Redox Titration Lab Question

Question

The reactions given are:
I2 + 2Na2S2O3 --> 2NaI + Na2S4O6
( I2 + 2e^- --> 2I- )
( 2S2O3^-2 --> S4O6^-2 + 2e^- )

Assume that a 2.130 g sample of bleach was weighed out. When a sodium thiosulfate solution containing 6.829g of Na2S2O3 x 5H2O in 250.0 mL was used to titrate the above bleach sample, it required 11.21 mL to reach the starch endpoint.

a) What is the molarity of the sodium thiosulfate pentahydrate solution?
b) What is the number of moles of Na2S2O3 in the 11.21 mL?
c) What is the number of moles of I2 that react with the Na2S2O3?
d) How many moles of NaOCl are in the bleach sample? How many grams?
e) What is the percent by mass of NaOCl in the bleach?

I'm not sure which numbers to use where. Any help is greatly appreciated! Thank you so much :)

Answer 1

One reaction which is missing is that involving NaOCl.  Presumably the lab experiment involves the iodometric determination of sodium hypochlorite:

 

OCl^- + 2H⁺ + 3I^-  -->  I₃^- + Cl^- + H₂O

 

I₃^- + 2S₂O₃^2-  -->  3I^- + S₄O₆^2- 

 

A) mol_Na2S2O3*5H2O = (6.829 g)/(.250 L) (1 mol/248.1841 g) = 0.1101 M

 

B)  mol_Na2S2O3 = 0.01121 L (0.1101 M) = 0.001234 mol

 

C)  mol I₂ = mol I₃^- = 0.001234 mol_NaS2O3(1 mol_I3- )/(2 mol_NaS2O3) = 6.169 x 10^-4 mol

 

D)  mol NaOCl = 6.169 x10^-4 mol_NaS2O3 (1 mol_NaOCl)/(1 mol_NaS2O3) = 6.169 x 10^-4 mol

 

g NaOCl = 6.169 x 10^-4 mol (74.4422 g)/(mol)  =  0.04592 g

 

E)  % = 100 (0.04592 g)(2.130 g sample) = 2.156%