# 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–.

Jun 30, 2018

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

#### Explanation:

$n \left({\text{S"_2 "O}}_{3}^{2 -}\right) = c \cdot V = 7.43 \times {10}^{- 4} \textcolor{w h i t e}{l} m o l$

$n \left(\text{I"_2):n("S"_2 "O"_3^(2-)) = 1:2 color(grey)(" from the second chemical equation}\right)$

$n \left(\text{OCl"^(-)):n("I"_2) = 2:1 color(grey)(" from the first chemical equation}\right)$

Thus

$n \left({\text{OCl}}^{-}\right)$
=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))))
$= 7.43 \times {10}^{- 4} \textcolor{w h i t e}{l} m o l \textcolor{w h i t e}{l} {\text{OCl}}^{-}$

$n \left({\text{NaOCl") = n("OCl}}^{-}\right) = 7.43 \times {10}^{- 4} \textcolor{w h i t e}{l} m o l$

$m \left(\text{NaOCl}\right) = n \cdot M$
$\textcolor{w h i t e}{m \left(\text{NaOCl}\right)} = 7.43 \times {10}^{- 4} \textcolor{w h i t e}{l} m o l \times 74.44 \textcolor{w h i t e}{l} g \cdot m o {l}^{- 1}$
$\textcolor{w h i t e}{m \left(\text{NaOCl}\right)} = 0.0553 \textcolor{w h i t e}{l} g$

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%