TEST THE TEST - hkasme.org20Kei%20Wan... · solution and dilute sulphuric acid added will be in...

Hong Kong Chemistry Olympiad for Secondary Schools (2013 - 2014) TEST THE TEST

p.1

SHAU KEI WAN GOVERNMENT

SECONDARY SCHOOL

筲箕灣官立中學

TEST THE TEST

Group Members:

LAM YAT LONG, ALEX 林日朗

LEUNG TSUM TONG, THOMAS 梁雋堂

WONG TING HEI, SHERMAN 黃庭熙

YIK KAI HEI, ANDREW易啟希

Teacher Adviser:

Mr Lau Chun-On

P.1

Hong Kong Chemistry Olympiad for Secondary Schools (2013 - 2014)

Presentation material: photoalbum.ppt in CD


p.2

Abstract:

To determine the concentration of sodium hypochlorite, iodometric method is used

currently. However, is it the only method? The goal of this study is to determine

whether two new-designed tests can be possible substitutes of the traditional test.

In the traditional method, chlorine is collected through

iodide solution. The resultant solution is then titrated

against sodium thiosulphate. The concentration of

sodium hypochlorite can be found quite accurately. In

our studies, this tradition test will be used as a reference

to compare the feasibilities of the proposed tests.

In the traditional methods, multiple steps involve and it

leads to errors. So, our first proposed test is a

gravimetric test which only based on one reaction, the

redox reaction between hydrogen peroxide and sodium

hypochlorite. Our 1st test can achieve a test result close

to that from the traditional method.

Our second proposed test is a redox back titration which

requires the addition of iron(II) sulphate and followed

by the titration with potassium dichromate. This

proposed test has similar numbers of steps to the

original one with the use of different reagents.

According to the experimental result, this test has a

larger % difference than our 1st proposed test.

SHAU KEI WAN GOVERNMENT SECONDARY SCHOOL

LAM YAT LONG, ALEX LEUNG TSUM TONG, THOMAS

WONG TING HEI, SHERMAN YIK KAI HEI, ANDREW


p.3

Contents

Introduction P. 4

Principles and theories P. 5-8

Experiments P. 9-22

Data Analysis P. 23-33

Discussion P. 34-37

Further study P. 38-39

Conclusion P. 40

Reference P. 41

Acknowledgement P. 42


p.4

Introduction:

Normally, to determine the sodium hypochlorite (NaClO) content in bleach, chemists

test for the amount of chlorine in the hypochlorite compound with acid, potassium

iodide (KI) and sodium thiosulphate(Na2S2O3) However, the test is actually quite

complicated. So, is there any simpler yet accurate quantitative test for sodium

hypochlorite?

Our experiment is designed to investigate whether a gravimetric analytical test(with

hydrogen peroxide),and a redox test (with iron(II) sulphate, potassium dichromate)

can test the concentration of sodium hypochlorite accurately and replace the

traditional test. In the experiment, the traditional test will be carried out, which the

result will act as a benchmark for comparison of the accuracy of the two new

designed tests.

Objectives:

1. Determine the sodium hypochlorite concentration in bleach using the

traditional test.

2. Test for the feasibilities of using the two new-designed tests, i.e. the

gravimetric analytical test with hydrogen peroxide and the redox test (with

iron(II) sulphate, potassium dichromate), to determine the sodium

hypochlorite concentration in bleach.


p.5

Principles and theories:

(1) Principles and theories of the traditional test

Determine the sodium hypochlorite concentration with sulphuric acid, potassium

iodide, sodium thiosulphate and starch

The amount of sodium hypochlorite(NaClO) ingredient available can be determined

by allowing a certain quantity of it to react with excess acidified potassium iodide(KI)

solution, the liberated iodine(the amount is determined by the concentration of sodium

hypochlorite), is titrated against the standardized sodium thiosulphate(Na2S2O3)

solution.

When hypochlorite ions mix with excess acidified potassium iodide solution, aqueous

iodine is produced:

OCl- + 2I

- + 2H

+ I2 + H2O + Cl

-.

Aqueous iodine reacts with thiosulphate ions to form iodide ions:

I2 + 2 S2O32-

S4O62-

+ 2 I-

, which the amount of iodide ions can help finding the amount of hypochlorite ions.

Starch is used as indicator for the titration and is added when the solution in the

conical flask turns pale yellow. When iodine is present, the starch-iodine complex

turns the solution dark blue. But the addition of excess sodium thiosulphate solution

will decolorize the solution, as the thiosulphate ions reduce all the iodine back into

iodide ions. So, the first drop of sodium thiosulphate solution causing decolorization

marks the end point of the titration.

Since sodium thiosulphate cannot be a primary standard, standardization is needed to

be performing simultaneously with this traditional test. It can be standardized with

known amount of iodine(I2), which is produced by the redox reaction of known

amount of potassium iodate(KIO3) and excess acidified potassium iodide solution.

When potassium iodate is mixed with acidified potassium iodide, iodine and water is

produced:

6H+

+ IO3- + 5 I

- 3I2 + 3H2O .

Known amount of potassium iodate will be added as the limited reagent, while the KI

solution and dilute sulphuric acid added will be in excess amount. So known amount

of iodine is produced, and is titrated against sodium thiosulphate solution, with starch

solution as the indicator. Every molecule of aqueous iodine reacts with two

thiosulphate ions to form iodide ions:

I2 + 2 S2O32-

S4O62-

+ 2 I- .

When iodine is present, the starch-iodine complex turns the solution dark blue. But


p.6

the addition of sodium thiosulphate solution will decolourize the solution, as the

thiosulphate ions reduce the iodine back into iodide ions. The decolorization of the

solution marks the end point of the titration. Therefore, knowing the amount of iodine

presence and the volume of sodium thiosulphate solution added, the molarity of the

sodium thiosulphate solution can be calculated.

(2) Principles and theories of our proposed tests

TEST 1 Gravimetric analytical method

Determine the sodium hypochlorite concentration with hydrogen peroxide

Sodium hypochlorite is a common oxidizing agent which tends to accept two

electrons in alkaline medium to produce sodium chloride.

The half equation goes like this:

NaClO + H2O + 2e- 2OH

- + NaCl (E°=+0.90).

