Volume = ___________1.5L______________ Volume = ____________2L_______________
Price = _______$12.9 / Bottle____________ Price = ________$10.9 / Bottle____________
Mole = _________0.753 Mol_____________ Mole = ___________1.152 Mol ___________
$/mole = _______$17.13 / Mol ___________ $/mole = __________$9.46 / Mol__________
Best buy is ________ Best Buy___________
Titration of Brand A against the standard sodium thiosulphate solution
Mean volume of titrant (ml) = __(25.90+25.40+23.50)÷3 = 25.1 ml__
Calculation:
The Result Of Brand A
Sodium chlorate(I) reacts with excess potassium iodide in the Bleaching solution, which is a acid medium. Iodine solution produced.
Following equation: 2H+-(aq) + OCl--(aq) + 2I--(aq) → Cl--(aq) + I2-(aq) + H2O-(l)
Secondly, the iodine solution is titrated with sodium thiosulphate solution
Following equation: I2(aq) + 2 S2O32- (aq) → S4O62- (aq) + 2 I-(aq)
The Molarity of Na2S2O3 :
= 0.1M
Number of mole of S2O3- reacted with I2: :
= Molarity X Volume
= 0.1 X (25.1÷1000) = 0.00251 mol
The Mole ratio of Na2S2O3:: I2= 2:1
Number of moles of I2 reacted with S2O3-
= (1÷2) X 0.00251 mol
= 0.001255 mol
The Mole ratio of NaOCl: I2= 1:1
Number of moles of NaOCl used in the titration
= I2
= 0.001255 mol
Number of moles of NaOCl in the 250ml volumetic flask
= 0.001255 mol X 10
= 0.01255 mol
Number of moles in 1.5L bleach solution
= 0.01255 mol ÷ 25 X 1500
= 0.753 mol
Price of OCl- per mole in brand A
= $12.9 ÷ 0.753mol
= $17.13 / mol
Results :
Titration of Brand B against the standard sodium thiosulphate solution
Mean volume of titrant (ml) = __(28.90+28.60+28.90)÷3 = 28.8 ml__
Calculation:
The Result Of Brand B
Sodium chlorate(I) reacts with excess potassium iodide in the Bleaching solution, which is a acid medium. Iodine solution produced.
Following ionic equation: 2H+-(aq) + OCl--(aq) + 2I--(aq) → Cl--(aq) + I2-(aq) + H2O-(l)
Secondly, the iodine solution is titrated with sodium thiosulphate solution
Following ionic equation: I2(aq) + 2 S2O32- (aq) → S4O62- (aq) + 2 I-(aq)
The Molarity of Na2S2O3 :
= 0.1M
Number of mole of S2O3- reacted with I2: :
= Molarity X Volume
= 0.1 X (28.8÷1000)= 0.00288 mol
The Mole ratio of Na2S2O3:: I2= 2:1
Number of moles of I2 reacted with S2O3-
= (1÷2) X 0.00288 mol
= 0.00144 mol
The Mole ratio of NaOCl: I2= 1:1
Number of moles of NaOCl used in the titration
= I2
= 0.00144 mol
Number of moles of NaOCl in the 250ml volumetic flask
= 0.00144 mol X 10
= 0.0144 mol
Number of moles in 2L bleach solution
= 0.0144 mol ÷ 25 X 2000
= 1.152 mol
Price of OCl- per mole in brand B
= $10.9 ÷ 1.152 mol
= $9.46 / mol
Questions :
1. Why must the KI be present in excess ? If less than the specified quantity of KI is added, what effect will this have on the results ?
The bleach solution contain sodium chlorate(I),which have ions OCl-.We can prepare the iodine solution by adding the sodium chlorate(I) to potassium iodide in a acidic medium. At first, the sodium chlorate(I) Is the limiting agent. When we add the excess potassium iodide into the bleach solution. Not only it will not effect the result, but also the calculation can be more accurate. All the ions OCl- can be completely reacted.
In addition, Iodine solution is only slightly soluble in water but it is very soluble in the solution, which contain I- ions.
2. What is the function of the sulphuric acid ?
The function of sulphuric acid is provide a acidic medium containing excess iodide to ionizes the iodine solution to triiodide ions. We must know that the Brown colour of iodine solution is cause for the triiodide ions(I3-).This colour in observation of titration is very important.
3. Bleaching solutions may deteriorate for 2 reasons :
(a) react with CO2 in the air according to the equation : 2 OCl- + CO2 → CO32-+ H2 + Cl2
(b) what is the other reason ?
It must be effected by light. It is because the The hypochlorite ions OCl-( will be decompose quickly under light Following equation:2NaOCl →2NaCl + O2, .While losing some OCl- ions, the result in calculation will not be accurate
- What should the starch indicator not be added too early ?
The starch solution turns the iodine to blue black because of the formation of starch-iodine complex. Also, the complex is not reversible when the concentration of iodine is high. If we add the starch solution early, the attraction of starch molecules and iodine molecules will attract so strongly. Although we have add standard sodium thiosulphate solution, but we can not do completely finish the real result of titration and effecting the calculation. The above-mentioned tell us that the starch solution should be added when only a few of iodine solution left, near the end point of the titration.
Conclusion:
The bleach solution of brand B is the best buy.
