Posted by Sara on Sunday, October 19, 2008 at 9:35pm.
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
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
Molarity = 5.86*10^-4 /0.1
Therefore, the concentration of KIO3 is: 5.86*10^-3
Average Volume of Na2S2O3 = 27.15 mL
Concentration of Na2S2O3 _______________M.
I am not sure how to calculate this
* 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 - DrBob222, Sunday, October 19, 2008 at 10:01pm
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 - Sara, Sunday, October 19, 2008 at 11:05pm
Oh okay i got confuse i didn't know you were referring to the significant digits.
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