When hydrogen peroxide(H2O2) is mixed with sodium hypochlorite, redox reaction

occurs. Hydrogen peroxide acts as the reducing agent, and is oxidized to release

oxygen gas. The half equation goes like this:

H2O2 O2 + 2H+ + 2e

- (E°=-0.70).

The overall equation is:

NaClO(aq) + H2O2(aq) H2O(l) + NaCl(aq) + O2(g)

As a result, when excess hydrogen peroxide is added to bleach sample (sodium

hypochlorite solution), oxygen gas is released, that is, the mass of the resultant

solution decreases. Therefore, by calculating the weight difference of the solution, the

amount of oxygen gas released can be found, so does the amount of sodium

hypochlorite in the solution.

TEST 2 Redox back titration

Determine the sodium hypochlorite concentration with iron(II) sulphate, acidified

potassium dichromate and diphenylamine

Sodium hypochlorite is a common oxidizing agent which tends to accept two

electrons in alkaline medium to produce sodium chloride. The half equation goes like

this:

NaClO + H2O + 2e- 2OH

- + NaCl (E°=+0.90).

While iron(II) ion (Fe2+

), in this case, acts as a reducing agent, which lose electrons to

form iron(III) ions(Fe3+

). The half equation goes like this:


p.7

Fe2+

Fe3+

+ e- (E°=-0.77).

When sodium hypochlorite is mixed with iron(II) sulphate(FeSO4) at an alkaline

medium, redox reaction occurs with a standard voltage of +0.13V, which iron(II) ions

will be oxidized to iron(III) ions, and sodium hypochlorite will be reduced to sodium

chloride.

The overall chemical equation:

NaClO(aq) + H2O(l) + 2FeSO4(aq) NaCl(aq) + 2Fe3+

(aq/s)# +

2OH-(aq/s)* + 2SO4

2-(aq)

# The state is either aqueous or solid due to formation of both iron(III) chloride and iron(III)

hydroxide.

* The state is either aqueous or solid due to formation of both sodium hydroxide and iron(III)

hydroxide.

As a result, when excess iron(II) sulphate of known concentration is added to bleach

sample (sodium hypochlorite solution), iron(III) ions are produced. If the amount of

iron(III) ions is found, the amount of sodium hypochlorite in the solution can be

known. Since excess known concentration of iron(II) sulphate is added, finding out

the iron(II) ions remained can let us calculate the amount of iron(III) ions. Therefore,

the next step is to find out the amount of iron(II) ions by back titration with acidified

potassium dichromate.

Potassium dichromate(K2Cr2O7) is a common oxidizing agent in the school laboratory,

which tends to accept six electrons in acidic medium to produce chromium(III)

ions(Cr3+)

.

The half equation goes like this:

14H+ + 6e

- + Cr2O7

2- 2Cr

3+ + 7H2O (E°=+1.33).

While iron(II) ion is a reducing agent, which lose electron to form iron(III) ion. The

half equation goes like this:

Fe2+

Fe3+

+ e- (E°=-0.77).

It will be oxidized to iron(III) ions when it is titrated against acidified potassium

dichromate solution, with a standard voltage of +0.56V. Potassium dichromate is

orange in color, but when it is added to the solution containing iron(II) ions, it

changes into green due to the presence of chromium(III) ion, produced in the

reduction of dichromate ions.

The chemical equation is:

14H+(aq) + Cr2O7

2-(aq) + 6Fe

2+(aq) 2Cr

3+(aq) + 6Fe

3+(aq) + 7H2O(l).


p.8

However, the colour change from orange to green of dichromate is not sharp for end

point detection. Diphenylamine is added as a redox indicator, with the chemical

formula: (C6H5)2NH, which turns purple when being oxidized.

The equation is 2(C6H5)2 24H20N2 + 2H+ + 2e- :

2 +2H++ 2e

-

It is used as redox indicator in this back titration. It will become oxidized (turns

purple) immediately by acidified potassium dichromate when acidified potassium

dichromate solution is added during the titration. But soon become colorless again

due the reduction by iron(II) ions. When the first drop of excess acidified potassium

dichromate is added, intense color is produced due to oxidation of diphenylamine, and

this signals the end point of the titration.

Therefore, by calculating the amount of acidified potassium dichromate added, the

amount of iron(II) ions can be known and so as the iron (III) ions produced by sodium

hypochlorite. So the amount of sodium hypochlorite can be found.

However, iron(II) sulphate cannot be a primary standard due to it reaction with

oxygen in air. So standardization is needed to be performing simultaneously with this

test. It can be standardized by titration with known amount of acidified potassium

dichromate solution, with diphenylamine as redox indicator. The titration principle is

actually same as the redox titration above.


p.9

Experiments:

(1) Traditional Test

Test for the amount of ClO- ions with Potassium iodide solution KI(aq), Sodium

thiosulphate solution Na2S2O3(s) by the following equations:

NaClO(aq) + 2HCl(aq) NaCl(s) + Cl2(g) + H2O(l)

Cl2(g) + 2KI(aq) 2KCl(aq) + I2(aq)

I2(aq) +2Na2S2O3(aq) 2NaI(aq) +Na2S4O6(aq)

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Part 1 Standardization of sodium thiosulphate solution Na2S2O3(aq) by the following

equation:

IO3- + 6H

+ + 5I

- 3I2 + 3 H2O

I2 + 2 S2O32-

S4O62-

+ 2 I-

Reagents:

~2M H2SO4solution

KI solid

NaIO3 solid

~0.01M Na2S2O3 solution

Starch solution

Apparatus:

100.0 cm3 and 250.0 cm

3 volumetric flasks

Burette

Magnetic stirrer

Conical flasks

Beakers

Electronic balance

25.0 cm3 pipettes

Stand and clamps

Procedures:

Preparation of KI solution

1. Using an electronic balance, weigh about 14 g of KI(s)

2. Dissolve the KI(s) in a beaker containing about 150 cm3

of distilled water

3. Pour all the KI solution into a 250.0 cm3 volumetric flask.


p.10

4. Add distilled water until the meniscus reaches the graduation mark.

5. Stopper the flask and turn it upside down for several times.

Preparation of NaIO3 solution

1. Using an electronic balance, weigh about 2 g of NaIO3(s), record the mass

weighted.

