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?
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.
Oh okay i got confuse i didn't know you were referring to the significant digits.
The "4 places" refers to the number of decimal places you should include in your calculation. In this case, the mass of KIO3 you weighed is 0.1255 g. The molar mass of KIO3 is 214.00 g/mol.
To find the molarity of KIO3, you need to calculate the number of moles of KIO3 in the solution.
Number of moles = mass / molar mass
Number of moles = 0.1255 g / 214.00 g/mol
Number of moles = 0.000585 or 5.85 x 10^-4 mol
Since you pipetted 25 mL of the KIO3 solution, you now need to calculate the mols of S2O3^-2 based on this.
Mols of S2O3^-2 = moles of KIO3 x (3 moles I2 / 1 mole KIO3) x (2 moles S2O3^-2 / 1 mole I2)
Next, you need to calculate the concentration of Na2S2O3 (sodium thiosulfate) solution.
Concentration = mols of S2O3^-2 / 0.02715 L (average volume of Na2S2O3 used in titration)
When calculating the molarity of the potassium iodate (KIO3) solution, you need to consider the number of significant figures. In this case, the mass of KIO3 is given as 0.1255 g. The molar mass of KIO3 is 214 g/mol, not 214. So the calculation should be:
0.1255 g / 214 g/mol = 0.000585794 mL
Now, let's calculate the mols of KIO3 in the 25 mL pipetted solution. Using the formula M x L = number of mols, we have:
Mols KIO3 = molarity KIO3 x volume KIO3 solution = 0.000585794 mol/L x 0.025 L = 1.4645 x 10^-5 mol
Next, we need to use the stoichiometric ratios from the given reactions to determine the mols of sodium thiosulfate (Na2S2O3) required to react with the mols of iodine (I2) produced:
1 mol KIO3 reacts with 3 mol I2
1 mol I2 reacts with 2 mol Na2S2O3
Therefore, the mols of Na2S2O3 used in the titration is calculated as follows:
Mols Na2S2O3 = (1.4645 x 10^-5 mol KIO3) x (3 mol I2/1 mol KIO3) x (2 mol Na2S2O3/1 mol I2)
= 8.787 x 10^-5 mol Na2S2O3
Finally, we can calculate the concentration of Na2S2O3 by dividing the mols of Na2S2O3 by the volume of Na2S2O3 solution used in the titration. It looks like the average volume of Na2S2O3 used is 27.15 mL or 0.02715 L. Therefore:
Concentration Na2S2O3 = (8.787 x 10^-5 mol Na2S2O3) / (0.02715 L Na2S2O3 solution)
= 3.23 x 10^-3 M
So, the concentration of Na2S2O3 is approximately 3.23 x 10^-3 M.