Question

Calculate the % w/v NaOCl in the sample of bleach ??

A 24.90-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 0.0458 M Na2S2O3 and required 16.23 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–

HINT: This is an example of an Iodometric method of analysis (slides 29 - 30 of Module 5 Lecture Notes). Iodometric methods are indirect titrimetric methods that require the use of standardized thiosulfate solutions. They do not involve a back-titration. In any indirect method of analysis, the key is to derive the overall stoichiometry between the analyte (in this case NaOCl) and the standard solution (in this case thiosulfate). Remember: the final answer must be expressed as % w/v NaOCl, not as % w/v OCl–.

Answer 1

`"%w/v" color(white)(l) "NaOCl" = 0.222 %`

Explanation:

`n("S"_2 "O"_3^(2-))=c * V = 7.43 xx 10^(-4) color(white)(l) mol`

`n("I"_2):n("S"_2 "O"_3^(2-)) = 1:2 color(grey)(" from the second chemical equation")`

`n("OCl"^(-)):n("I"_2) = 2:1 color(grey)(" from the first chemical equation")`

Thus

`n("OCl"^(-))`
`=7.43 xx 10^(-4) color(white)(l) mol color(white)(l) color(red)(cancel(color(black)("S"_2 "O"_3^(2-)))) * (1 color(white)(l) color(red)(cancel(color(black)("I"_2))))/(color(purple)(cancel(color(black)(2))) color(white)(l) color(red)(cancel(color(black)("S"_2 "O"_3^(2-))))) * (color(purple)(cancel(color(black)(2))) color(white)(l) "OCl"^(-))/(1 color(white)(l) color(red)(cancel(color(black)("I"_2))))`
`=7.43 xx 10^(-4) color(white)(l) mol color(white)(l) "OCl"^(-)`

`n("NaOCl") = n("OCl"^(-)) = 7.43 xx 10^(-4) color(white)(l) mol`

`m("NaOCl") = n * M`
`color(white)(m("NaOCl")) = 7.43 xx 10^(-4) color(white)(l) mol xx 74.44 color(white)(l) g * mol^(-1)`
`color(white)(m("NaOCl")) = 0.0553 color(white)(l) g`

The hypochlorite solution, therefore, has the mass concentration (in grams per milliliter)

`"%w/v" color(white)(l) "NaOCl" = 100% xx m/V`
#color(white)("%w/v" color(white)(l) "NaOCl" ) = 100% xx (0.0553 color(white)(l) g)/(24.90 color(white)(l) ml) #
`color(white)("%w/v" color(white)(l) "NaOCl" ) = 0.222%`