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Topic: Stoichiometry problem (titration)  (Read 2274 times)

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Offline aaazureee

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Stoichiometry problem (titration)
« on: November 22, 2015, 11:30:07 AM »
I am currently having this problem on reaction kinetics / titration stochiometry
Reaction: H2O2 + 2H+ + 2I- -> I2 + 2H2O
In an experiment, 25.0 cm3 of a 1.5 mol dm-3 hydrogen peroxide, 25.0cm3 of a 1.5 mol dm-3 of sulfuric acid and 50.0 cm3 of a 0.04 mol dm-3 potassium iodide were mixed. 25.0cm3 of the reaction mixture was pipetted out at regular time intervals, quenched and titrated with 0.0200 mol dm-3 sodium thiosulfate
Calculate the volume of sodium thiosulfate which would be needed to react with the 25.0 cm3 sample when the reaction reached completion?
So I calculate the mol first and found that mol of I- = 50/1000x0.04 = 2x10^-3 mol so mol of I2 = 10^-3 mol?
And I am able to write the next reaction which is :
I2 + 2S2O3 2- -> 2I- + S4O6 2-

So im confused and dont know how to calculate the volume of sodium thiosulfate (S2O3 2-)?
Here was my solution but the answer i got was wrong:
mol of S2O3 2- = 2 x mol I2 = 2x10^-3 mol => Volume of S2O3 = mol of S2O3 / conc. of S2O3 = 2 x 10^-3 / 0.02 = 0.1dm3?

Offline Hunter2

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Re: Stoichiometry problem (titration)
« Reply #1 on: November 22, 2015, 02:36:39 PM »
The mistake is: 1 mol S2O32- = 2 mol I2,

it is vise versa its only 0,5 mol

Offline aaazureee

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Re: Stoichiometry problem (titration)
« Reply #2 on: November 23, 2015, 12:45:16 AM »
Can you explain to me why is that possible? I thought the coefficient of S2O3 2- was 2 and that of I2 was 1?

Offline Hunter2

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Re: Stoichiometry problem (titration)
« Reply #3 on: November 23, 2015, 12:49:15 AM »
I2 + 2 S2O3 2- -> 2 I- + S4O6 2-

or devided by two

0.5 I2 +  S2O3 2- ->  I- + 0.5 S4O6 2-

Offline aaazureee

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Re: Stoichiometry problem (titration)
« Reply #4 on: November 23, 2015, 03:15:24 AM »
I still dont understand your point? I was able to write the equation like u and if u divide by 2 the coefficient of S2O2 2- is still doubled that of I2 , doesnt that mean the mole is doubled?

Offline Hunter2

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Re: Stoichiometry problem (titration)
« Reply #5 on: November 23, 2015, 04:36:15 AM »
You want to know how much Thiosulfate you need.
You wrote
Quote
mol of S2O3 2- = 2 x mol I2 = 2x10^-3 mol => Volume of S2O3 = mol of S2O3 / conc. of S2O3 = 2 x 10^-3 / 0.02 = 0.1dm3?

But the reaction says 2 S2O3 2- => I2 , so you have to devide by 2.

Offline aaazureee

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Re: Stoichiometry problem (titration)
« Reply #6 on: November 23, 2015, 05:00:25 AM »
From what I've learnt, the coefficient indicates the rate of number of moles?
From reaction 1 , I found out that no. of moles of I2 = 10^-3 mol
Substitute in the next reaction we have I2 + 2 S2O3 2- -> 2 I- + S4O6 2- , therefore no. of mol of S2O3 2- = 2 x moles of I2 since the coefficient of S2O3 2- is 2 while that of I2 is 1. I dont know how is this wrong?

Offline Borek

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Re: Stoichiometry problem (titration)
« Reply #7 on: November 23, 2015, 06:18:56 AM »
Here was my solution but the answer i got was wrong:
mol of S2O3 2- = 2 x mol I2 = 2x10^-3 mol => Volume of S2O3 = mol of S2O3 / conc. of S2O3 = 2 x 10^-3 / 0.02 = 0.1dm3?

Don't worry about what Hunter2 wrote, your calculation of the thiosulfate volume is correct.

However, you have solved the wrong problem. Question is not "what volume of thiosulfate is needed to titrate TOTAL INITIAL amount if iodine".
« Last Edit: November 23, 2015, 06:57:30 AM by Borek »
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline aaazureee

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Re: Stoichiometry problem (titration)
« Reply #8 on: November 23, 2015, 08:06:11 AM »
I got the correct answer. Thank you so much Mr.Borek :D

Offline Hunter2

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Re: Stoichiometry problem (titration)
« Reply #9 on: November 23, 2015, 08:23:08 AM »
And that is what now?

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