Answer:
0.0325M S₂O₃²⁻
Explanation:
Based on the reaction:
S²⁻ + ZnCO₃ → ZnS(s) + CO₃²⁻
<em>1 mole of S²⁻ reacts per mole of ZnCO₃</em>
when the reaction occurs, the sulfide ions are precipitated keeping in solution just S₂O₃²⁻ ions.
Then, these ions are titrated with iodine, thus:
2 S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻
That means 2 moles of thiosulfate react per mole of iodine.
Moles of iodine spent are:
5.20x10⁻³L ₓ (0.01000 mol / L) = 5.20x10⁻⁵ moles of I₂
5.20x10⁻⁵ moles of I₂ ₓ (2 moles S₂O₃²⁻ / 1 mole I₂) =
1.04x10⁻⁴ moles of S₂O₃²⁻
As dilution factor of the S₂O₃²⁻ solution was 50.0mL / 20.0mL = 2.5. Moles of the initial solution X are:
1.04x10⁻⁴ moles of S₂O₃²⁻ ₓ 2.5 = 1.60x10⁻⁴ moles of S₂O₃²⁻. In 20.0mL:
1.60x10⁻⁴ moles of S₂O₃²⁻ / 0.0200L =
<em>0.0325M S₂O₃²⁻</em>
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