25.0 mL of a solution of .200 M NaIO3 was added to acidified iodide ions, the iodine produced reacted with 20.3 mL of sodium thiosulfate (Na2S2O3). Calculate the concentration of sodium thiosulfate solution given the redoc equation:
IO3 + 5I + H -----> 3I2 + 3H20
I2 + 2S2O3 ------> 2I + S4O6
Thanks!
2007-10-22 17:57:05 · 3 answers · asked by techie08 2 in Science & Mathematics ➔ Chemistry
Digging through all the words, the question is that you start with 0.025ml * 0.2M IO3- = 0.005 M IO3-. The product of its reaction with iodide and H+ leads to elemental iodine. Then, the I2 is reduced to Iodide (-1) by reaction with [O=S-O-S=O]-2, which is oxidized to the S4O6-4 ion.
With that in mind, the initial iodate (IO3-) reaction forms 0.015 g-atoms of iodine (as the dimer). This is triple what you start with, but you added alot of iodide to get this far. According to the second reaction, it takes 0.03 M of S2O3= ion to react with it. So.....
moles= molarity x volume and 0.03= M x 0.0203 L. Solving, M = 1.5 molar (appx)
2007-10-22 18:24:15 · answer #1 · answered by cattbarf 7 · 0⤊ 0⤋
mL x Molarity = millimoles
25.0 mL x 0.200 M = 5.00 millimoles of IO3-
5.00 millimoles of IO3- yield 3x5.00 millimoles of I2
15.00 millimoles of I2 will react with 30 millimoles of S2O3-2
20.3 mL x Molarity = 30.00 millimoles
Molarity of thosulfate solution = 1.478
2007-10-22 18:14:14 · answer #2 · answered by skipper 7 · 0⤊ 0⤋
I got .296M, not sure if that's right though
2007-10-22 18:05:23 · answer #3 · answered by commie Panda 2 · 0⤊ 0⤋