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
To determine if the H2O2 is up to the full strength as an antiseptic solution, we need to calculate the percentage of H2O2 by mass.
First, let's balance the chemical equation for the reaction between H2O2 and I^-:
H2O2(aq) + 2H^+(aq) + 2I^-(aq) → 2H2O(l) + I2(s)
Now, let's calculate the number of moles of I2 produced from the titration with Na2S2O3:
The volume of Na2S2O3 used in the titration is 12.40 mL, which is equal to 0.0124 L (since 1 mL = 0.001 L).
The concentration of Na2S2O3 is 0.1025 M, which means there are 0.1025 moles of Na2S2O3 in 1 liter of the solution.
Therefore, the number of moles of Na2S2O3 used in the titration is 0.0124 L x 0.1025 moles/L = 0.00127 moles.
Since the balanced equation shows that 1 mole of I2 is produced for every 2 moles of Na2S2O3, the number of moles of I2 produced is 0.00127 moles x (1 mole I2 / 2 moles Na2S2O3) = 0.000635 moles.
Now let's calculate the number of moles of H2O2 present in the 2.00 mL sample:
The density of the H2O2 solution is given as 1.00 g/mL. Therefore, the mass of the 2.00 mL sample is 2.00 g.
To calculate the number of moles of H2O2, we need to know the molar mass of H2O2, which is 34.02 g/mol.
So, the number of moles of H2O2 is 2.00 g / 34.02 g/mol = 0.0588 moles.
Now we can calculate the percentage of H2O2 by mass:
Percentage of H2O2 = (mass of H2O2 / mass of solution) x 100
The mass of H2O2 is 0.0588 moles x 34.02 g/mol = 1.999 g (approximately 2.00 g).
The mass of the solution is the sum of the mass of the H2O2 and the mass of water produced in the reaction, which is 2.00 g of H2O + 2.00 g of H2O2 = 4.00 g.
Therefore, the percentage of H2O2 by mass is (2.00 g / 4.00 g) x 100 = 50%.
Since the required strength for an antiseptic solution is 3% H2O2 by mass, we can conclude that the H2O2 solution is not up to the full strength.