*For transferring or carrying different solutions, it was reminded that to repeat the step 1 to step 3.For burette or pipette, it was rinsed with the solutions which would be transferred. Besides, the conical flask should be swirled throughout the titration.
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Data recording
The molarity of the standard sodium thiosulphate: 0.05018M
Calculation
ClO-(aq) +2H+(aq) + Cl-(aq) → Cl2(g) + H2O(l)------------(1)
Cl2(g) + 2I- (aq) +2H+ (aq)→ I2(aq) +H2O(l) + 2Cl- (aq)—(2)
I2(aq) + 2S2O32-(aq) → 2I-(aq) +S4O62-(aq)-------------------(3)
Average volume of Na2S2O3 used to titrate with the dilute bleaching solution= [(29.05 + 28.75 + 29.00)/3] /1000
=0.02893dm3
In reaction (3), it is known: no. of mole of I2 in the diluted bleaching solution = (1/2) X no. of mole of S2O32
= (1/2) X (0.02893 X 0.05018)
= 7.259 X 10-4
No. of mole of I2 in the original solution = 7.259 X 10-4 X (250.0/25.0)
=7.259 X 10-3
In reaction (2), it is given that: no. of mole of I2 = no. of mole of Cl2
=7.259 X 10-3
Available chlorine in a bleaching solution = mass(g) of Cl2 / volume(dm3) of bleaching solution
= 7.259 X 10-3 X (35.5+35.5) / (25/1000)
= 20.62 g dm-3
*Relative atomic masses were taken to be: H=1.0; O=16.0; Na=23.0 S=32.1; Cl=35.5; I=126.9
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Discussion
ClO-(aq) +2H+(aq) + Cl-(aq) → Cl2(g) + H2O(l)------------(1)
Cl2(g) + 2I- (aq) +2H+ (aq)→ I2(aq) +H2O(l) + 2Cl- (aq)—(2)
I2(aq) + 2S2O32-(aq) → 2I-(aq) +S4O62-(aq)-------------------(3)
- It is NOT necessary to measure the accurate amount of potassium iodide and ethanoic acid. Why?
Both potassium iodide and ethanoic acid act as excess reagent, so that it is not necessary to know the accurate amount, what are the most concerned are that the accurate amount of the limiting reagents. For a more clear explanation, it could be accounted for by these ways:
First, it is unnecessary to measure the accurate amount of potassium iodide because:
Iodide ions must be in excess amount so as to ensure the chlorine gas evolved in reaction (1) to be reduced to chloride ions in reaction (2) provided that chlorine gas evolved in reaction(1) is easy to escape from the solution in the conical flask. Thus, in the excess amount of potassium iodide, it can reduce the maximum amount of chlorine gas. What is more is that; potassium ion is a extremely weak oxidizing agent which would not partake any reaction in this experiment and so would not make any inaccuracies in the experiment even they are in the excess.
Second, it is also not necessary to measure the accurate amount of ethanoic acid because:
Ethanoic acid is needed to be excess for speeding up the completion of reaction (1) and (2). In reaction (1) and reaction (2), both need H+ ions for the completion of the redox reactions. Thus, ethanoic acid is needed to be speed up the two reactions or it will make the great errors for the experiment. Apart from it, chlorine gas evolved in reaction (1) is easy to escape from the solution in the conical flask, which needs a short time for reaction to minimize the escaping of chlorine gas. In this way, the acid in excess amount can achieve this. Furthermore, NO3- , which is a very weak oxidizing agent compare to chlorine gas, so that it will not make any inaccuracy for the experiment though in excess amount.
- What is the function of starch solution? Why we should NOT add the starch solution at the beginning of the titration?
The starch solution acts as indicator, whose end point is to indicate the completion of reaction. In fact, there is the color change from iodine to iodide. It, however, cannot be used to accurately detect the end point (the change in color ‘brown→yellow→colorless’ is very difficult to observe), thus, starch (preserved with salicylic acid) is used as the indicator.
Starch + iodine Blue ‘complex’
Starch solution should not be added at the beginning of the titration because:
Since starch irreversibly combines with iodine at a high concentration of I2(aq) ( so that I2 will not be released from starch at the end point), the starch solution should be added at the later stage of the titration (when the solution just turns from brown to pale yellow). After the addition of starch, the mixture turns deep blue. Then the end point is shown by the complete decolorization of the blue color.
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(3)Sometimes black precipitate may appear in the brown solution.
- What is the precipitate?
- Why it is formed?
- What should we do if the black precipitate appears?
(a)The black precipitate should be the solid I2 as I2 is insoluble or only slightly soluble in water; the solution is iodide.
(b)As I2 is insoluble or only slightly soluble in water while I2 is soluble in iodide solution, in small amount of potassium iodide, a iodine solid, a black precipitate will be formed rather than fully dissolved in the solution.
(c)So when the black precipitate appears, it is the signal to imply the potassium iodide is not enough to let all the iodine being soluble in the iodide solution. To solve this problem, the more or excess amount of potassium iodide until the black precipitate disappears.
