a 25.00 ml sample of a liquid bleach was fully diluted to 100 in a volumetric fl

Question

a 25.00 ml sample of a liquid bleach was fully diluted to 100 in a volumetric flask a 25 ml portion of the diluted sample was transferred into a Erlenmeyer flask containing an excess of KI which reduced the OCl- and produced I3-
OCl- + 3I- + H2O <--> Cl- + I3- + 2OH- the liberated I3- was determined by titration with 0.06523M Na2S2O3z it required 9.56 mL of titrant to reach the starch indicator end point what is the % w/v of NaOCl in the original sample of bleach
S2O3^2- + I3- <----> S4O6^2- + 3I-
NaOCl=74.44g/mol a 25.00 ml sample of a liquid bleach was fully diluted to 100 in a volumetric flask a 25 ml portion of the diluted sample was transferred into a Erlenmeyer flask containing an excess of KI which reduced the OCl- and produced I3-
OCl- + 3I- + H2O <--> Cl- + I3- + 2OH- the liberated I3- was determined by titration with 0.06523M Na2S2O3z it required 9.56 mL of titrant to reach the starch indicator end point what is the % w/v of NaOCl in the original sample of bleach
S2O3^2- + I3- <----> S4O6^2- + 3I-
NaOCl=74.44g/mol a 25.00 ml sample of a liquid bleach was fully diluted to 100 in a volumetric flask a 25 ml portion of the diluted sample was transferred into a Erlenmeyer flask containing an excess of KI which reduced the OCl- and produced I3-
OCl- + 3I- + H2O <--> Cl- + I3- + 2OH- the liberated I3- was determined by titration with 0.06523M Na2S2O3z it required 9.56 mL of titrant to reach the starch indicator end point what is the % w/v of NaOCl in the original sample of bleach
S2O3^2- + I3- <----> S4O6^2- + 3I-
NaOCl=74.44g/mol

Explanation / Answer

The reaction taking place between Sodium Thiosulfate and Iodine is

           2S2O32- + I3- = S4O62- + 3I-

Number of moles of Thiosulfate used is (0.06523 M * 0.00956 L) = 0.0006235988 mol

So, number of moles of I3- present in 25 mL of diluted solution is (0.0006235988/2) mol = 0.0003117994 mol

Thus number of moles of OCl- present in 25 mL of diluted solution is 0.0003117994 mol

In original 25 mL solution amount of OCl- is (0.0003117994*4) mol = 0.0012471976 mol

Hence, number of moles of NaOCl is 100 mL original solution is (0.0012471976 *4) mol = 0.0049887904 mol

Molar Mass of NaOCl is 74.44 g/mol

Therefore, (w/v) % of the solution is 0.0049887904 mol * 74.44 g/mol = 0.3714 %

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