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.
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
To determine if the H2O2 antiseptic solution is up to full strength, we need to calculate the amount of H2O2 in the solution and compare it to the desired 3% concentration.
First, we balance the chemical equations:
1. H2O2 (aq) + 2H+ + 2I- (aq) ---> 2H2O (l) + I2
2. 2S2O3^2- + I2 ---> S4O6^2- (aq) + 2I- (aq)
Next, we write the information given under equation 2 to calculate the number of moles of iodine:
2S2O3^2- + I2 ---> S4O6^2- (aq) + 2I- (aq)
0.0124 L of iodine = half the volume of 2S2O3^2-
Concentration of 2S2O3^2- = 0.1025 M
Number of moles = concentration × volume
= 0.1025 M × 0.0062 L
= 0.0006355 moles
Since the iodine is present in a 1:1 ratio in both equations, the number of moles of iodine is also the number of moles of H2O2.
Now, we can calculate the mass of H2O2:
Mass = Molar mass × moles
= 34.02 g/mol × 0.0006355 moles
= 0.022 g of H2O2
We are told that the full strength of the solution is 3% H2O2 by mass.
To check if the solution is up to full strength, we calculate the maximum mass of H2O2 that should be present:
0.03 × 2 mL (or 2 g) = 0.06 g of H2O2
Comparing the calculated mass (0.022 g) with the maximum allowed mass (0.06 g), we find that the calculated mass does not exceed the full strength limit. Therefore, we can conclude that the H2O2 antiseptic solution is not up to full strength.