July 08, 2020, 06:51:20 PM
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### Topic: Confusion about the Initial Concentration of Iodine Clock Reaction  (Read 323 times)

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#### usmelmaz

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• Mole Snacks: +0/-0 ##### Confusion about the Initial Concentration of Iodine Clock Reaction
« on: February 25, 2020, 01:54:01 PM »
S2O82- + 2I- 2SO4-2 + I2
I2 + 2S2O32- 2I- + S4O6-2

I've performed this iodide clock reaction to determine the reaction rate in the PChem Lab. The procedure was like this:

1. Mix 50 mL of 0.036 M K2S2O8 and 50 mL of KI.
2. After 5 minutes, take 10 mL from this solution and put it into 200 mL cold distilled water.
3. Titrate it with 0.02 M Na2S2O3 until its light yellow. Add starch and continue to titrate until the solution is colorless.

Now, I understand that I use the volume of titrant to find the number of moles of S2O32- so that I can find these:

n(S2O32-) =  2n (I2) and n (I2) = n(S2O82-)

My lab manual claims that the volume of the titrated analyte is V=10 mL so that,

[S2O82-] (reacted) = n (S2O82-) / 10 mL

[S2O82-]{initial} - [[S2O82-]{reacted} = [S2O82-]{unreacted}

Then it says to proceed by plotting [S2O82-]{unreacted} vs. time graph etc.

I have no problem with using M1*V1 = M2*V2 but, I can't figure out the initial concentration of S2O82-. Here are some of the thoughts that I'm considering:

I started out with 50 mL of 0.036 M S2O82-. However, when I mixed them with 50 mL of KI, the total volume changed to 100 mL so there should be a dilution. By 50 * 0.036 = 100 *M2, I believe the new concentration of S2O82- is 0.018 M. That feels like correct up to now and I understand why I need to do this.

But then after 5 minutes, I took 10 mL of this solution and put it into 200 mL of distilled water to titrate. So by mixing 10 mL of 0.018 M S2O82- with 200 mL water,
10 * 0.018 = 210 * M2 , the initial concentration became 0.0008571? I'm lost after this part. Not sure if this is needed for initial concentration because the reaction was already started.

After all, I'm doing my calculations based on the titration data which is performed with 210 mL solution. I feel like either here that total volume should be 210 mL and the manual got it wrong:

[S2O82- reacted] = n ([S2O82-) / 10 mL

or

I should consider this 210 mL dilution while calculating initial concentration.

I don't also fully understand the idea behind [[S2O82-]{initial} - [[S2O82-]{reacted} = [S2O82-]{unreacted}
Shouldn't it be  Δ[S2O82-] = [S2O82-]{final} - [S2O82-]{initial} so that I can use the rate equation?

If anybody could help me to clarify where (or why) I'm getting lost, it would be really great. Thanks a lot!

#### Borek ##### Re: Confusion about the Initial Concentration of Iodine Clock Reaction
« Reply #1 on: February 25, 2020, 05:53:20 PM »
But then after 5 minutes, I took 10 mL of this solution and put it into 200 mL of distilled water to titrate. So by mixing 10 mL of 0.018 M S2O82- with 200 mL water,
10 * 0.018 = 210 * M2 , the initial concentration became 0.0008571?

General hint: there are moments when you should not worry about concentrations, more about amounts of the substance present in the sample, as that doesn't change during dilution.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info