2. Dissolve all the NaIO3(s) in a beaker containing about 150 cm3

of distilled water

3. Pour all the NaIO3 solution into a 250.0 cm3 volumetric flask.

4. Wash the inner of the beaker and pour all the washing into the volumetric flask for

several times.



7. Using a 25.0 cm3 pipette, collect 25.0 cm

3 of the NaIO3 solution from the

250.0 cm3 volumetric flask to a 100.0 cm

3 volumetric flask for dilution.



Titration

1. Using a measuring cylinder, collect about 15 cm3 of KI solution to a conical flask.

2. Using a 25.0 cm3 pipette, collect 25.0 cm

3 of the NaIO3 solution to the conical flask

3. Using a measuring cylinder, collect about 5cm3 of the H2SO4 solution to the conical

flask.

4. The color of the solution in the conical flask should have turned from colorless to

deep brown.

5. Place the conical flask on a magnetic stirrer, turn on the stirrer.

6. Titrate the solution in the conical flask against the Na2S2O3 solution, record the

initial reading of the burette.

7. When the color of the solution in the conical flask turns to pale yellow, add starch

as indicator, the color of the solution should turn into deep blue

8. Titrate until the color of the solution in the conical flask turns from deep blue to

colorless.

9. Record the final reading

10. Repeat the steps and perform the titration for four times.


p.11

Part 2

Test for the amount of NaClO(aq) with Potassium iodide solution KI(aq), Sodium

thiosulphate solution Na2S2O3(aq) by the following equations:



I2(aq) + 2Na2S2O3(aq) 2NaI(aq) + Na2S4O6(aq)

Experimental Set-up:

Explanation of the set-up:

Acidified KI solution will be transferred to the filtering flask that contains diluted

bleach solution by the syringe so that all Cl2 gas produced can immediately react with

the surrounding KI solution, which prevent the leakage of Cl2 gas which may make

the result of the test under-estimated. The closed system is to ensure that although

there may be some Cl2 gas that cannot react with the KI solution fast enough, it can

still be transferred to the boiling tube containing KI solution and react with the KI

solution, which can minimize the error of the test.

Reagents:

Bleach (NaClO solution)

~2M HCl solution

KI solid

~0.07M standardized Na2S2O3 solution

Starch solution

Fig.1


p.12

Apparatus:

100 cm3 syringe

Delivery tubes

Stopper with a hole

Stopper with two holes

Burette

Magnetic stirrer

Conical flasks

Boiling tubes

25.0 cm3 pipettes

Stand and clamps

250.0 cm3 volumetric flasks

Procedures:

1. Using an electronic balance, weight about 45 g of KI(s).

2. Dissolve the KI(s) in a beaker containing about 150 cm3

of distilled water.

3. Pour all the KI solution into a 250.0 cm3 volumetric flask.



6. Use a pipette to transfer 25.0 cm3of the bleach (NaClO solution) to another

250.0 cm3 volumetric flask.

7. Add water into the volumetric flask until the meniscus reaches the graduation mark.

8. Prepare the set up (see Fig. 1).

9. Use a pipette to transfer 25.0 cm3of the diluted NaClO solution from the volumetric

flask to the filtering flask of the set up.


p.13

10. Add about 30 cm3 1M KI solution into the boiling tubes of the set up.

11. Stopper the filtering flask and the boiling tube of the set up.

12. Mix about 50cm3 1M KI solution with ~50cm

3 2M HCl in a beaker.

13. Using the syringe, collect the KI solution prepared in step 12, connect the rubber

tubing and add the solution to the filtering flask containing bleach of the set up

slowly.

14. The color of the solution in the filtering flask should become reddish brown in

color.

15. Collect all the solution in a 250.0 cm3 volumetric flask, use distilled water to wash

the container and pour all the washing into the 250.0 cm3 volumetric flask.

16. Add water into the volumetric flask until the meniscus reaches the

graduation mark.

17. Use a 25.0 cm3 pipette twice to collect 50.0 cm

3 of the solution from the

volumetric flask to a small conical flask.

18. Place the conical flask on a magnetic stirrer, turn on the stirrer.

19. Titrate the solution in the conical flask against the standardized Na2S2O3 solution,

record the initial reading of the burette.


p.14

20. When the color of the solution in the conical flask turns to pale yellow, add starch

as indicator, the color of the solution should turn into deep blue.

21. Titrate until the color of the solution in the conical flask turns from deep blue to

colorless.

22. Record the final reading.

23. Repeat the steps and perform the titration for four times.


p.15

(2) Proposed test



Gravimetric Analysis of ClO¯ ion with H₂O₂ solution by the following equation:

NaClO(aq) + H2O2(aq) NaCl(aq) + O2(g) + H2O(l)

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━


Apparatus:

two electrical balances

beakers

25.0 cm3

pipette

Reagents:

Bleach (NaClO solution)

~ 0.5M Hydrogen Peroxide solution

Procedures:

1. Set up the apparatus (refer to Fig.2) .

2. Weigh an empty clean beaker with an electrical balance. Then, transfer 25.0 cm3

bleach solution to the beaker that put on the electrical balance, record the weight.

3. Pour 100 cm3 hydrogen peroxide into another beaker.

Fig. 2


p.16

4. Put the hydrogen peroxide-containing beaker on the electrical balance, record the

reading.

5. Add the hydrogen peroxide solution to the bleach-containing beaker on the

electrical balance, with a glass rod guiding the liquid.

6. Weigh the emptied hydrogen peroxide-containing beaker with an electrical

balance after pouring out the hydrogen peroxide.

