Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Sis290025 on April 02, 2006, 09:44:46 AM
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What is the chemical formula for a potassium iodide-iodate solution? I know that potassium iodide is KI and potassium iodate is KIO3.
Thank you for any help.
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Iodide is oxidized to elementar iodine with iodate.
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From that, is this setup even remotely correct?
KI + KIO3 --> I2
(I'm trying to understand the oxidation in this situation. Thanks.)
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From that, is this setup even remotely correct?
KI + KIO3 --> I2
(I'm trying to understand the oxidation in this situation. Thanks.)
Yeah! Remotely correct! Try to write two half-reactions!
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I am assuming that this solution is in a basic solution and not acidic:
IO3- + I- --> I2
a. Oxidized: IO3- --> I2
b. Reduced: I- --> I2
a. IO3- --> I2
12OH- + 12H+ + 2IO3- --> I2 + 6H2O
10e- + 12H2O + + 2IO3- --> I2 + 6H2O + 12OH-
10e- + 6H2O + 2IO3- --> I2 + 12OH-
b. I- --> I2
2I- --> I2
2I- --> I2 + 2e-
5(2I- --> I2 + 2e-)= 10I- --> 5I2 +10e-
10e- + 6H2O + 2IO3- --> I2 + 12OH-
10I- --> 5I2 +10e-
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6H2O + 2IO3- + 10I- --> I2 + 12OH- + 5I2 ???
Thanks.
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I am assuming that this solution is in a basic solution and not acidic:
WHY?!
IO3- --> I2
This reaction takes place in acidic conditions.
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Why does it take place under acidic conditions? I don't understand that part.
Thanks again.
Now that this is the method for a potassium iodide-iodate solution, how do I write the equation for the reaction between this iodide-iodate solution with sulfuric acid, H2SO4?
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I read your previous post again. The equation WAS balanced also under basic conditions. However, when I performed this very titration, I carried it out under acidic condition. Moreover, it's easier to balance reactions this way. ;)
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Now that this is the method for a potassium iodide-iodate solution, how do I write the equation for the reaction between this iodide-iodate solution with sulfuric acid, H2SO4?
I think it's just potassium iodide which reacts with sulfuric acid.
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Oxyacids oxidation potential is higher in low pH, that's whay they are used in solutions of strong acids. Cl- gets easily oxidized to Cl2, so sulfuric acid is preferred over hydrochloric for solution acidification.
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Basically, I'm trying to find the molarity of a titrant sodium thiosulfate solution (containing NaOH and Na2S2O3) through standardization in which a mixture of H2SO4 and the potassium iodide-iodate solution, along with starch indicator, are involved.
The setup of equations is giving some difficulty. Can anyone direct me, or give me hints, on how to write out the equations involved, so I can eventually find the molarity?
Thanks for any help.
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IO3- + I- + H+ -> I2 + H2O
I2 + S2O32- -> I- + S4O62-
Note these are not balanced.
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To balance them, how can I tell if it is a basic or an acidic condition?
To find the molarity, I would use the ratio of mol of IO3- to mol S2O3(2+)?
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Both go in low pH, but it doesn't matter, just balance them.
Amount od iodine produced is defined by mass of iodate used, rest is simple stoichiometry.
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For second part,
I2 --> 2I-
2e- + I2 --> 2I-
2S2O3(2-) --> S4O6(2-) + 2e-
2e- + I2 --> 2I-
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2S2O3(2-) + I2 --> 2I- + S4O6(2-)
First part:
using change-in-oxidation method-
IO3- + 5I- +6H+ --> 3I2 + 3H2O ???
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Both OK :)
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Now that the equations are balanced, I am still unsure of how to use the stoichiometry since there are two separate equations :-[. How do I connect them?
Thanks again.
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You should know mass of iodate used - use it to calculate how much iodine was produced in the first reaction. This is the amount used in the second part.
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An already prepared 0.00125 M potassium iodide-iodate solution was used. Does this mean that the mass is 0.00125 mol of iodate to start off?
Thanks again.
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n=CV
Is it 0.00125 iodide, 0.00125 iodate, or both? Usually iodide is used in small excess.
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Well, I suppose it is both since the chemical bottle is labelled potassium iodide-iodate, but I don't know how many grams of iodate or iodide was used to create the solution's molarity.
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I think you should assume it is 0.00125 in both. atlhough that's probably not true. Still, n=CV holds.
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So to find the more accurate number of moles, I would let
n = 0.00125 M*.2 L of iodide-iodate solution used = mol of IO3- instead of mol IO3- = 0.00125 M?
Then, mol of IO3-(3 mol I2/1 mol IO3-)*(2 mol S2O3 (2-)/1mol I2) = mol of S2O3 (2-)
M of sodium thiosulfate = mol of S2O3 (2-)/ L of titrant used
Thanks again.
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Is the above method correct?
Any help is greatly appreciated.
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Seems OK. Where did you get 0.2L from - I don't recall this value posted earlier. Is it the volume of the iodide/iodate solution used?
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Yes, 200 mL of the solution was used. As a matter of fact, it was 20 mL of 0.0125 M of potassium iodide-iodate solution plus 180 mL of H2O to create 0.00125 M of potassium iodide-iodate . . .
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IO3- + I- + H+ -> I2 + H2O
I2 + S2O32- -> I- + S4O62-
Note these are not balanced.
How did you setup, or get, these equations?
Thanks again.
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They are published in most analytical chemistry books, they can be also googled in less then 30 seconds.