Can somebody help me out to solve this as well,it is an iodimetric method.?

A 22.31-mL sample of liquid bleach (contains NaOCl; 74.44 g/mol) was diluted to 1000.0 mL in a volumetric flask. A 25.00-mL aliquot of the diluted sample was transferred into an Erlenmeyer flask and treated with excess KI to oxidize the OCl- to Cl2 and producing I2. The liberated I2 was determined by titrating with 0.0990 M Na2S2O3 and required 8.23 mL to reach a starch indicator endpoint.

Calculate the % w/v NaOCl in the sample of bleach. Give your answer to 2 places after the decimal point.

Reactions: 2OCl– + 2I– + 4H+ → I2 + Cl2 + 2H2O

                I2 + 2S2O32– →    2I– + S4O62–

1 Answer
Michael
Apr 16, 2018

5.44 %

Explanation:

I disagree with the 1st equation. Chlorine(I) is reduced right down the chlorine(-1) and not free chlorine. If free chlorine were to be produced, the titration result would be affected.

I get: #sf(ClO^(-)+2I^(-)+2H^(+)rarrI_2+Cl^(-)+H_2O)#

The liberated iodine is titrated against thiosulfate:

#sf(I_2+2S_2O_3^(2-)rarr2I^(-)+S_4O_6^(2-))#

#sf(n=cxxv)#

#:.##sf(n_(S_2O_3^(2-))=0.0990xx8.23/1000=8.1477xx10^(-4))#

From the 2nd equation you can see that the number of moles of iodine must be half of this:

#sf(n_(I_2)=(8.1477xx10^(-4))/2=4.07385xx10^(-4))#

From the 1st equation you can see that the no. moles of #sf(OCl^-)# must be the same:

#sf(n_(OCl^-)=4.07385xx10^(-4))# in 25.00 ml

This means that the no. of moles of #sf(OCl^(-))# present in the original 1000ml

#sf(=(4.073585xx10^(-4))/(25.00)xx1000=0.0162854)#

#sf("Mass"_(NaOCl)=nxxM_r=0.0162954xx74.44=1.21303color(white)(x)g)#

#:.##sf(%w/v=1.21303/(22.31)xx100=5.44)#

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