7. Record the mass of the mixture every 30 seconds in the first 20 minutes.

8. After 20 minutes, record the weight of the mixture-containing beaker every five

minutes

until a steady reading is received.


p.17




Redox Back-Titration of NaClO with FeSO4, acidified K2Cr2O7 and (C6N5)2NH by the

following reactions:

1. Reduction of NaClO with FeSO4

NaClO(aq) + 2FeSO4(aq) + H2O(l)

NaCl(aq) + 2Fe3+

(aq) + 2SO42+

(aq) + 2OH-(aq)

2. Oxidation of Fe2+

ion with K2Cr2O7

14H+(aq)

+ K2Cr2O7(aq) + 6FeSO4(aq)

2Cr3+

(aq) + 6Fe

3+(aq)

+ 6SO4

2-(aq) + 2K

+(aq) + 7H2O(l)

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━


Fig. 3


p.18

Apparatus Required:

250 cm3 conical flasks

a 50 cm3 burette

a stand and burette clamp

10.0 cm3 and 25.0 cm

3pipettes

a 250.0 cm3 volumetric flask

Beakers

Reagent Required:

Household bleach( same bottle as for other tests)

FeSO4 solution,

i.e. dissolving about 20g FeSO4 •7H2O into approximately 500 cm3 water

~0.014M K2Cr2O solution(acidified),

i.e. dissolving about 1g anhydrous K2Cr2O7 into a 250.0 cm3 volumetric flask with


p.19

~100 cm3 deionized water and ~150 cm

3 2M H2SO4 solution

Diphenylamine (as a redox indicator, melted in hot concentrated sulphuric acid)

Procedure:

Part 1: Standardization of FeSO4

1. Setup the apparatus (refer to Part 1 of Fig.3) .

2. Transfer 18.0 cm3 FeSO4 into the conical flask.

3. Add a few drops of redox indicator into the solution .


p.20

4. Record the initial reading of the acidified K2Cr2O7 solution.

5. Titrate the mixture against the acidified K2Cr2O7 until the solution just change

from green to dark purple.

6. Record the amount of acidified K2Cr2O7 solution added.

7. Repeat the procedure at least three times to obtain an accurate data of titrant

required. Repeat the procedures 1-7 with 10.0 cm3 FeSO4 used in step 2.

Part 2: Reduction of NaClO with FeSO4

1. Wash a 25.0 cm3 pipette first with distilled water and then with FeSO4.

Wash a 25.0 cm3 pipette first with distilled water and then with bleach.

2. Transfer 25.0 cm3 bleach solution into a 250.0 cm

3 volumetric flask.



5. Transfer 3.4 cm3

diluted bleach into a conical flask using a 10.0 cm3 pipette that

was washed with the diluted bleach.

6. Transfer 25 cm3 FeSO4 Solution into the same conical flask.


p.21

7. Wait 5 minutes for reaction to occur until red precipitate formed (i.e. Fe (OH)3).

Part 3: Oxidation of Fe2+

ion with K2Cr2O7 solution:

1. Wash the burette first with deionized water, subsequently with acidified K2Cr2O7

solution.

2. Set up apparatus (refer to the part 3 of Fig.3)

3. Place the conical flask of resultant mixture under the burette and add a small

amount of redox indicator into the mixture


p.22

4. Record the initial reading of the acidified K2Cr2O7 solution

5. Titrate the mixture against the acidified K2Cr2O7 until the mixture just change

from green to dark purple .

6. Record the amount of acidified K2Cr2O7 solution added

7. Repeat the procedure at least three times to obtain an accurate data of titrant

required


p.23

Data and data analysis:

(1) Traditional test

Determine the sodium hypochlorite concentration with sulphuric acid, potassium

iodide, sodium thiosulphate and starch

Equation involves:




━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Part 1 Standardization of sodium thiosulphate solution Na2S2O3(aq) by the following

equation:

IO3- + 6H

+ + 5I

- 3I2 + 3 H2O

I2 + 2 S2O32-

S4O62-

+ 2 I-

Mass of NaIO3(s) used /g 2.078

Number of mole of NaIO3= Mass of NaIO3(s) used / Molar mass of NaIO3(s)

= (2.078) / (197.8924)

=0.010500655mol

All NaIO3(s) is dissolved in 250.0 cm3 solution.

Molarity of the NaIO3 solution = Number of mole of NaIO3 / Volume

= (0.010500655) / (250/1000)

= 0.042002623 mol dm-3

25.0 cm3 of the NaIO3 solution is diluted to 100.0 cm

3

Number of mole of NaIO3 in the diluted solution=Molarity of the initial NaIO3

solution × Volume used

=(0.042002623) × (25/1000)

=1.050065591×10-3

mol

Molarity of the diluted NaIO3 solution= Number of mole of NaIO3 used for dilution /

Volume

= (1.050065591×10-3

) / (100/1000)

=0.010500655 mol dm-3


p.24

25.0 cm3 of the diluted solution is used for titration

Number of mole of NaIO3 used = Molarity of the diluted NaIO3 solution × Volume

used

= (0.010500655) × (25/1000)

= 2.625163978×10-4

mol

Titration data

Volume of Na2S2O3 used Trial 1st

2nd

3rd

Initial reading /cm3

1.60 23.50 3.70 12.40

Final reading /cm3

23.50 45.40 25.60 34.30

Difference /cm3

21.90 21.90 21.90 21.90

By IO3- + 6H

+ + 5I

- 3I2 + 3 H2O

Number of mole of I2 formed = Number of mole of NaIO3 × 3

= 2.625163978×10-4

× 3

=7.875491934×10-4

mol

By I2 + 2 S2O32-

S4O62-

+ 2 I-

Number of mole of Na2S2O3 reacted = Number of mole of I2×2

= 7.875491934×10-4

×2

=1.575098387×10-3

mol

Molarity of Na2S2O3 solution = Number of mole of Na2S2O3 / volume used in titration

= (1.575098387×10-3

) / (21.90/1000)

= 0.0719223 mol dm-3

Molarity of Na2S2O3 solution: 0.0719223 mol dm-3


p.25

Part 2

Test for the amount of NaClO(aq) with Potassium iodide solution KI(aq), Sodium

thiosulphate solution Na2S2O3(aq) by the following equations:

NaClO(aq) + 2HCl(aq) NaCl(s) + Cl2 (g) + H2O(l)



━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Titration data

Volume of Na2S2O3 used Trial 1st

2nd

3rd

Initial reading /cm3

1.50 10.10 18.70 27.30

Final reading /cm3

10.10 18.70 27.30 35.90

Difference /cm3

8.60 8.60 8.60 8.60

Reagent data

Molarity of Na2S2O3 solution: 0.0719223 mol dm-3

Number of mole of Na2S2O3 used in titration = Volume used in titration × Molarity of

Na2S2O3 solution

= (8.60 / 1000) × (0.0719223)

= 6.185317866 × 10-4

mol

Let MNaClO be the molarity of NaClO in bleach.

