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A 150 mL of solution of I(2) is divided ...

A 150 mL of solution of `I_(2)` is divided into two unequal parts. I part reacts with hypo solution solution in acidic medium. `15 mL of 0.4 M hypo` was consumed. II part was added with `100 mL of 0.3 M NaOH ` solution. Residual base required `10 mL 0.3 M H_2SO_4 ` solution for complete neutralization. What was the initial concentration of `I_(2)` ?

A

1

B

2

C

3

D

4

Text Solution

Verified by Experts

The correct Answer is:
B
First part: mmoles of `Na_(2)S_(2)O_(3)` used
`=8xx2xx1 ("n-factor")=16`
`I_(2)+2Na_(2)S_(2)O_(3)to2NaI+Na_(2)S_(4)O_(6)`
m" mol of "`I_(2)` used `=(1)/(2)` m" mol of "`Na_(2)S_(2)O_(3)=(16)/(2)=8` ..(i)
Second Part:
`3I_(2)+6NaOHto5NaI+NaIO_(3)+3H_(2)O`
mmoles of `H_(2)SO_(4)=` excess `NaOH=30xx0.1xx2`
`xx("n-factor")=6`
m" mol of "total `NaOH=300xx0.1xx1("n-factor")=30`
m" mol of "NaOH `used=30-6=24`
m" mol of "`I_(2)` used `=(1)/(2)` m" mol of "NaOH used
`=(24)/(2)=12 m" mol of "I_(2) used`
Total m" mol of "`I_(2)` used `=` part I`+` part II
`=8+12=20mmol`
`M of I_(2)=(mmol)/(V_(mL))=(20)/(200)=0.1M`
20 times the initial `M_(I_(2))=0.1xx20=2`
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