Question

1. The determination of the amount of hypoclorite in bleach is accomplished indirectly:

OCl^- + 3 I^- + 2 H⁺ ---> I₃^-  + Cl^- + H₂O

I₃^- + 2 S₂O₃^2- ---> 3 I^- + S₄O₆^2-

a. What is the net overall reaction between hypochlorite and thiosulfate? Identify the oxidizing and the reducing agents in the overall reaction.

b. Why can't the titration of hypochlorite be done directly with thiosulfate?

c. What is the purpose of each step and how do these two reactions permit the determination of the amount of hypochlorite?

d. What indicator is used and what change indicates the end point of the reaction?

e. Based on the volume of a known concentration of thiosulfate, write the stoichiometric expression that yields the moles of hypochlorite at the end point.

Answer 1

B) By the each addition of thiosulfate, the concentration of I₂ decreases, and the brown colour will fade to a pale yellow colour then to colourless so the detection of the end point will be difficult .
So, a small amount of starch solution is added near the end point of the titration, when the solution is pale yellow.

C) The concentration of sodium hypochlorite in bleach solutions can be determined by titration. A desirable
method would be to find a titrant that reacts with NaOCl to form a colored product. But there are no
known titrant-indicator systems that work well. Therefore, we must use a multi-step method to titrate
sodium hypochlorite.
In the first step sodium hypochlorite, hydrochloric acid, iodide ion, and starch are combined to form a
starch–triiodide complex. In this step there are three reactions that take place:
(1) Hydrochloric acid reacts with sodium hypochlorite to form hypochlorous acid:
NaOCl(aq)+HCl(aq) HOCl(aq) + NaCl(aq)
(2) Hypochlorous acid reacts with iodide when the solution is acidic:
HOCl (aq)+ HCl (aq)+ 3I (aq) I (aq) +2Cl (aq)+H O(l) 
Tridiodide, I3-, is a dark red complex. A dark blue complex is formed when triiodide is combined with Istarch.
(3) triiodide triiodide starch ion complex (dark blue)
I + starch [I ][starch]
The result of these three reactions is that when sodium hypochlorite is present the starch-triiodide complex is produced. This is useful because the result of these three reactions is the formation of a dark
blue complex that has a concentration that is proportional to the amount of sodium hypochlorite in the
solution.
In the next step, the starch-triiodide product is titrated by sodium thiosulfate to form a colorless solution
of iodide, dithionate, and uncomplexed starch:
(4)iodideion triiodidestarch thiosulfate dithionate colorless complex (dark blue)

D) Starch serves as anindicator because it forms a dark blue complex with iodine, but that complex disappears (turning the blue solution colorless) when all the iodine is used up.

E) amount of thiosulfate has been added to the flask.
2 S2O32- + I2  2 I- + S4O62-
From the above equations it can be seen that 1 mole of ClO-
reacts to form 1 mole I2, which
consumes 2 moles of S2O32-
. Hence, 1 mole of hypochlorite is equivalent to 2 moles of thiosulfate.
C(hypochlorite) x V(hypochlorite) = C(thiosulfate) xV(thiosulfate)/2