Data of results
Titration table of KAO bleach with KI and H2SO4 against standard 0.07676M sodium thiosulphate
Average volume of sodium thiosulphate used
= (19.9 + 20.0 + 19.9) ÷ 3
= 19.93 cm3
20.1 cm3is rejected because this is just a trial.
Titration table of Best Buy bleach with KI and H2SO4 against standard 0.07676M sodium thiosulphate
Average volume of sodium thiosulphate used
= (14.1 + 14.2 + 14.2) ÷ 3
= 14.167 cm3
14.5 cm3 is rejected because this is just a trial.
Calculation
KAO
In the titration, iodine in the reaction mixture react with the sodium thiosulphate as the following equation: I2 + 2S2O32- → 2I- + S4O62-
Mole ratio of I2 to S2O32- is 1: 2
As 19.93 cm3 of sodium thiosulphate is used,
No. of mole of sodium thiosulphate = 0.07676M x 0.01993 dm3 = 0.0015298268 mol.
No. of mole of iodine = 0.0015298268 ÷ 2 = 0.0007649134 mol.
Iodine present in the reaction mixture is come from the reaction between bleach and potassium iodide and dilute sulphuric acid as the following equation:
OCl- + 2I- +2H+ → I2 + H2O + Cl-
Mole ratio of OCl- to I2 is 1:1
No. of mole of diluted sodium hypochlorite (25 cm3) = 0.0007649134 mol.
No. of mole of diluted sodium hypochlorite (250 cm3) = 0.0007649134 x 10
= 0.007649134 mol.
No. of mole of original sodium hypochlorite (10 cm3) = 0.007649134 mol.
Molarity of the original bleach = 0.007649134 ÷ 0.01 = 0.7649134 M
= 0.769 M (corr. to 3 sig. fig.)
Mass of the sodium hypochlorite in 10 cm3 bleach = 0.007649134 x (23+16+35.5)
= 0.569860483 g
Strength of the original bleach = 0.569860483 ÷ 0.01
= 56.9860483 g dm-3
= 57.0 g dm-3 (corr. to 3 sig. fig.)
Available bleach present in one bottle = 0.569860483 x 150
= 85.47907245 g
= 85.5 g (corr. to 3 sig. fig.)
Cost of bleach per gram = $13.9 ÷ 85.47907245 g = $ 0.163 (corr. to 3 sig. fig.)
Best Buy
In the titration, iodine in the reaction mixture react with the sodium thiosulphate as the following equation: I2 + 2S2O32- → 2I- + S4O62-
Mole ratio of I2 to S2O32- is 1: 2
As 14.167 cm3 of sodium thiosulphate is used,
No. of mole of sodium thiosulphate = 0.07676M x 0.014167 dm3
= 0.00108745892 mol.
No. of mole of iodine = 0.00108745892 ÷ 2 = 0.00054372946 mol.
Iodine present in the reaction mixture is come from the reaction between bleach and potassium iodide and dilute sulphuric acid as the following equation:
OCl- + 2I- +2H+ → I2 + H2O + Cl-
Mole ratio of OCl- to I2 is 1:1
No. of mole of diluted sodium hypochlorite (25 cm3) = 0.00054372946 mol.
No. of mole of diluted sodium hypochlorite (250 cm3) = 0.00054372946 x 10
= 0.0054372946 mol.
No. of mole of original sodium hypochlorite (10 cm3) = 0.0054372946 mol.
Molarity of the original bleach = 0.0054372946 ÷ 0.01 = 0. 54372946 M
= 0.544 M (corr. to 3 sig. fig.)
Mass of the sodium hypochlorite in 10 cm3 bleach = 0.0054372946 x (23+16+35.5)
= 0.405078447 g
Strength of the original bleach = 0.405078447 ÷ 0.01
= 40.50784477 g dm-3
= 40.5 g dm-3 (corr. to 3 sig. fig.)
Available bleach present in one bottle = 0.405078447 x 200
= 81.0156894 g
= 81.0 g (corr. to 3 sig. fig.)