25 cm3 of bleach solution is collected and diluted to 250 cm

3

Number of mole of NaClO in the diluted bleach solution = molarity of NaClO in

bleach × Volume used

= MNaClO × (25/1000)

= 0.025 MNaClO

Molarity of NaClO in the diluted bleach solution = Number of mole of NaClO /

Volume

= 0.025 MNaClO / (250/1000)

= 0.1 MNaClO

25 cm3 of the diluted solution is used for reaction

Number of mole of NaClO reacted = Molarity of NaClO in the diluted bleach solution

× Volume used

= 0.1 MNaClO × (25/1000)

= 0.0025 MNaClO


p.26

By NaClO (aq) + 2HCl (aq) NaCl(s) + Cl2 (g) + H2O (l)

NaClO is reacted to Cl2

Number of mole of Cl2 formed = Number of mole of NaClO reacted

= 0.0025 MNaClO

By Cl2(g) + 2KI(aq) 2KCl(aq) + I2(aq)

Number of mole of I2formed = Number of mole of Cl2

= 0.0025 MNaClO

The solution of I2 formed was diluted to 250 cm3 for titration, 50 cm

3 of the diluted

solution was used for titration each time.

Molarity of the diluted solution of I2 = Number of mole of I2 formed / Volume

= 0.0025 MNaClO / (250 / 1000)

= 0.01 MNaClO

Number of mole of I2 used in each titration = Molarity of the diluted solution of I2 ×

Volume used

= 0.01 MNaClO × (50 / 1000)

= 0.0005 MNaClO

By I2(aq) + 2Na2S2O3(aq) 2NaI(aq) + Na2S4O6(aq)

Number of mole of Na2S2O3 used in titration = Number of mole of I2 × 2

= (0.0005 MNaClO ) × 2

= 0.001 MNaClO

∵As calculated before, number of mole of Na2S2O3 used in titration = 6.185317866 ×

10-4

mol

∴ 0.001 MNaClO = 6.185317866 × 10-4

MNaClO = 0.618531786 mol dm-3

∴ Molarity of NaClO in bleach = 0.618531786 mol dm-3

Molarity of NaClO in bleach : 0.618531786 mol dm-3


p.27

(2) Our proposed tests



H2O2(aq) + NaClO(aq) H2O(l) + NaCl(aq) + O2(g)

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Experimental Data

Mass of the beaker with H2O2/g 202.682

Mass of the empty beaker after pouring H2O2/g 100.859

Mass of 100 cm3 H2O2 solution/g 101.823

Mass of the empty beaker/g 108.629

Mass of the beaker with bleach/g 133.763

Mass of 25 cm3 bleach /g

25.134g

Sum of the mass of the above reactants with the

beaker/g

235.586


p.28

Time / s Mass of the reacting

mixture/g

Time / s Mass of the reacting

mixture/g

0 235.586 1680 235.469

30 235.574 1800 235.460

60 235.57 1920 235.453

90 235.567 2040 235.446

120 235.565 2160 235.44

150 235.562 2280 235.434

180 235.56 2400 235.429

210 235.558 2520 235.423

240 235.556 2640 235.419

270 235.554 2760 235.413

300 235.552 2880 235.407

330 235.55 3000 235.401

360 235.548 3300 235.39

390 235.547 3600 235.375

420 235.546 3900 235.362

450 235.544 4200 235.351

480 235.543 4500 235.292

510 235.541 5100 235.265

540 235.54 5700 235.247

570 235.538 6300 235.22

600 235.532 7500 235.19

720 235.526 7800 235.174

840 235.517 8100 235.16

960 235.51 8400 235.151

1080 235.503 9000 235.136

1200 235.496 9900 235.109

1320 235.488 10500 235.088

1440 235.481 10800 235.077

1560 235.474


p.29

By NaClO + H2O2 NaCl + H2O + O2

Change in mass of the solution = Mass of O2 produced

Mass of O2 produced = 235.586 – 235.077

= 0.509 g

Number of mole of O2 produced = Mass of O2 produced / Molar mass of O2

= (0.509) / (32.0)

= 0.01590625 mol

∴ Number of mole of NaClO = Number of mole of O2 produced

= 0.01590625 mol

Molarity of NaClO = Number of mole of NaClO / Volume of the NaClO solution used

= (0.01590625) / (25/1000)

= 0.63625 mol dm-3



p.30




Equation involves:

1. Reduction of NaClO with FeSO4


NaCl(aq) + 2Fe3+

(aq) + 2SO42+

(aq) + 2OH-(aq)

2. Oxidation of Fe2+

ion with K2Cr2O7

14H+(aq) + K2Cr2O7(aq) + 6FeSO4(aq)

2Cr3+

(aq) + 6Fe

3+(aq)

+ 6SO4

2-(aq) + 2K

+(aq) + 7H2O(l)

14H+(aq) + Cr2O7

2-(aq) + 6Fe

2+(aq) 2Cr

3+(aq) + 6Fe

3+(aq) + 7H2O(l)

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Part 1 Standardization of iron(II) sulphate solution

19.993g of iron(II) sulphate (heptahydrate) was dissolved into 500 cm3 of water, and

1.017g of anhydrous potassium dichromate was dissolved into 250 cm3 of acidic

solution.

Molarity of acidified potassium dichromate solution = (1.017/ 294.185)/0.25

W

= 0.013828033 mol dm-3

First set of data collection

18 cm3 of iron(II) sulphate solution was titrated against 0.013828033 M acidified

potassium dichromate solution.

