When the NaOCl found in a dilute solution of bleach is reacted with I^- the solu

Question

When the NaOCl found in a dilute solution of bleach is reacted with I^- the soluble I_3^- is produced according to the equation below. 3I^- + OCl^- + 2H^+ rightarrow I^- + Cl^- + H_2 O A sample of 1.75 g of bleach purchased from Publix Supermarket was reacted with excess potassium iodide and the reaction solution titrated with 0.05500 M sodium thiosulfate, using starch as an indicator. The titration consumed exactly 41.78 mL of the thiosulfate solution. I_3^- + 2S_2 O_3^2- rightarrow I^- + S_4 O_6^2- How many moles of I_3^- reacted? How many moles of OCl^- reacted? What is the percentage of NaOCl in the bleach sample?

Explanation / Answer

from the equations given

3 mol I^- = 1 mol Bleach (OCl-) = 1 mol I3^-

1 mol I3^- = 2 mol S2O3^2-

No of mol of Thiosulphate(S2O3^2-) = 0.055*0.04178 = 0.0023 mol

No of mol of i3- reacted = 0.0023/2 = 0.00115 mol

No of mol of NaOCl reacted = 0.00115 mol

Molarmass of NaOCl = 74.4422 g/mol

mass of NaOCl reacted = 0.00115*74.4422 = 0.086 grams

Percentage of NaOCl in bleach = 0.086/1.75*100 = 4.914%

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