Answer:
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
