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Notice that the effective analytical connection for analysis of hypochlorite anion by the thiosulfate anion in this experiment is the sum of the two reactions: 1) hypochlorite anion with iodide anion to form iodine, and 2) thiosulfate anion with the iodine formed in reaction one. In your laboratory notebook sum these two reactions to find the stoichiometric factor that relates moles of thiosulfate anion needed to react with each mole of hypochlorite anion in the bleach sample. Then select the correct response. a. One mole of thiosulfate anion needed to react with one mole of hypochlorite anion. b. Two moles of thiosulfate anion needed to react with one mole of hypochlorite anion. c. Three moles of thiosulfate anion needed to react with one mole of hypochlorite anion.

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Answer:

b. Two moles of thiosulfate anion needed to react with one mole of hypochlorite anion.

Explanation:

1. Hypochlorite with iodide

                      2I⁻ ⟶ I₂ + 2e⁻

ClO⁻ + 2H⁺ + 2e⁻ ⟶ Cl⁻ + H₂O      

2I⁻ + ClO⁻ + 2H⁺ ⟶ I₂  + Cl⁻ + H₂O

2. Thiosulfate with iodine

     2S₂O₃²⁻ ⟶ S₄O₆²⁻ + 2e⁻

      I₂ + 2e⁻ ⟶ 2I⁻                  

2S₂O₃²⁻ + I₂ ⟶ S₄O₆²⁻ + 2I⁻

3. Sum of the reactions

                2I⁻ + ClO⁻ + 2H⁺ ⟶ I₂  + Cl⁻ + H₂O

                      2S₂O₃²⁻ + I₂ ⟶ S₄O₆²⁻ + 2I⁻                    

2I⁻ + ClO⁻ + 2S₂O₃²⁻ + 2H⁺ ⟶ I₂  + Cl⁻ + S₄O₆²⁻ + H₂O

4. Molar ratio

S₂O₃²⁻:ClO⁻ = 2:1

b. When Two moles of thiosulfate anion needed to react with one mole of hypochlorite anion. The molar ratio is = S₂O₃²⁻:ClO⁻ = 2:1

What is Hypochlorite?

1. Hypochlorite with iodide

  Then, 2I⁻ ⟶ I₂ + 2e⁻

After that, ClO⁻ + 2H⁺ + 2e⁻ ⟶ Cl⁻ + H₂O    

Now, 2I⁻ + ClO⁻ + 2H⁺ ⟶ I₂  + Cl⁻ + H₂O

2. When Thiosulfate with iodine

2S₂O₃²⁻ ⟶ S₄O₆²⁻ + 2e

I₂ + 2e⁻ ⟶ 2I⁻                  

Then 2S₂O₃²⁻ + I₂ ⟶ S₄O₆²⁻ + 2I⁻

3. Now the Sum of the reactions is:

So, 2I⁻ + ClO⁻ + 2H⁺ ⟶ I₂  + Cl⁻ + H₂O

Then 2S₂O₃²⁻ + I₂ ⟶ S₄O₆²⁻ + 2I⁻                    

After that 2I⁻ + ClO⁻ + 2S₂O₃²⁻ + 2H⁺ ⟶ I₂  + Cl⁻ + S₄O₆²⁻ + H₂O

4. The Molar ratio

Therefore, S₂O₃²⁻:ClO⁻ = 2:1

Find more information about Hypochlorite here:

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