When you standardized the Na2S2O3, what molarity of Na2S2O3 did you obtain? V(Na2S2O3): 6.5 ml

V(KIO3)L: 5.0 ml

stoichiometric factor ( number of moles of Na2S2O3 reacting with one mole of KIO3): 6

0.46M NaS₂O₃ (Assuming KIO₃ solution with a concentration of 1.0M)

Explanation:

Based on the reaction:

6 Na₂S₂O₃ + KIO₃ + 5 KI + 3 H₂SO₄ → 3 Na₂S₄O₆ + 3 H₂O + 3 K₂SO₄ + 6 NaI

6 moles of  Na₂S₂O₃ react per mole of KIO₃

Assuming the molarity of the KIO₃ solution is 0,1M:

Moles of KIO₃: = 5.0x10⁻³L ₓ (0.1 mol / L) = 5.0x10⁻⁴ moles

As 6 moles of thiosulfate reacted per mole of iodate:

5.0x10⁻⁴ moles KIO₃ ₓ (6 moles Na₂S₂O₃ /  1 mole KIO₃) =

3.0x10⁻³ moles of Na₂S₂O₃. In 6.5mL (6.5x10⁻³L):

3.0x10⁻³moles Na₂S₂O₃ / 6.5x10⁻³ L = 0.46M NaS₂O₃

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the answer is:  

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