i did the last part differently:
d= 1 g/ i mL
mass of H2O2(aq) = 2 mL * 1 g/ 1mL = 2 g
therefore:
0.022 g / 2 g * 100% = 1.081 %
which is not up to full strength
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)
I posted this up before and got some responses however, the answer i got is different from the answer posted up...can someone please check and let me know if this is accurate or if i made a mistake somewhere..
Thank YOu in ADVANCE..
1. So first you start off by balancing the equations:
H2O2 (aq) + 2H + 2I^ - (aq) ---> 2H20 (l) + I2
2S203^2- + I2 --- >S406 ^2- (aq) + 2I^- (aq)
2. Write all the given information out under equation number 2 as you have enough information there to solve the #mols of Iodine
2S203^2- + I2 --- >S406 ^2- (aq) + 2I^- (aq)
0.0124L Iodine has half of that of 2S203^2-
0.1025M =0.0006355 mols
#mols = 0.001271mols
3. The Iodine in both equations is same
4.H2O2 must be the same to 1:1 ratio.
now i can calculate the grams of H2O2
mass = Molarmass*moles
=34.02Molarmass*0.0006355mols
=0.022g
5.We are told that full strength is 3% H2O2 by mass)
0.03*2ml (or 2g)
=0.06g H2O2
6.Since your answer in step 4 doesn't exceed the full strength limit in step 5, you can conclude that the h2o2 antiseptic solution is not up to full strength.
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