Question

The peroxydisulfate ion (S₂O₈^2-) reacts with the iodide ion in aqueous solution via the reaction:

S₂O₈^2- (aq) + 3I^- → 2SO₄^2- + I₃^- (aq)

An aqueous solution containing 0.050 M of S₂O₈^2-ion and 0.072 M of I^-is prepared, and the progress of the reaction followed by measuring [I^-].The data obtained is given the table below.

Time (s)	0.000	400.0	800.0	1200.0	1600.0
[I-] (M)	0.072	0.057	0.046	0.037	0.029

The concentration of S₂O₈^2- remaining at 800 s is ________ M.

1) 0.015

2) 0.041

3) 4.00 ⋅ 10-3

4) 0.046

5) 0.076

Answer 1

It has to:

[I-] 800 s = 0.072 - 3 * X = 0.046

It clears X = 0.009 M

It has to:

[S2O8-2] 800 s = 0.05 - X = 0.05 - 0.009 = 0.041 M

Option 2.

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