Experiment 3

Experiment 3 Topic Purpose

Sample report

: Volumetric analysis − Acid base and redox : To determine the mass of sodium ethanedioate used to prepare a solution containing sodium ethanedioate and hydrated ethanedioic acid

Materials : KA 1 is a solution containing hydrated ethanedioic acid, H2C2O4.2H2O, and sodium ethanedioate KA 2 is a solution containing 3.5 g potassium manganate(VII) per dm3 KA 3 is a solution containing 1.7 g hydroxyl ions per dm3 KA 4 is 1.0 mol dm-3 sulphuric acid Phenolphthalein as indicator Apparatus : Two 25 cm3 pipettes and pipette fillers Two 50 cm3 burettes Six titration flasks Two retort stands and clamps One 50 cm3 measuring cylinder Two white tiles One wash bottle filled with distilled water One thermometer 0 °C–100 °C (by 1.0 °C) Procedure : (1) 25.0 cm3 of KA 1 is pipetted into a titration flask. (2) Two or three drops of phenolphthalein indicator is added and this solution is titrated with KA3. (3) The reading is recorded in the table below. (4) The titration is repeated to achieve accurate results.

Results

:

Titration

Rough

Final reading/cm3

Accurate 1

2

16.30

16.30

16.40

Initial reading/cm3

0.00

0.00

0.00

Volume of KA 3/cm3

16.30

16.30

16.40

3 -

(i) 25.0 cm3 of KA 1 required 16.35 cm3 of KA 3 for a complete reaction. (ii) Calculate your average titre value showing the suitable titre values that you use. 16.30 + 16.40 Average titre value = 2 = 16.35 cm3

Procedure : (a) 25.0 cm3 of KA 1 is pipetted into a titration flask. (b) 25 cm3 of KA 4 into the titration flask. (c) This solution is heated about 60 °C and titrated with KA 2 until a faint pink colour persists in the solution. (d) Record your readings in the table below. (e) The titration is repeated to achieve accurate results. Results

:

Titration

Rough

Final reading/cm3

Accurate 1

2

24.80

24.80

24.90

Initial reading/cm3

0.00

0.00

0.00

Volume of KA 1/cm3

24.80

24.80

24.90

3 -

(i) 25.0 cm3 of KA 1 required 24.85 cm3 of KA 2 for a complete reaction. (ii) Calculate your average titre value showing the suitable titre values that you use.

24.80 + 24.90 2 = 24.85 cm3

Average titre value =

Questions : (e) Calculate the concentration, in mol dm-3, of hydrated ethanedioic acid in solution KA 1. H2C2O4 + 2OH-
C2O42- + 2H2O 1 mole of ethanedioic acid reacts with 2 moles of hydroxyl ions ma va a = mb vb b while ma = concentration of ethanedioic acid in mol dm-3 mb = concentration of hydroxyl ions in mol dm-3 va = volume of ethanedioic acid in cm3 vb = volume of hydroxyl ions in cm3 a=1 b=2

Using the formula

ma × (25.00) 1 = 0.100 × 16.35 2 ma = 3.070 × 10−2 mol dm -3 ∴ The concentration of hydrated ethanedioic acid is 3.070 × 10−2 mol dm -3 .

Therefore,

(f) Calculate the mass of ethanedioate ions, C2O42-, in 1 dm3 of KA 1. 5C2O42- + 2MnO4- + 16H+
10CO2 + 2MN2+ + 8H2O 5 moles of C2O42- ions reacts with 2 moles of MnO4- ions mx v x x = my vy y while mx = concentration of C2O42- ions in mol dm-3 my = concentration of MnO4- ions in mol dm-3 vx = volume of C2O42- ions in cm3 vy = volume of MnO4- ions in cm3 x=5 y=2

Using the formula

3.5 = 2.215 × 10−2 mol dm -3 158 mx × (25.00) 5 Therefore, = −2 2.215 × 10 × 24.85 2 mx = 5.504 × 10−2 mol dm -3 ∴ The concentration of hydrated ethanedioic acid is 5.504 × 10−2 mol dm -3 .

my =

The mass of ethanediote ions in 1 dm3 of KA l = mx x molar mass of ethanediote ions = 5.504 × 10−2 × 88.0 = 4.844 g

∴ The mass of ethanediote ions in 1 dm3 of solution is 4.844 g (g) Calculate the concentration, in mol dm-3, of ethanedioate ions which originated from the sodium ethanedioate salt. The concentration of ethanediote ions from sodium ethanediote = mx − ma = (5.504 × 10−2 ) − (3.270 × 10−2 ) = 2.234 × 10−2 mol dm -3 ∴ The concentration of ethanediote ions from the sodium ethanediote salt is 2.234 × 10−2 mol dm -3 (h) Calculate the mass of sodium ethanedioate present in 1 dm3 of solution KA 1. The number of moles of Na2C2O4 = The number of moles of C2O42- in Na2C2O4 = 2.234 × 10−2 mol dm -3 The mass of Na2C2O4 = 2.234 × 10 −2 × molar mass of Na 2 C2 O 4 = 2.234 × 10−2 × 134.2 = 2.998 g

(i) Calculate the percentage of sodium ethanedioate in solution KA 1. The mass of H2C2O4.2H2O = 3.270 × 10−2 × molar mass of H 2 C 2 O 4 .2H 2 O = 3.270 × 10−2 × 126.0 = 4.120 g The percentage of Na2C2O4 in KA 1 solution 2.998 = × 100% = 42.12% 2.998 + 4.120 ∴ The percentage of sodium ethanediote is 42.12% .

(j) Why was solution KA 4 added to solution KA 1 before the titration? KA 4 solution was added to KA 1 solution to acidify the aqueous potassium manganate(VII) so that it can act as an oxidising agent. Conclusion

: The mass of sodium ethanediote used to prepare a solution containing sodium ethanediote and hydrated ethanedioic acid is 2.998 g.