Question

An excess of zinc carbonate was added to a 20.0 mL solution X that contains sulfide and thiosulfate ions. Reaction occured according to the net ionic equation below:

S2- + ZnCO3 → ZnS + CO32-

Upon completion of the reaction, the solution was filtered into a 50.0 mL volumetric flask and diluted to the mark.

20.0 mL of the diluted filtrate was then withdrawn and titrated with 5.20 mL of 0.01000 M standard solution of iodine.

What is the concentration (in molarity) of thiosulfate ion in the initial solution X.

Answer 1

0.0325M S₂O₃²⁻

Step-by-step explanation:

Based on the reaction:

S²⁻ + ZnCO₃ → ZnS(s) + CO₃²⁻

1 mole of S²⁻ reacts per mole of ZnCO₃

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 =

0.0325M S₂O₃²⁻