IO3^- + 5I^- + 6H^+ ==> 3I2 + 3H2O
Then you titrate the liberated I2 with the thiosulfate.
2S2O3^2- + I2 ==> 2I^- + S4O6^2-
How much KIO3 did you weigh out? It was supposed to be weighed accurately but weigh out about 1 g. I will assume you weighed out exactly 1.00 g KIO3. You placed in a 250 mL volumetric flask and transferred exactly 50 mL into the titration flask so the amount of KIO3 in the titration is 1 g x (50/250) or 0.2000 g.
mols KIO3 in 0.2g = g/molar mass KIO3 or about 0.00093 mols KIO3
Convert that to mols I2 released when adding the KI. Look at the equation and that will be mols KIO3 x 3 or about 0.0028.
Then the titration step. Look at the coefficients in the 2S2O3^2- + I2 equation. 0.0028 mols I2 will take twice that or 0.0056 mols thiosulfate.
M thiosulfate = mols/L or rearrange to L thiosulfate = mols/M = 0.0056/M
You must know the M of the thiosulfate; divide that into the 0.0056 to find L. Convert to mL. If the thiosulfate is about 0.1M it will take about 56 mL.
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 buret several times with small portions of the thiosulfate solution.
Fill the buret with the thiosulfate titrant solution.
Add with a pipet a 50.00 mL aliquot of the KIO3 solution to a 250 mL Erlenmeyer flask. This is the titration flask.
Add ca. 2 g of KI to the titration flask and swirl to dissolve.
Add, with rapid mixing, 5 mL of dilute H2SO4.
Titrate immediately with thiosulfate solution.
Pre-lab calculation: Calculate the approximate amount of titrant that will be required for this titration, using reasonable assumptions.
1 answer
1 answer