL

Final volume of NaOH = 26.5 mL

Volume of NaOH delivered = (Final volume of NaOH) - (Initial Volume of NaOH)

= 26.5 - 18.7

= 7.8 mL

= (7.8/1000) L

= 0.0078 L

Molarity of NaOH = (0.0052/0.0078) -----using equation(a)

= 0.666666667

= 0.6667 M

FOR TRIAL TWO

Volume of oxalic acid = 10.2 mL = (10.2/1000) L -----(Given)

= 0.0102 L

Concentration of oxalic acid (Molarity) = 0.2567 M -----(Given)

No. of moles of oxalic acid = (0.0102).(0.2567) -----using equation(a)

= 0.00261834

= 0.0026 moles

Balanced chemical equation for oxalic acid and NaOH:-

H2C2O4(aq) + 2 NaOH(aq) -----> Na2C2O4(aq) + 2 H2O

It is evident from the balanced chemical equation that for 1 mole of oxalic acid, 2 moles of NaOH are required for titration.

Hence, for 0.0026 moles of oxalic acid:-

No. of moles of NaOH required = 2(0.0026)

= 0.0052 moles

Initial volume of NaOH = 26.5 mL

Final volume of NaOH = 34.6 mL

Volume of NaOH delivered = (Final volume of NaOH) - (Initial Volume of NaOH)

= 34.6 - 26.5

= 8.1 mL

= (8.1/1000) L

= 0.0081 L

Molarity of NaOH = (0.0052/0.0081) -----using equation(a)

= 0.6419753086

= 0.6420 M

FOR TRIAL THREE

Volume of oxalic acid = 10.1 mL = (10.1/1000) L -----(Given)

= 0.0101 L

Concentration of oxalic acid (Molarity) = 0.2567 M -----(Given)

No. of moles of oxalic acid = (0.0101).(0.2567) -----using equation(a)

= 0.00259267

= 0.0026 moles

Balanced chemical equation for oxalic acid and NaOH:-

H2C2O4(aq) + 2 NaOH(aq) -----> Na2C2O4(aq) + 2 H2O

It is evident from the balanced chemical equation that for 1 mole of oxalic acid, 2 moles of NaOH are required for titration.

Hence, for 0.0026 moles of oxalic acid:-

No. of moles of NaOH required = 2(0.0026)

= 0.0052 moles

Initial volume of NaOH = 34.6 mL

Final volume of NaOH = 42.6 mL

Volume of NaOH delivered = (Final volume of NaOH) - (Initial Volume of NaOH)

= 42.6 - 34.6

= 8.0 mL

= (8.0/1000) L

= 0.008 L

Molarity of NaOH = (0.0052/0.008) -----using equation(a)

= 0.65

= 0.65 M

Average molarity of NaOH = (Molarity from trial 1 + Molarity from trial 2 + Molarity from trial 3)/3

= (0.6667 + 0.6420 + 0.6500)/3

= 0.6529

= 0.65 M

Average Molarity of NaOH is 0.65 M.

B) Acetic acid and NaOH

FOR TRIAL ONE

Volume of acetic acid = 10.0 mL = (10.0/1000) L -----(Given)

= 0.01 L

Average Molarity of NaOH = 0.65 M -----(taken from the above calculation of table 1)

Initial volume of NaOH = 2.8 mL

Final volume of NaOH = 15.65 mL

Volume of NaOH delivered = (Final volume of NaOH) - (Initial Volume of NaOH)

= 15.65 - 2.8

= 12.85 mL

= (12.85/1000) L

= 0.01285 L

No. of moles of NaOH deliverd = (0.01285).(0.65) -----using equation(a)

= 0.0083525

= 0.00835 moles

Balanced chemical equation for acetic acid and NaOH:-

C2H4O2(aq) + NaOH(aq) -----> C2H3O2Na(aq) + H2O(l)

It is evident from the balanced chemical equation that 1 mole of NaOH titrates 1 mole of acetic acid.

Hence, 0.00835 moles of NaOH titrates:-

No. of moles of acetic acid = 0.00835

= 0.00835 moles

Molarity of acetic acid = (0.00835/0.01) -----using equation(a)

= 0.835

= 0.8350 M

FOR TRIAL TWO

Volume of acetic acid = 10.0 mL = (10.0/1000) L -----(Given)

= 0.01 L

Average Molarity of NaOH = 0.65 M -----(taken from the above calculation of table 1)

Initial volume of NaOH = 15.65 mL

Final volume of NaOH = 28.45 mL

Volume of NaOH delivered = (Final volume of NaOH) - (Initial Volume of NaOH)

= 28.45 - 15.65

= 12.80 mL

= (12.80/1000) L

= 0.0128 L

No. of moles of NaOH deliverd = (0.0128).(0.65) -----using equation(a)

= 0.00832

= 0.0083 moles

Balanced chemical equation for acetic acid and NaOH:-

C2H4O2(aq) + NaOH(aq) -----> C2H3O2Na(aq) + H2O(l)

It is evident from the balanced chemical equation that 1 mole of NaOH titrates 1 mole of acetic acid.

Hence, 0.0083 moles of NaOH titrates:-

No. of moles of acetic acid = 0.0083

= 0.0083 moles

Molarity of acetic acid = (0.0083/0.01) -----using equation(a)

= 0.83

= 0.8300 M

FOR TRIAL THREE

Volume of acetic acid = 10.0 mL = (10.0/1000) L -----(Given)

= 0.01 L

Average Molarity of NaOH = 0.65 M -----(taken from the above calculation of table 1)

Initial volume of NaOH = 28.45 mL

Final volume of NaOH = 41.2 mL

Volume of NaOH delivered = (Final volume of NaOH) - (Initial Volume of NaOH)

= 41.2 - 28.45

= 12.75 mL

= (12.75/1000) L

= 0.01275 L

No. of moles of NaOH deliverd = (0.01275).(0.65) -----using equation(a)

= 0.0082875

= 0.0083 moles

Balanced chemical equation for acetic acid and NaOH:-

C2H4O2(aq) + NaOH(aq) -----> C2H3O2Na(aq) + H2O(l)

It is evident from the balanced chemical equation that 1 mole of NaOH titrates 1 mole of acetic acid.

Hence, 0.0083 moles of NaOH titrates:-

No. of moles of acetic acid = 0.0083

= 0.0083 moles

Molarity of acetic acid = (0.0083/0.01) -----using equation(a)

= 0.83

= 0.8300 M

Average molarity of acetic acid = (Molarity from trial 1 + Molarity from trial 2 + Molarity from trial 3)/3

= (0.8350 + 0.8300 + 0.8300)/3

= 0.8317

= 0.83 M

Average Molarity of acetic acid is 0.83 M.

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