Question

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–

Answer 1

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)`