25ml of a solution containing HCl was treated with excess of M/5 KIO3 and KI solution of unknown concentration where I2 liberated is titrated against a standard solution of 0.021M Na2S2O3 solution whose 24ml were used up. Find the strength of HCl and volume of KIO3 solution consumed.

Asked by abhishekangadi2002 | 2nd Apr, 2020, 12:46: PM

Expert Answer:

straight I subscript 2 space plus 2 Na subscript 2 straight S subscript 2 straight O subscript 3 space rightwards arrow with blank on top 2 NaI space plus space Na subscript 2 straight S subscript 4 straight O subscript 6
KIO subscript 3 space plus space 5 KI space plus space 6 HCl space rightwards arrow with blank on top 6 HCl space plus space 3 straight H subscript 2 straight O space plus space 3 straight I subscript 2
From space above space equations comma
6 space mole space of space HCl space equals space 3 space mole space of space straight I subscript 2 space equals space 6 space mole space Na subscript 2 straight S subscript 2 straight O subscript 3

1 space mole space of space HCl space equals space 1 space mole space of space Na subscript 2 straight S subscript 2 straight O subscript 3
25 cross times straight M equals 24 cross times 0.021
therefore straight M equals fraction numerator 24 cross times 0.021 over denominator 25 end fraction
therefore straight M equals 0.020 space straight M

Answered by Ramandeep | 2nd Apr, 2020, 07:51: PM