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A 2.00 ml sample of an aqueous solution of hydrogen peroxide, H2O2 (aq) is treated with an excess of Kl (aq). The liberated I2 requires 12.40 mL of 0.1025 M Na2S2O3, for its titration. Is the H2O2 up to the full strength (3% H2O2 by mass) as an antiseptic solution? Assume the density of the aqueous solution of H2O2 ( aq) is 1.00 g/ mL.

H2O2(aq) + H + I^-(aq) ---> H20(l)+I2 (not balanced)

S203^2- + I2 ----> S406 ^2- (aq) + I^- (aq) (not balanced)

H2O2 + 2H^+ + 2I^- ==> I2 + 2H2O check me out.
2S2O3^-2 + I2 ==> 2I^- + S4O6^-2
check me out.
Convert 12.40 mL x 0.1025 M S2O3^- to mols.
Using the coefficients in the balanced equations, convert mols S2O3^-2 to mols I2 and from there to mols H2O2.
Convert mols H2O2 to grams.
Then calculate percent H2O2.
%H2O2 = (mass H2O2/mass solution)*100 =

How many mol of S2O3^-2?
12.4*10^-3 L * .1025 M/L = 1.27*10^-3 M of S2O3^-2

balance
2 S2O3^-2 + I2 = 2I^-1 + S4 ...
so
one mol of S2O3 for every mol I (not I2)
so
1.27*10^-3 mol of I
so
1.27*10^-3 mol of KI
so
balance
H2O2+ 2H^1+ 2I^-1 = 2H2O + I2
so one mol H2O2 for every 2 mol of I
so
.6355 *10^-3 mol of H2O2 in 2*10^-3 L
or
.317 mol of H2O2 in a Liter of solution
what is the mass of that?
.3178(2+32) = 10.8 grams/liter =.0108 g/mL
It is supposed to be .03 g/mL
so it is weak
check my arithmetic!

Thank YOu soo mUch

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