Sodium thiosulfate doesn't react with potassium iodate.
KIO3 is added to KI in one reaction to obtain I2. The KIO3 is a primary standard. Then the I2 is titrated with Na2S2O3 to standardized the Na2S2O3.
Many profs/sites show the iodine as I2 and many show it as [I3]^- because of the reaction of I2 with KI; i.e., I2 + I^- ==> [I3]^-. Please rephrase your question and I shall be happy to answer. Thank you.
What is reaction equation of sodium thiosulphate and potassium iodate when standardized? I have seen many different equations online and was confused.
4 answers
How is the standardization of sodium thiosulfate best shown in an equation ? Is I^- included on the reactant side because I have seen it on both reactant and product on some sites or just on products? I have also seen a difference between 2 moles of sodium thiosulfate and 6 moles of sodium thiosulfate on the reactant side, which is correct?
All of what you say is correct. I prefer this one.
IO3^- + 5I^- 6H^+ ==> 3I2 + 3H2O
Then the titration step is
I2 + 2S2O3^2- ==> S4O6^2- + 2I^-
in which 1 mol IO3^- = 3 mols I2 and 1 mol I2 = 2 mols S2O3^2- or 1 mol IO3^- = 6 mols S2O3^-.
Technically, the other one written is a little more correct (the answer is the same for both) but I3^- is the correct species in solution and not I2.
IO3^- + 8I^- + 6H^+ ==> 3I3^- + 3H2O
Then the Na2S2O3 step is this:
I3^- + 2S2O3^2- ==> S4O6^2- + 3I^-
So 1 mol IO3^- = 3 mols I3^- and 1 mol I3^- = 2 mols Na2S2O3 or 1 mol IO3^- = 6 mols S2O3^-. Answer for both is
1 mol IO3^- = 6 mols S2O3^-. I hope this clears things up for you. I prefer the first one (my training) when everyone wasn't so nit-picky but the second method is better chemistry.
IO3^- + 5I^- 6H^+ ==> 3I2 + 3H2O
Then the titration step is
I2 + 2S2O3^2- ==> S4O6^2- + 2I^-
in which 1 mol IO3^- = 3 mols I2 and 1 mol I2 = 2 mols S2O3^2- or 1 mol IO3^- = 6 mols S2O3^-.
Technically, the other one written is a little more correct (the answer is the same for both) but I3^- is the correct species in solution and not I2.
IO3^- + 8I^- + 6H^+ ==> 3I3^- + 3H2O
Then the Na2S2O3 step is this:
I3^- + 2S2O3^2- ==> S4O6^2- + 3I^-
So 1 mol IO3^- = 3 mols I3^- and 1 mol I3^- = 2 mols Na2S2O3 or 1 mol IO3^- = 6 mols S2O3^-. Answer for both is
1 mol IO3^- = 6 mols S2O3^-. I hope this clears things up for you. I prefer the first one (my training) when everyone wasn't so nit-picky but the second method is better chemistry.
Explain why is important to add starch indicator while k2 is pale yellow not deep brown
2 answers