CHemistry Please Help. DR. BOBB

posted by .

Standardization of the Sodium Thiosulfate Solution

Make up a standard solution of potassium iodate by accurately “weighing by difference” about 0.1 g of KIO3 and placing it in your 100 mL volumetric flask. Fill the flask to mark with distilled water. Potassium iodate reacts with excess KI in acid solution according to the following reaction:

IO3- + 5I- + 6H^+ ----> 3I2 + 3H2O

RECALL:

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

To perform then standardized, first prepare the buret with the given solution of Na2S2O3. Then dispense 25 mL of KIO3 solution by pipet into a 250 mL Erlenmeyer flask, add 25 mL of 2% (v/v) KI solution and then add 10 mL of 1M HCL, and mix swirling. A deep brown colour should appear, indicating the presence of iodine. Titrate immediately with the Na2S2O3 solution in the buret until the brown fades to a pale yellow and then add 5 mL of starch indicator solution and continue the titration until the deep blue colour of the starch indicator disappears. Record the volume of the sodium thiosulafte solution used in the titration. Determine the stoichiometric ratio between the iodate and thiosulafte and use this to calculate the accurate concentration if the Na2S2O3 solution.

A.Standardization of Na2S2O3 ( Sodium thiosulfate)

Mass of KIO3 in 100 mL - 0.1255g

Find Molarity of potassium Iodate: (PLEASE CHECK IF CORRECT)

n= 0.1255/ 214 = 5.86*10^-4
100ml/1000= 0.1L

Molarity = 5.86*10^-4 /0.1
= 5.86*10^-3

Therefore, the concentration of KIO3 is: 5.86*10^-3

Titration:

Average Volume of Na2S2O3 = 27.15 mL

Concentration of Na2S2O3 _______________M.

I am not sure how to calculate this

Responses

* CHemistry Please


DrBob222, Sunday, October 19, 2008 at 4:55pm

Your calculation for the molarity of KIO3 is ok except that you didn't carry it out far enough. You have four places in the mass of 0.1255 g KIO3 you weighed, (and the molar mass is 214.00 and not 214) so you need at least 4 places in the molarity.
The procedure calls for you to pipet 25 mL of this KIO3 solution; therefore, the mols will be M x L = ?? Now use the equations to convert mols KIO3 to mols S2O3^-2. mols KIO3 x (3 mols I2/1 mol KIO3) x (2 mols S2O3^-2/1 mol I2) = xx mols S2O3.
Then M S2O3^-2 = xxmols/0.02715 L.


Hey DrBobb i have a question...

I am sure what you mean by 4 places... since they are asking for the mass of KIO3 in 100 mL do we not assme its before we add it together with the other solutions?

  • CHemistry Please Help. DR. BOBB -

    By four places I mean four significant figures. You have only 3 s.f. in 5.86 x 10^-3.
    0.1255/214.00 = 0.000586448 mols KIO3 (which is too many places but let's keep them since they are already in the calculator and we have other calculations to do). Now molarity = mols/L so 0.000586448 mols/0.1 L = 0.00586448 M which I would round to 0.005864 M or if you want to express it in exponential form that would be 5.864 x 10^-3 M. I'm not sure what you mean by adding it in to the other solution but the 100 mL (0.1 L) IS before we pipet 25 mL of the solution into the titration flask. Since they asked for the molarity of the KIO3 solution, they want the molarity of the 0.1255 g in the 100 mL volumetric flask before the 25 mL aliquot is taken for titration.

  • CHemistry Please Help. DR. BOBB -

    Oh okay i got confuse i didn't know you were referring to the significant digits.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. CHemistry Please Help..............LAB

    Standardization of the Sodium Thiosulfate Solution Make up a standard solution of potassium iodate by accurately “weighing by difference” about 0.1 g of KIO3 and placing it in your 100 mL volumetric flask. Fill the flask to mark …
  2. DrBobb222 : Chemistry HElp...Please Check

    Standardization of the Sodium Thiosulfate Solution Make up a standard solution of potassium iodate by accurately “weighing by difference” about 0.1 g of KIO3 and placing it in your 100 mL volumetric flask. Fill the flask to mark …
  3. Chemistry Help Please

    Titration of a Sample of Household bleaches: The oxidizing agent in a household bleach is determined by placing 100 mL of distilled water in a 250 mL Erlenmeyer flask and adding 10 mL of the 10% (w/w) KI solution, swirling the contents …
  4. Chemistry: please double check...

    Please double check one last time have to hand it in tomorrow.Thank-You in Advance. This was the lab experiment: ------------------------------------------- A. Standardization of the Sodium Thiosulfate Solution Make up a standard solution …
  5. chemistry

    Transfer the tablet to a clean 125-mL Erlenmeyer flask. Add 10-mL of 1M NaOH and heat the mixture to a gentle boil. Once the solution has cooled to room temperature, transfer it to a 100-mL vol. flask. Using the diluted solution from …
  6. Chemistry

    How can i explain the key features of the periodic table relate to the conclusions after a titration experiment The experiment was the preperation of a standard solution of sodium carbonate and titration for the equation between hydrchloric …
  7. Chemistry

    Accurately weigh 0.9-1.1 g into a 200 mL beaker. Dissolve the KIO3 in a small amount of distilled water. Quantitatively transfer, with rinsing, to a 500 mL volumetric flask. Dilute to the calibration mark and shake well. Rinse the …
  8. ChemistryPRELABS

    The mass of potassium iodate (KIO3) contained in an impure sample was determined by titration with sodium thiosulfate (0.1005 M). The impure KIO3 was dissolved in 50 cm 3 of water, an excess of potassium iodide (KI) and 5 cm 3 of dilute …
  9. chemistry

    A standard sodium carbonate solution is prepared in a 250 cm^3 volumetric flask during a titration,20 cm^3 of a 0,1 mol.dm^-3 nitric acid solution neutralises 25 cm^3 of the above standard solution according to the following balanced …
  10. Chemistry

    The mass of potassium iodate (KIO3) contained in an impure sample was determined by titration with sodium thiosulfate (0.1005 M). The impure KIO3 was dissolved in 50 cm 3 of water, an excess of potassium iodide (KI) and 5 cm^3 of dilute …

More Similar Questions