Analysis of Two Brands of Commercial Bleaches

Brand 2                Trade Name : LION                Price : $6.67per dm3   ($10/1500mL)

Questions

1. For each brand, calculate        (a)        the amount of the active ingredient available in g dm-3.

2.  the cost per gram of this compound.

1a) For KAO Bleach,
Averge volume of sodium thiosulphate solution used=(11.5+11.6)/2=11.55cm3
∴no. of mole of S2O32-=(11.55/1000)(0.0992)
                  =0.00114576mol.

∵no. of mole of I2 =(no. of mole of S2O32- )/2

                =0.00057288mol

∴no. of mole of OCl-  in10 cm3 solution = no. of mole of I2  x10

                                =0.00057288x10

                                =0.0057288mol

∴Mass of OCl-  contained in 10 cm3 solution=0.0057288x(16+35.45)
                                    =0.29474676g

∴Concentration of OCl-   (active ingredient available)   =0.29474676/(10/1000)=29.47gdm-3

For LION Bleach,
Averge volume of sodium thiosulphate solution used=(14.1+14.1)/2=14.10cm3
∴no. of mole of S2O32-=(14.10/1000)(0.0992)
                  =0.00139872mol.

∵no. of mole of I2 =(no. of mole of S2O32- )/2

                =0.00069936mol

∴no. of mole of OCl-  in10 cm3 solution = no. of mole of I2  x10

                                =0.00069936x10

                                =0.0069936mol

∴Mass of OCl-  contained in 10 cm3 solution=0.0069936x(16+35.45)
                                    =0.35982072g

∴Concentration of OCl-   (active ingredient available) =0.35982072/(10/1000)=35.98gdm-3

1b)For KAO Bleach,
 Cost per one gram of OCl-  =  11 /[29.474676x(1500/1000)]
                        =$0.277/g

  For LION Bleach,

   Cost per one gram of OCl-  =  10 /[35.982072x(1500/1000)]

                        =$0.185/g

2. Based on the calculated results, decide which of the two brands of bleach is a better buy.

When compring the costs per one gram of OCl-   of KAO(0.277/g) and LION($0.185/g), as we can pay a lower price ($0.185)to buy the same weight of active ingredient (OCl- ) in LION bleach, LION bleach is a better buy.

3. Explain why potassium iodide and sulphuric acid must be in excess.

ClO-(aq) + 2I-(aq) + 2H+(aq) → I2(aq) + H2O(l) + Cl-(aq)
From the equation above, iodide ions and hydrogen ions react with Hypochlorite ions to form iodine ,water and chloride ions.To determine the actual number of mole of ClO- ,all ClO-    ions sholud react to give iodine. KI (I-) and H2SO4 (H+) must be in excess to ensure all ClO- has been converted into Cl- ions. Then the number of mole of I2 produced is calculated by titrating against sodium thiosulphate and hence the number of mole of ClO- ions can be calculated by ratio:

                  No. of moles of ClO- = No. of moles of I2

In other words, excess potassium iodide and sulphuric acid are used for ensuring all ClO-  ions are reacted react to give iodine. Moreover,dilute sulphuric acid is added to increase the rate of reaction.

If KI and H2SO4 are not in excess, some of the ClO- ions are not converted into Cl- .Consequently the calculated number of mole of ClO- ions present would be smaller than the actual one since the number of mole of ClO- ions which are not converted into Cl- are not taken into account.

4. Bleaching solution may deteriorate for two main reasons. One is the attack by carbon dioxide in air according to the equation:

2ClO-(aq) + CO2(aq) → CO32-(aq) + Cl2(aq)

        What is the other possible reason?

Under normal condition,Hypochlorite ions decompose to give oxygen and chloride ions.
Ionic equation:
ClO-(aq) → O2(g) + 2Cl-(aq)

Chemical equation:
NaClO(aq) → O2(g) + 2NaCl(aq)
The rate of the reaction above can been increased at high temperature or under sunlight.

When sodium hypochlorite is heated or evaporated, sodium chlorate and sodium chloride will be formed.

Chemical equation:
3 NaOCl(aq) → NaClO3 (aq)+ 2 NaCl(aq)

5. The starch indicator should not be added too early. Why?
   
    is often used in chemistry as an  for  where  is present as starch forms a very dark blue-black  with iodine. Moreover,the indicator is not affected by the presence of  (I-). In order for the the starch solution to change colour, both starch and iodide must be present. Starch forms an unstable complex which is blue coloured in low concentrations of . However, this complex is in high concentrations of iodine . If starch solution is added too early, the conical flask still contains high concentration of iodine.As a result, highly stable complex is formed,the starch indicator then cannot be decolourised and the complex will spoil the reaction medium. Consequently, the titration cannot continue. Therefore the iodine must be diluted first with the titrant ,sodium  to ensure no higly stable complex is formed.After that the starch can be added.

The End