Asked by Andrew
Sodium thiosulfate, Na2S2O3, is an important reagent for titrations. Its solutions can be standardized
by titrating the iodine released when a weighed amount of potassium hydrogen iodate, KH(IO3)2 (389.912
g/mol), is allowed to react with excess potassium iodide in acidic solutions. The net ionic equations are:
production of iodine from KH(IO3)2: IO3
–
+ 5 I–
+ 6 H+
→ 3 I2 + 3 H2O
titration of iodine: I2 + 2 S2O3
2– → 2 I–
+ S4O6
2–
What is the molarity of a sodium thiosulfate solution if 37.76 mL are required to titrate the iodine releasedfrom 0.1650 g of KH(IO3)2? I got 0.01345. I really need to know of I am even remotely close asap
by titrating the iodine released when a weighed amount of potassium hydrogen iodate, KH(IO3)2 (389.912
g/mol), is allowed to react with excess potassium iodide in acidic solutions. The net ionic equations are:
production of iodine from KH(IO3)2: IO3
–
+ 5 I–
+ 6 H+
→ 3 I2 + 3 H2O
titration of iodine: I2 + 2 S2O3
2– → 2 I–
+ S4O6
2–
What is the molarity of a sodium thiosulfate solution if 37.76 mL are required to titrate the iodine releasedfrom 0.1650 g of KH(IO3)2? I got 0.01345. I really need to know of I am even remotely close asap
Answers
Answered by
DrBob222
The digits are right but you are off by a factor of 10. It should be 0.1345 M.
Answered by
( ͡° ͜ʖ ͡°)
Looks like youve got his handled drbob
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