Asked by Anonymous
What is reaction equation of sodium thiosulphate and potassium iodate when standardized? I have seen many different equations online and was confused.
Answers
Answered by
DrBob222
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.
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.
Answered by
Anonymous
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?
Answered by
DrBob222
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.
Answered by
HEMED
Explain why is important to add starch indicator while k2 is pale yellow not deep brown