Ozone Results
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 10-3 molar.

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, Na2S2O3.5H2O (FM = 248.18) in water to make 1.000 L.

Titration of Sodium Thiosulfate Solution.  (Bruce) Exactly 10.0 mL of the 4.13 x 10-3 M KIO3 solution was titrated with the approximate 0.005 M Na2S2O3 solution using several drops of a starch indicator solution.    The reaction is:

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 10-5 mol/L).  The absorbance was measured every 30 seconds, giving these results:

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 10-4t, 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 10-4t) 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))