The bottle of bleach lists the percentage of sodium hypochlorite as 6.0%. If the density of commercial bleach is 1.084 g/mL, how many mL of .150 M sodium thiosulfate is required to reach the endpoint in a titration if a student analyzed a 2.0 mL sample of bleach.
The first thing I would do is calculate the molarity of the bleach.
That is 1.084 g/mL x 1000 mL x 0.06 = 65.04 g and that divided by the molar mass NaOCl = 65.04/74.44 = 0.874M. Therefore, a 2 mL sample will have mols = M x L = 0.874 x 0.002L = 0.00175 mols.
Are you treating the NaOCl with I^- to oxidize it to I2 then titrating the liberated I2 with thiosulfate?
OCl^- + 2I^- ==>I2 + Cl^-
2S2O3^2- + I2 ==>2I^- + S4O6^2-
1 mol OCl^- = 1 mol I2 = 2 mol S2O3
Therefore 1/2 mol NaOCl = 1 mol S2O3
We have 0.00175 mol NaOCl. That will use 1/2 that of thiosulfate.
M thiosulfate = mols/L
You know M and mols; solve for L and convert to mL.posted by DrBob222