Problem 7.28 The concentration of an aqueous solution of NaOCl (sodium hypochlor
ID: 528633 • Letter: P
Question
Problem 7.28
The concentration of an aqueous solution of NaOCl (sodium hypochlorite; the active ingredient in household bleach) can be determined by a redox titration with iodide ion in acidic solution: OCl(aq)+2I(aq)+2H+(aq)Cl(aq)+I2(aq)+H2O(l)Assume that the blue spheres in the buret represent I ions, the red spheres in the flask represent OCl ions, the concentration of the Iions in the buret is 0.180 M , and the volumes in the buret and the flask are identical. (Figure 1)
Part A
What is the concentration of NaOCl in the flask?
SubmitMy AnswersGive Up
Part B
What percentage of the I solution in the buret must be added to the flask to react with all the OCl ions?
Express your answer using two significant figures.
SubmitMy AnswersGive Up
Problem 7.28
The concentration of an aqueous solution of NaOCl (sodium hypochlorite; the active ingredient in household bleach) can be determined by a redox titration with iodide ion in acidic solution: OCl(aq)+2I(aq)+2H+(aq)Cl(aq)+I2(aq)+H2O(l)Assume that the blue spheres in the buret represent I ions, the red spheres in the flask represent OCl ions, the concentration of the Iions in the buret is 0.180 M , and the volumes in the buret and the flask are identical. (Figure 1)
Part A
What is the concentration of NaOCl in the flask?
[NaOCl] = MSubmitMy AnswersGive Up
Part B
What percentage of the I solution in the buret must be added to the flask to react with all the OCl ions?
Express your answer using two significant figures.
%SubmitMy AnswersGive Up
Explanation / Answer
The figure is not mentioned .I assumed that there are 12 blue spheres and 4 red spheres. [resolve using your respective number of spheres]
OCl-(aq) + 2 I-(aq) + 2 H+(aq) ---> Cl-(aq) + I2 (aq)+ H2O (l)
Part A)
There are 12 blue spheres in the buret. There are 4 red spheres in the flask. So the I- ions in the buret are 3 times as concentrated as the OCl- ions in the flask.
Supposing complete ionization of the NaOCl:
(0.180 M OCl-) / 3 = 0.06 M NaOCl
Part B)
It takes two I- ions to react with one OCl- ion.
So it takes 8 I- ions to react with the 4 OCl- ions in the flask.
8/12 = 0.667 x100 % = 66.7% of the I- ions must be added.