(4)Suggest the possible sources of errors in this experiment.
In this experiment, apart from the careless errors, (e.g. the solution of burette was not all run into the conical flask) and the errors due to precision of apparatus (i.e. reading errors of burette.)there may still have been some possible sources of errors:
The iodine solution should be used immediately because its molarity changes with time because:
(i) Iodine is volatile which means that I2 can escape from the solution, causing the decrease of [I2]* with time.
(ii)Iodine can also oxidize most organic substances which also cause the decrease of [I2] with time.
(iii)Iodine can be oxidized by air( promoted by acids, heat &light) which also cause the decrease of [I2] with time: 4I-(aq) + O2(g) +4H+(aq) →2I2(aq) +2I-(aq)
*[I2] means the molarity of the iodine solution.
As time must be required, errors may be occurred from at least one of the above.
Besides, thiosulphate solution is unstable in acidic medium, the presence of microorganism, Cu(II) or sunlight, it must be standardized with standard iodine solution before use. Thus, it would contribute to errors if the thiosulphate solution was not properly kept.
As the thiosulphate solution is not the primary standard, the solution should be standardized beforehand or it would make other possible errors. The Na2S2O3(aq) can be prepared by dissolving Na2S2O3●5H2O crystals in recently boiled distilled water [to ensure no dissolved carbon dioxide gas, otherwise Na2S2O3 it can be decomposed by acids, e.g.H2CO3]. Na2S2O3 can be added to the solution to keep the solution alkaline. Some HgI2 is also added to suppress bacterial action.
And the solution then titrated with standard solution of iodine.
Likewise, as starch irreversibly combines with iodine at a high concentration of I2(aq) ( so that I2 will not be released from starch at the end point), the starch solution should be added at the later stage of the titration (when the solution just turns from brown to pale yellow). After the addition of starch, the mixture turns deep blue. Hence if the end point was only detected by the color of iodine in iodide (the change in color ‘brown→yellow→colorless’, there would be a large percentage error, provided that this kind of end point is very difficult to observe.
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(5)Why the available chlorine of old bleaching solution is lower than that of a new one?
Available chlorine of old bleaching solution is lower than that of a new one because:
First, bleaching solution may deteriorate due to the attack or decomposition by carbon dioxide in air according to the equation: 2ClO-(aq) + CO2(aq) → CO32-(aq) +Cl2(g). Thus the active ingredient, ClO- will so decompose into Cl2 by time resulting in the decrease in available chlorine as the chlorine gas can be more easily to escape from the solution.
Second, the bleaching solution will also be decomposed by hydrogen positive ions to chlorine gas provided that hypochlorous acid is present and also may react with other acid in the bleaching solution. [i.e. ClO-(aq) +2H+(aq) + Cl-(aq) → Cl2(g) + H2O(l)]
By this way, the active ingredient, ClO- will so decompose into Cl2 by time resulting in the decrease in available chlorine as the chlorine gas can be more easily to escape from the solution..
Third, the bleaching solution, will be decomposed by sunlight.[i.e. 2OCl- (aq)→ 2Cl-(aq) + O2(g)]By this way, the active ingredient, ClO- will so decompose into Cl2 by time resulting in the decrease in available chlorine as the chlorine gas can be more easily to escape from the solution..
Above all, these happen gradually by time, so the older of the solution, the lower available of the chlorine is.
(6) Suggest another application of iodometry besides analysis of bleach.
Besides the analysis of chlorine bleach, it also can be used to determine the copper content in an alloy-a physically mixture of different atoms at least one in which is metal. It can be used by this way:
The certain amount of copper-containing alloy was dissolved by the sample in noitric acid. After the excess nitric acid is removed by boiling, an excess of potassium iodide crystals are added to the resultant copper(II) nitrate solution. The liberated I2 then is titrated with standard sodium thiosulphate solution until the end point. (i.e. Starch is used as indicator.)
The equations for the reactions are:
Cu→ Cu2+ + 2e ---(1)
2Cu2+(aq) +4I-(aq) → 2CuI(s) +I2(aq)------(2)
I2(aq) + 2S2O32-(aq) → 2I-(aq) +S4O62-(aq)-(3)
By comparing the coefficients of those reactant, the content can be found.
Apart from it, it also can be used to determine the oxidizing agents(e.g. Fe3+). In this experiment, excess NaI(aq) is added to an oxidizing agent, Fe3+ without unknown molarity. The I2 will be generated quantitatively.*
2Fe3+(aq) + 2I-(aq) → 2Fe2+(aq) + I2(aq)
oxidizing agent (stronger than I2)
The iodine liberated is then determined by the titration with standard thiosulphate solution to a starch end point.
[i.e. I2(aq) + 2S2O32-(aq) → 2I-(aq) +S4O62-(aq)].The end point is indicated when the blue-black color disappears. Then use the data marked and then calculate the required molarity.
By Kong Siu Wai
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