Cost of bleach per gram = $ 11.9 ÷ 81.0156894 g = $ 0.147 (corr. to 3 sig. fig.)
Conclusion
Best buy is cheaper than KAO as cost of bleach per gram of Best Buy is $ 0.147 while cost of bleach per gram of KAO is $ 0.163.
Discussion
- Why potassium iodide is added to the bleach solution before dilute sulphuric acid?
If dilute sulphuric acid is first added to the bleach solution, the hypochlorite ions in the bleach solution will react with the hydrogen ions in the dilute sulphuric acid to form
Study Questions
- Why the KI has to be present in excess? If a student use much less than the specified quantity, what effect would there be in his results?
KI has to be present in excess because we need to make sure that all the sodium hypochlorite is reacted with potassium iodide to give out iodine. If a student use much less than the specified quantity, potassium iodide may not be excess in the reaction and not all the sodium hypochlorite is reacted. Therefore the number of mole of sodium hypochloite presents cannot be calculated from the number of mole of iodine formed.
- What is the function of the dilute sulphuric acid?
The equation of the reaction between the sodium thiosulphate and the potassium iodide is ClO- + 2 I- + 2 H+ → I2 + H2O + Cl-.
The equation shows that the reaction between sodium thiosulphate and potassium iodide need acid medium to proceed. H+, act as a catalyst, speed up the reaction.
-
Bleaching solution may be deteriorated for 2 reasons. One is the attack by CO2 in air according to the reaction: 2ClO- + CO2 → CO32- + H2O + Cl2
What is the other possible reason?
Sunlight is the other possible reason that makes bleaching solution deteriorated. Bleach solution contains hypochlorite ions (OCl-) as the active ingredient. The hypochlorite ions in bleach solution is not very stable and slowly decomposes:
2OCl- → 2Cl- + O2
The decomposition is speeded up by sunlight.
- Why the starch indicator should not be added too early?
Starch will react with iodine to form a dark blue aqueous complex. If starch is added too early, starch may be combined with iodine too strongly. It is because the time for the combination of starch molecule with iodine molecule is longer, and the complex ion will be more stable. As a result, the iodine is combined too strongly with starch and cannot react with sodium thiosulphate to form iodine and indicate the end point.
- Why are crystals of hydrated sodium thiosulphate though obtainable in a state of high purity, not used as a primary standard?
Primary standard is a solution of known concentration by dissolving known amount of solute in a given volume of solvent. It is because crystals of hydrated sodium thiosulphate are not stable under sunlight. Features of a primary standard include: high , low , low and , high (if used in and high . In other words, primary standard solution will remain standard after a reasonable time elapsed. Crystals of hydrated sodium thiosulphate are not used as a primary standard because they are efflorescent (i.e. to release the water of crystallization in dry air) and deliquescent (i.e. to absorb water from moist air and to dissolve in the water absorbed). In other words, it is necessary to standardize the Na2S2O3 solution. If the concentration of Na2S2O3 is calculated based on the mass of hydrated sodium thiosulphate crystals and the volume of the solution, a significant error would be caused.
-
As the brown I2/I- solution becomes colourless at the end-point, why do we bother to add starch indictor in the titration?
The addition of starch indicator at the end-point is to check that whether there is any iodine still present in the solution. When this titration is near to the end-point, the I2/I- solution will become very pale yellow that we may wrongly judge it as colourless. The addition of starch solution can clearly show the presence or absence of iodine in the solution as starch will form a dark blue aqueous complex with iodine.
-
Why do thiosulphate ions only reacts with I2 and not H+ in the experiment?
Thiosulphate ions only react with I2 and not H+ in the experiment because I2 is a stronger oxidizing agent than. Therefore thopsulphate ions is more likely to react with I2 than H+.
- Explain the action of bleaching.
Colour in most and is produced by , such as , which contain . Household bleach solution is a kind of oxidizing bleach due to the hypochlorite ions. They react with many organic and inorganic compounds. An oxidizing bleach works by breaking the that make up the chromophore. This changes the molecule into a different substance that either does not contain a chromophore, or contains a chromophore that does not absorb .
End of Report