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The peroxydisulfate ion (S2O82–) reacts with the iodide ion in aqueous solution via the reaction:
S2O82–(aq) + 3 I–(aq) 2 SO4(aq) + I3–(aq)
An aqueous solution containing 0.050 M of S2O82– ion and 0.072 M of I– is prepared, and the progress of the reaction followed by measuring [I–]. The data obtained is given in the following table:
Time (s) | 0.0 | 400.0 | 800.0 | 1200.0 | 1600.0 |
[I–] (M) | 0.072 | 0.057 | 0.046 | 0.037 | 0.029 |
The concentration of S2O82– remaining at 400 s is ________ M
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