A 9.00 mL sample of an aqueous solution of H2O2 is treated with an excess of KI(aq). The liberated I2 requires

Question

A 9.00 mL sample of an aqueous solution of H2O2 is treated with an excess of KI(aq). The liberated I2 requires 44.93 mL of 0.5022 M Na2S2O3 for its titration. Is the H2O2(aq) up to full strength (3% H2O2 by mass) as an antiseptic solution? Assume that the density of the H2O2(aq) is 1.01 g/mL.

H2O2(aq) + H+ (aq) + I -(aq) ---> H2O(l) + I2(s) (not balanced)

I2(s) + S2O32-(aq) ---> S4O62-(aq) + I - (aq) (not balanced)

Answer 1

The balanced equations are:
H2O2(aq) + 2H+ (aq) + 2I -(aq) ---> 2H2O(l) + I2(s)
I2(s) + 2[S2O3]2-(aq) ---> [S4O6]2-(aq) + 2I-
The overall results can be considered as:
H2O2(aq) + 2H2[S2O3](aq) ---> H2[S4O6](aq) + 2H2O(l)
Hence:
0.00900*[H2O2] = 0.04493*0.5022 /2
[H2O2] = 1.25M <==> 42.64 g H2O2 in 1L <==> 42.64 g H2O2 in 1010 g solution <==> 4.22% H2O2 by mass.

Answer 2

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