Please Be detailed When 25.0 mL of a solution of 0.200 M NalO3 was added to acidified iodide ions, the iodine producedreacted with 20.3 mL of sodium thiosulfate (Na2S203). Calculate the concentration of sodium thiosulfatesolution given the REDOX equations below. (4)103 + 51- + 6H* → 3l2 + 3H20I2 + 2S2032- → 21- + S4O62-

solution ; following redox reaction - IO3-+5I-+5I- +6H+ → 3I2 +3H2O,I2 +2S2O32_ → 2I- +S4…

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Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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When 25.0 mL of a solution of 0.200 M NalO3 was added to acidified iodide ions, the iodine produced
reacted with 20.3 mL of sodium thiosulfate (Na2S203). Calculate the concentration of sodium thiosulfate
solution given the REDOX equations below. (4)
103 + 51- + 6H* → 3l2 + 3H20
I2 + 2S2032- → 21- + S4O62-

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Step 1

solution ;

following redox reaction -

$I {O_{3}}^{-} + 5 I^{-} + 5 I^{-} + 6 H^{+} \to 3 I_{2} + 3 H_{2} O, I_{2} + 2 S_{2} {O_{3}}^{2_} \to 2 I - + S_{4} {O_{6}}^{2 -},$

$G i v e n, v o l u m e o f N a I O_{3} = 25 m L = 0.025 L c o n c e n t r a t i o n o f N a I O_{3} = 0.200 M v o l u m e o f N a_{2} S_{2} O_{3} = 20.3 m L$

$N o w; c a l c u l a t e m o l e s o f N a I O_{3} = v o l u m e \times m o l a r i t y m o l e s o f N a I O_{3} = 0.025 L \times . 200 M = 0.005 m o l e s o, m o l e s o f I_{2} = 3 \times m o l e o f N a I O_{3} = 3 \times 0.005 = 0.015 m o l e$

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