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Rinse pipette with bleach sample, and then use it to transfer 10cm3 of the bleach to the volumetric flask.
- Add deionized water into the flask until the level is within 1cm of the graduation mark. Then use a dropper to bring the water level to the mark
- Insert stopper and invert flask for about ten times to ensure thorough mixing.
Titration of diluted bleach solution
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Rinse the pipette with diluted bleach solution, use the pipette and pipette filter to transfer 25cm3 of the solution to a conical flask.
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Measure approximately 10cm3 0.5M potassium iodide, 10cm3 2M sulphuric acid using the measuring cylinder, then add them to the diluted bleach solution.
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Rinse the burette by 0.0562M sodium thiosulphate (Na2S2O3), then fill it with the same solution. Mark the initial reading.
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Start to titrate the bleach solution with Na2S2O3 until the colour of bleach turns to pale yellow. (it means that end-point is near)
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Add about 1cm3 of starch indicator to the solution, and the colour of solution turns dark blue.
- Continue titration until the blue colour discharges completely. Mark the final burette reading.
- Repeat the process from step 1 for at least 4 times.
Data
Experiment Data
Mean: 26.455cm3
Statistics Data of the three bands of beaches:
Calculation
Equations:
OCl－(aq) + 2I－(aq) + 2H＋(aq) → I2(aq) + H2O(l) + Cl－(aq)
I2(aq) + 2 S2O32－(aq) → 2 I－(aq) + S4O62－(aq)
a) Concentration of each bleach (by Statistics data):
Kao:
Volume of Na2S2O3 used: 25.50cm3 / 1000 = 0.02550dm3
No. of mole of Na2S2O3 used: 0.0562M x 0.02550dm3 = 0.0014331mol
No. of mole of I2 = No. of mole of Na2S2O3 / 2
= 0.0014331 / 2
= 0.00071655mol
No. of mole of OCl－ in 25cm3 diluted bleach = No. of mole of I2
= 0.00071655 mol
Conc. of original bleach = 0.00071655mol x (250/25) / 0.01dm3
= 0.71655mol dm－3
NoFrills:
Volume of Na2S2O3 used: 26.00cm3 / 1000 = 0.02600dm3
No. of mole of Na2S2O3 used: 0.0562M x 0.02600dm3 = 0.0014612mol
No. of mole of I2 = No. of mole of Na2S2O3 / 2
= 0.0014612 / 2
= 0.0007306mol
No. of mole of OCl－ in 25cm3 diluted bleach = No. of mole of I2
= 0.0007306mol
Conc. of original bleach = 0.0007306mol x (250/25) / 0.01dm3
= 0.7306 mol dm－3
Brand X:
Volume of Na2S2O3 used: 25.30cm3 / 1000 = 0.02530dm3
No. of mole of Na2S2O3 used: 0.0562M x 0.02530dm3 = 0.00142186mol
No. of mole of I2 = No. of mole of Na2S2O3 / 2
= 0.00142186 / 2
= 0.00071093mol
No. of mole of OCl－ in 25cm3 diluted bleach = No. of mole of I2
= 0.00071093mol
Conc. of original bleach = 0.00071093mol x (250/25) / 0.01dm3
= 0.71093 mol dm－3
b) Concentration of “NoFrills”bleach (by own experiment data)
Volume of Na2S2O3 used: 26.455cm3 / 1000 = 0.026455dm3
No. of mole of Na2S2O3 used: 0.0562M x 0.026455dm3 = 0.0014868mol
No. of mole of I2 = No. of mole of Na2S2O3 / 2
= 0.001486 / 2
= 0.00074339mol
No. of mole of OCl－ in diluted bleach = No. of mole of I2
= 0.00074339mol
Conc. of original bleach = 0.00074339mol x (250/25) / 0.01dm3
= 0.74339 mol dm－3
c) cost per mole of sodium hypochlorite
Kao:
The price = $11.9
No. of mole of NaOCl per package = concentration x volume
= 0.71655 x 1.5L
= 1.0748 mol
cost per mole = 11.9 / 1.0748
= $11.072
NoFrills:
The price = $9.9
No. of mole of NaOCl per package = concentration x volume
= 0.7306 x 2L
= 1.4612 mol
cost per mole = 9.9 / 1.4612
= $6.775
Band X:
The price = $10
No. of mole of NaOCl per package = concentration x volume
= 0.71093 x 1.5L
= 1.0664 mol
cost per mole = 10 / 1.0664
= $9.377
d) conclusion:
From the calculation above, we can easily notice that “NoFrills” cost least for every mole of sodium hypochlorite.
“NoFrills” is the “BEST BUY” between these commercial bleach solutions.
Discussion
Using the method above can determine the amount of sodium hypochlorite in solution.
Sulphuric acid is added since that it made the KI solution release iodine.
We add KI to the solution so that the OCl－ can react with iodide in it and turn the colourless iodide into brown iodine, which was then changed back into iodide, so that we could know that the titration is complete.
KI solution has to present in excess volume so that it won’t limit the number of atoms of sodium hypochlorite reacted with it.
There are possible error in the experiments.
We take the end point as the equivalence point, but there are difference between them. Moreover, we might have add starch indicator too early, then we would need more Na2S2O3 to titrate the solution.