Volume of K2Cr2O7 solution used Trial 1st 2

nd 3

rd 4

th

Initial burette reading /cm3 1.00 6.00 7.10 9.20 8.60

Final burette reading /cm3 31.20 35.50 36.10 38.50 37.80

Difference /cm3 30.20 29.50 29.00 29.30 29.20

Average volume of acidified potassium dichromate solution added

= (29.5+29+29.3+29.2)/4

= 29.25 cm3

= 0.02925 dm3


p.31

Number of moles of potassium dichromate used

= 0.013828033 × 0.02925

= 0.0004044699653 mol

Number of moles of iron(II) sulphate in 18 cm3 of iron(II) sulphate solution

= 0.0004044699653 × 6

= 0.002426819792 mol

Therefore,

Molarity of iron(II) sulphate solution

= 0.002426819792 / 0.018

= 0.134823321 mol dm-3

Second set of data collection

10 cm3 of iron(II) sulphate solution was titrated against 0.013828033 M acidified

potassium dichromate solution.

K2Cr2O7 solution used Trial 1st 2

nd 3

th

Initial burette reading /cm3 2.50 2.80 19.30 3.10

Final burette reading /cm3 19.40 19.30 35.10 19.70

Difference /cm3 16.90 16.50 16.80 16.60


= (16.5+16.8+16.6) / 3

= 16.633333 cm3

= 0.016633333 dm3


= 0.013828033 × 0.016633333

= 0.0002300062822 mol

Number of moles of iron(II) sulphate in 18 cm3 of iron(II) sulphate solution

= 0.0002300062822 × 6

= 0.001380037693 mol

Therefore,


= 0.001380037693 / 0.010

= 0.138003769 mol dm-3


p.32

Finally, take the average of the two results


= (0.134823321+ 0.138003769) / 2

= 0.136413545 mol dm-3

Part 3 Back titration of NaClO

3.4 cm3 of bleach (10 times diluted) was added to 25 cm

3 iron(II) sulphate solution.

Then, the solution was titrated with 0.013828033 M acidified potassium dichromate

solution.

K2Cr2O7 solution used Trial 1st 2

nd 3

th 4

th 5

th

Initial burette reading /cm3 3.40 2.10 2.30 1.50 0.70 0.60

Final burette reading /cm3 40.20 38.90 38.90 37.90 37.30 37.20

Difference /cm3 36.80 36.80 36.60 36.40 36.60 36.60


= (36.8+36.6+36.4+36.6+36.6) / 5

= 36.6 cm3

= 0.0366 dm3


= 0.013828033 × 0.0366

= 0.0005061060078 mol

Number of moles of iron(II) sulphate remained after mixing with NaClO

= 0.0005061060078 × 6

= 0.003036636047 mol

Number of moles of iron(II) sulphate in 25 cm3 iron(II) sulphate solution

originally

= molarity × volume

= 0.136413545 × 0.025

= 0.003410338625 mol

Number of moles of iron(II) sulphate being oxidized

= 0.003410338625 – 0.003036636047

= 0.000373702578 mol


p.33

Since 1 hypochlorite ion was reacted with 2 iron(II) ions,

Number of moles of sodium hypochlorite in 3.4 cm3 10-times-diluted bleach

= 0.000373702578 / 2

= 0.000186851289 mol

And so,

Molarity of sodium hypochlorite in 10-times-diluted bleach

= 0.000186851289 / 0.0034

= 0.054956261 mol dm-3

By M1V1=M2V2

Molarity of NaClO in the original bleach : 0.54956261 mol dm-3

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

As reference to the traditional test using by chemists, we can calculate the percentage

differences of the two tests that we proposed by comparing with the traditional test.

Traditional test result:


Test 1 result:


Test 2 result


Percentage difference of test 1

= ( Test 1 result - Traditional test result ) / Traditional test result × 100%

= ( 0.63625 - 0.618531786 ) / (0.618531786) × 100%

≈ 2.864 %

Percentage difference of test 2

= ( Test 2 result - Traditional test result ) / Traditional test result × 100%

= ( 0.54956261 - 0.618531786 ) / (0.618531786) × 100%

≈ -11.150 %


p.34

Discussions:

(1)Discussion of TEST 1 Gravimetric Analytical method

Determine the sodium hypochlorite concentration with hydrogen peroxide.

The result of the gravimetric test of the NaClO(~0.636M) is similar to the one in the

traditional test (~0.619M).

Throughout the observation and data recording, there is first a higher rate of mass fall

till 1000s.And the fall become less significant at the end of the measurement as the

rate of gas evolvement is as reflected from the mass lost and the instantaneous slope

of the graph. Such pattern implies that oxygen can be readily evolved when H2O2

react with NaClO and the rate of reaction slow down as time passes due to the

decrease in concentration of reactants. The lost mass can quite accurately reflect the

molarity of NaClO ion, when compared to the traditional test.

Furthermore, H2O2 react rigorous once it mixed with NaClO, as stated in the first 10

min on the graph. Therefore, we are required to measure the mass of solution used

separately before the mixing of reactants, in order to reduce the error in mass. By that,

two electronic balances are utilized from that consideration. We can therefore

accurately obtained the initial mass of reaction mixture.


p.35

Possible Error in the test

Concerning the gravimetric method, the error or differences could be resulted from

the prolonged observation interval of the experiment. Water in the mixture can easily

evaporate for the time being while oxygen evolved can re-dissolve in the water. This

may explain the sudden change in evolvement for the interval at about time t=4000s.

(2)Discussion of TEST 2 Redox Back titration



Comparing to the traditional test for ClO- , the test result over the concentration of

NaClO was significantly differs and have a result of lower concentration

(~0.619M>~0.550M) In terms of chemical principles applied, redox, same principle

applied as in traditional test. This can be resulted from the release of Cl2 during the

reaction between the diluted bleach and FeSO4. More iron(II) ion would be left and

concentration of bleach would be underestimated although we already allowed the

reaction to be performed under alkaline medium in order to reduce this error.

Choice of reagent:

Originally, MnO4- ion (EOA

o= +1.51)(in acidic medium) will be used as the oxidizing

agent to give a clear end point when compared to Cr2O7- ion. However, MnO4

- will

react with the product Cl- ion from the reduction of Cl2 (ERA

o= +1.36), as of the high

oxidation potential between the Cl- ion and MnO4

- (EOX

o= EOA

o-ERA

o=+0.15). This

will overestimate the use of permanganate as it reacts with both the excess FeSO4 and

the Cl- ion. Another drawback is that we have to standardize the permanganate

solution as it easily decompose into manganese(IV) oxide while sodium salt of

dichromate is so stable that it can be used as a primary standard.

