Creighton University Part 3. Ozone + Potassium Iodide

Study of the iodide reaction with ozone works well in this trial shown here. The solution of KI is starting to form aqueous I2. We use a solution that is 0.0010 M KI and 0.10 M H2SO4.

Yellow iodine appears after just a few minutes.

We next add starch and titrate with standardized thiosulfate solution.

31 May 2002

Here we are getting ready for the reaction between
ozone
and KI(aq). (Tere page 13) The ozone generator utilizes Pt and
grapite
and 3 M H2SO4(aq)
in a Beral pipette. The stem of the pipet is drawn out to make a
delivery tube about 30 cm in length. Ozone generator is cooled in an
ice
bath. Here you can see that we are testing for bubbles while
purging
the air from the generator.

Here moist starch-iodide indicator paper detects
ozone.

A view of the bubbles being generated.

We added a drop starch to the solution to detect
iodine.
We then titrate with thiosulfate until the blue disappears. We
titrated
every 10 minutes and could do so without stopping the ozone
generator.
As you can see here, the ozone generator is still delivering ozone to
the
flask.

me titrating...

The results:

Tere has repeated the experiment for a period of 2
hours
and the rate of ozone generation remains constant. As you can see
from the calculations below, we are making 785 nanomoles ozone per
minute
and the gas produced is 0.25% O3
and (presumably) 99.75% O2:

Preparation of Potassium Iodate Solution. This solution is used for standardization of sodium thiosulfate solution. Exactly 0.0883 g of KIO3 (analytical reagent) was dissolved in water to make 100.00 mL solution. Given the molar mass of KIO3 to be 214.0 g/mol, the concentration of this solution is exactly 4.13 x 10

Preparation of 0.005 M Sodium Thiosulfate Solution. A solution of sodium thiosulfate with an approximate molarity of 0.005 mol/L was prepared by dissolving 1.24 g sodium thiosulfate pentahydrate, Na2S2O3

Titration of Sodium Thiosulfate Solution. (Bruce) Exactly 10.0 mL of the 4.13 x 10

6 H^{+}
+ IO3^{-} + 6 S2O3^{-2}
---> I^{-} + 3
S4O6^{-2}
+ 3 H2O

The titration (until the blue
colour
of I3^{-}/starch
disappeared)
took 49.0 mL of the sodium thiosulfate solution. Given that 4.13
x 10^{-5} mol KIO3
was used and the above reaction requires 6 mol thiosulfate for each 1
mol
iodate, the 49.0 mL of sodium thiosulfate solution contained 6 x 4.13 x
10^{-5} mol =
2.48 x
10^{-5} mol
sodium thiosulfate.
Thus, the concentration of the sodium thiosulfate solution is
calculated
to be 5.06 x 10^{-3}
M.

We also prepared a dilute
solution
that was 5.06 x 10^{-4}
M for use in some experiements.

Here are Tere's results with the Pb/C electrodes. (Tere page 16) Her rate of ozone generation is 14% larger than with the Pt electrode (897 nanomoles ozone per minute). Given that the rate of gas generation was 9.0 mL/minute, the percent ozone was about the same as with Pt. 0.24%.

Comparison of results between the iodide reaction (analytical technique known to be quantitative) and the reaction between ozone and potassium indigo trisulfonate (PIT). In this experiment, we first reacted ozone with PIT (25.004 g of 6.29 x 10

The value at t = 90 s was non-linear because the blue turns to a funny green before it goes clear, so that value (0.099) was not included in the following graph:

With this we used the equation
to
determine that A = 0 at t = 89.4 s. Thus, the 1.57 x 10^{-6}
mol PIT (6.29 x 10^{-5}
mol/L X 0.025 L) reacts completely after 89.4 s. To relate this
to
the moles of ozone generated in this amount of time, we used the same
generator
and same conditions to react ozone with excess KI for the same period
of
time (90 s) while titrating with thiosulfate. The titration was
complete
with 11.8 mL of 5.06 x 10^{-4}
M sodium thiosulfate solution, or 5.97 x 10^{-6}
mol S2O3^{-2}.
According the the reaction stoichiometry (earlier), 1 mol ozone
corresponds
to 2 mol S2O3^{-2
}so
that we have produced 2.99 x 10^{-6} mol O3
in 90 s. __These numbers allow us to indirectly compare moles
of
ozone produced in 90 s with moles of PIT reacted in 90 s. The number of
moles of ozone is about twice the number of moles of PIT.__ (It
is known with indigo reactions that these are often nonstoichiometric.)

**Results of 24 October 2002**

The following results were obtained by Scot Eskestrand on 24 october 2002. (He has been doing this reaction for a few weeks and these are his best data.) The slope of the line gives

vol thiosulfate solution used = 0.6495t, where volume is in mL and t is in minutes

Converting volume into L gives:

vol thiosulfate solution used
=
6.495 x 10^{-4}t,
where
volume is in L and t is in minutes

Given that the molar
concentration
of thiosulfate is 5.06 x 10^{-3}
mol/L, and that moles = molarity X volume, we get:

moles = molarity X volume

moles = (5.06 x 10^{-3}
mol/L) X (6.495 x 10^{-4}t)
where t is in minutes.

Therefore, we are using 3.29 x
10^{-6}
mol thiosulfate per minute. From the equations above (on my
handwritten
sheet), we know that 2 moles thiosulfate corresponds to 1 mol I2
and that 1 mole I2
corresponds
to 1 mole O3,
therefore:
2 moles thiosulfate corresponds to 1 mole O3

The amount of ozone generated
is
1.65 x 10^{-6}
mol O3
per minute. This value is about twice what Tere and I got last
summer.
The flow of total gas is also larger: 10.5 mL/minute. This
corresponds
to 4.29 x 10^{-4}
mol
gases/min. As a percentage, ozone represents:

Percent Ozone = 100% X (1.65 x
10^{-6}
mol O3 per minute)/(4.29
x 10^{-4}
mol gases/min)

= 0.38 % O3 (0.38 mol O3 per 100 mol total (H2 + O2 + O3))