A 150 ml solution of `I_(2)`, is divided into two unequal parts. First part reacts with hypo solution in acidic medium. 15 ml of 0.4M hypo was consumed. Second part was added with 100 ml of 0.3 M NaOH solution. The residual base required 10 ml of `0.3 M H_(2)SO_(4)` solution for complete neutralization. What was the initial concentration of `I_(2)`?

Options

A. 0.08 M
B. 0.1 M
C. 0. 2M
D. None of these

Answer

B

Explanation

`I_(2)+2Na_(2)S_(2)O_(3)to2NaI+Na_(2)S_(4)O_(6)" "......(i)`
m-moles of `Na_(2)S_(2)O_(3)=15xx0.4`
=6 m-mole
m-moles of `I_(2)` consumed =3 m-mole
`3I_(2)+6NaOHto5NaI+NaIO_(3)+3H_(2)O" ".......(ii)`
m-mole of `I_(2)` reacted with NaOH are `=(30-2xx3)/(2)=12m`
-mole (As residual base was 6 mmol)
Total m-mole of `I_(2)` consumed in the reaction
(i)and (ii) =3+12=15m- mole
Molarity of `I_(2)=(15)/(150)=0.1M`