As a result, we decided to utilize the Cr2O7- ion (EOA

o= +1.33) to act as the titrant in

the redox titration of Fe2+

ion (ERAo= -0.44) in FeSO4, as it will not react

spontaneously with the Cl- ion as of the small standard electrode potential

(EOX

o=

EOAo-ERA

o=-0.03). Such figures ensure the oxidation of FeSO4 by K2Cr2O7

Preparation of redox indicator:

In addition to the pre-test preparation of FeSO4, Diphenylamine, as known as

(C6N5)2NH), the redox indicator, have to be dissolved in concentrated sulphuric acid

instead of water due reagent’s low solubility in polar solvent. Concentrated sulphuric

acid can increase the solubility of diphenylamine so as to enhance the color intensity


p.36

for the end point given by the reduction of Cr2O7- ion. Sulphuric acid can also further

ensure the acidic environment for K2Cr2O7 to involve in oxidation.

Possible Error in the test

The major error that led to the underestimation of NaClO could be the formation of

Cl2 during mixing NaClO with FeSO4. During to the formation of iron(II)/(III)

hydroxide, the decrease in free OH- would decrease the alkalinity of the solution.

Upon mixing with NaClO, the ClO- ion may form HClO and may be reduced by the

Fe2+

to form Cl2 according to the following side reaction pathway:

2H+(aq) + 2HOCl(aq) + FeSO4(aq)

Fe3+

(aq/s) + SO42-

(aq) + Cl2(g) + 2H2O(l)

(Eo = ERA

o-EOA

o= +1.67-(+0.77) = +0.9)

If the Cl2 can be conserved completely in the container and react completely with the

iron(II) ion,

2Fe2+

+ Cl22Cl- + 2Fe

3+

the no. of moles of iron(II) ions left over the reaction would be the same as the

proposed pathway, i.e.


NaCl(aq) + 2Fe3+

(aq) + 2SO42+

(aq) + 2OH-(aq)

But in fact, the reaction of gaseous chlorine under a not so alkaline environment may

not be that efficient and it causes errors. Some Cl2 was formed and lost in the process,

allowing less FeSO4 to involve in the intended reaction

There is also an error, however, could lead to overestimation. The reduction of Fe2+

by

oxygen in air could result in experimental error as expected before experimental

work.

Before the test over NaClO, the FeSO4 utilized in the experiment requires

standardization. As FeSO4 in the solution will be oxidized by the oxygen in air readily

to form Fe2O3 (insoluble) and Fe2(SO4)3 (soluble), it will reduce the FeSO4 available

for subsequent reduction of NaClO and the state of excessing reactant for NaClO.

By that, the number of excess of FeSO4 is reduced and thus less K2Cr2O7 will be used

to titrate excess FeSO4. It will then led to overestimation of FeSO4 involved

this concern was stressed from the change in color from pale green to yellow for the

FeSO4 solution preparation.

As a result, for every FeSO4 solution preparation prior the test, approximately 15

-20ml of FeSO4 was titrated against K2Cr2O7 before the experiment twice in order to


p.37

obtain an accurate concentration for FeSO4 solution. However, from the result and

observation during solution preparation, the further standardization of FeSO4 can’t

remedy the fact of oxidization of FeSO4 under the presence of air. This led to a lower

amount of FeSO4 in excess and thus lowers the amount of K2Cr2O7 used. As a result,

it could result in overestimation.

Quantitative test which involves redox titration should utilize chemically more stable

standard solution, for instance, low reactivity with oxygen. The choice of iron(II) may

lead to some limitations for our 2nd

proposed test.

(3)Discussion on all test methods – feasibility of application of gravimetric

methods in every daily life

To summarize, gravimetric method for NaClO tends to have a greater accuracy and

ease on qualitative analysis, when compared to the redox method in our proposed test

and traditional test respectively.

Major advantage of the gravimetric test proposed is the minimized experimental

procedure and utilization of reagents. In fact, gravimetric analysis involves less

experimental preparation and procedure when compared to the redox tests tested

previously, say standardization is not required for the reagent involved, nor it required

multiple redox titrations. Fewer reagents and procedure reduce the amount of error

resulted from experimental estimation of apparatus. Multiple measurements and

reading taking can increase the experimental error significantly.

Another advantage of gravimetric test would be the ease of application. The reagent

utilized, H2O2 can be easily obtained when compared to other reagent in other test,

say Diphenylamine. This can allow easier and greater application of the test method

within daily life. Household electrical balance might be utilized as average

experimental electronic balance for gravimetric analysis.


p.38

Further study:

(1)Further improvements of our tests

In our experiments, the two new-designed tests could not get the exact value of

sodium hypochlorite concentration that found by the traditional test.

In the hydrogen peroxide test, evaporation of water and solvation of oxygen are

inevitable in our set-up. Although they seems are minor problems that will not hinders

the experimental operation, they will lower the accuracy of the test. To solve the

problem, it is suggested to cover the beaker with cotton wool that soaked in sodium

hydroxide solution, and place the experimental set-up in environment of dehydrated

air. Cotton wool that soaked in sodium hydroxide solution can absorb water that

evaporated from the mixture solution of sodium hypochlorite and hydrogen peroxide,

so that evaporated water will not escape from the system; placing the experimental

set-up in environment of dehydrated air can ensure the water absorbed is not from the

water vapor in air. These can help to eliminate the experimental error of mass loss due

to evaporation of water. Yet, it is quite difficult to conduct in ordinary laboratory.

Another solution is that we can increase the volume of reagent used. The percentage

of decrease due to mass of water evaporated would be much smaller than the decrease

of mass due to oxygen escaped.

In the potassium dichromate test, the result, i.e. sodium hypochlorite concentration, is

smaller than the sodium hypochlorite concentration that found by the traditional test.

A reasonable explanation is that there was loss of chlorine during the reaction of

sodium hypochlorite solution and iron(II) sulphate solution in a neutral or even

slightly acidic medium. Iron(II) ions will actually provide acidic environment by

absorbing hydroxide ion (forming Fe(H2O)5(OH)3+

complex). When excess amount of

iron(II) sulphate solution was mixed with slightly alkaline sodium hypochlorite

solution, a neutral or slightly acidic medium might be created. Hypochlorite ions

might had been turned into chlorine. Although chlorine can also reaction the same

way, some of them might bubble out from the solution. These will affect the

experimental result quite badly. To solve the problem, it is suggested to put a lid

or a stopper that perfectly fit the conical flask*, onto the conical flask that contain

bleach and iron(II) sulphate solution, so that the chlorine gas cannot escape from the

conical flask. After waiting for the reaction, put the liquid-tight conical flask upside

down for a few times and each for one minute. So the remaining chlorine gas can

react and the chlorine in hypochlorite can react completely.

*make sure the lid is really tight so that it won't bump out by the air pressure from the

flask inside. Also note that the flask should be strong enough to withstand the


p.39

increasing air pressure.

(2)Another new-designed test that have not been proposed

–the polarimetric analysis

Beside the two tests that we have conducted, there is still a test that is possible to test

for the sodium hypochlorite concentration. However, it cannot be conducted by us

since there is no pure (+) butan-2-ol or pure (-) butan-2-ol in our school laboratory.

This test involves the study of polarimetry. A chiral compound, butan-2-ol can be used

in this test. Structures of two enantiomers of butan-2-ol are shown below.

CH3CH3

OH

CH3CH3

OH

When polarized light passes through an optical isomer, also known as the enantisomer,

rotation of plane of polarized light occurs. When passing though the (+) form of an

enantisomer, the rotation is in clockwise direction. While passing though the (-) form

of an enantisomer, there will be anticlockwise rotation. When butan-2-ol is mixed

with sodium hypochlorite, it will be oxidized to butanone. Since butanone have no

enantisomer (the central carbon atom is only attaching three different groups, which

lack chirality due to the presence of plane of symmetry), it will not cause rotation of

plane of polarized light.

Structure of butanone:

CH3CH3

O

The experiment goes like this: prepare solutions of pure (+) butan-2-ol or pure (-)

butan-2-ol (can only choose one of them) of different concentrations, and then find

their 'angle of rotation of plane of polarization of light' with a polarimeter. Plot a

calibration curve of angles of rotation against concentrations of (+) / (-) butan-2-ol

times a constant, i.e. α = k c ( α is angle, k is constant , c is concentration). Then let a

known amount of butan-2-ol (in excess) to react with sample of sodium hypochlorite

solution (bleach). Then check for the angle of rotation of the butan-2-ol remains. By

referring to the graph plotted, the amount of butan-2-ol remained after reaction can be

found, and so as the butan-2-ol reacted.

Since C4H9OH + ClO- C4H8O + Cl

- + H2O ,

the amount of sodium hypochlorite can be calculated.

The idea is originally test the sodium hypochlorite concentration using glucose as


p.40

D-glucose is easy to get. D-glucose is an enantisomer of glucose which is favorable

for the polarimetric test. Glucose will react with sodium hypochlorite to form

gluconic acid. Yet, gluconic acid will also cause rotation of plane of polarization of

light. To solve this problem, two graph have to be plotted, one for different

concentrations of glucose and one for that of gluconic acid. But, another problem

appears, there is no pure gluconic acid in our school laboration for setting the

calibration curve. So we have to synthesis a solution of gluconic acid with known

concentration, by reacting known amount of glucose with some oxidizing agents.

However, the final problem appears, there is a chance that gluconic acid may be

further oxidized to dioic acid and not all oxidations stay at the gluconic level. This

will affect the result seriously, so the use of glucose was rejected at last. As

replacement of glucose-- pure (+) / (-) butan-2-ol cannot be found in our lab, this test

was finally abandoned.

Conclusion:

As calculated before, percentage difference of test 1 is around 2.864 % while

percentage difference of test 2 is around -11.150 %

By comparing the percentage difference of the two tests to the traditional, we can

conclude that the result of test one is nearer to the result of the traditional test. Test 1

may be a better substitute for determining the concentration of ClO- ions in solution.

However, test 2, comparing to test1, has a much higher percentage difference.

Therefore, it can’t be used to find the concentration of ClO- ions in solution unless

some improvements has been made.

Test 1 is gravimetric analysis, which has lesser procedures comparing to the

traditional test. Besides, the Na2S2O3 solution used in the traditional test needed to be

standardized. On the other hand, the H2O2 solution use in test 1 does not need to be

standardized, adding excess of it is praised instead. Not only can this enhance the

convenience of the test, but it can also avoid doing titration, which increases the

possible error. Therefore, we suggest that test 1 can replace the traditional test of

finding the amount of ClO- ions in solution.


p.41

References:

1. Walther F. Goebel, ON THE OXIDATION OF GLUCOSE IN ALKALINE

SOLUTIONS OF IODINE,J. Biol. Chem. 1927, 72:801-807,

http://www.jbc.org/content/72/2/801.full.pdf

2. C. Parameshwara Murthy, “University Chemistry, Volume 1” (2002), New

Age International, P.629-642

3. http://staff.buffalostate.edu/nazareay/che112/chromate.htm

4. http://wwwchem.uwimona.edu.jm/lab_manuals/c10expt31.html

5. www.hach.com/asset-get.download.jsa?id=7639984029

6. http://en.wikipedia.org/wiki/Hypochlorite


p.42

Acknowledgement:

We would like to express our heartfelt gratitude to Mr Lau Chun-On ,Chemistry

Teacher of SGSS and the teacher-advisor for our project, who advice and instruct us at

times of difficulties and during experimental set-back. We would also like to thank Mr.

Fung Wai-Ching, the Laboratory Technician of SGSS, for preparing the apparatus and

reagents for our experiment. At last, we were grateful to our school, SGSS that

provide us a superb site for scientific work and chance to participate in this

competition on behalf of the school

TEST THE TEST - hkasme.org20Kei%20Wan... · solution and dilute sulphuric acid added will be in...

Documents

Transcript of TEST THE TEST - hkasme.org20Kei%20Wan... · solution and dilute sulphuric acid added will be in...