Question

In iodometric titrations, an oxidizing agent such as `KNnO_(4), K_(2)Cr_(2)O_(7),CuSO_(4),H_(2)O_(2)` is allowed to react in neutral medium or in acidic medium with excess of potassium iodide to liberate free iodine `Kl+` oxidizon agent `to l_(2)`
Free iodine is titrated against stanard reducing agent usually with sodium thiosulphate i.e.,
`K_(2)Cr_(2)O_(7)+6Kl+7H_(2)SO_(4)toCr_(2)(SO_(4))_(3)+4K_(2)SO_(4)+7H_(2)O+l_(2)`
`2CuSO_(4)+4Kl to Cu_(2)l_(2)+2K_(2)SO_(4)+l_(2)`
`l_(2)+Na_(2)S_(2)O_(3)to 2Nal+Na_(2)S_(4)O_(6)`
In iodometric titrations, starch solution is used as an indicator. Starch solution gives blue or violet colour with free iodine. At the end point, blue or violet colour disappear when iodine is completely changed to iodide.
What volume of `0.40M Na_(2)S_(2)O_(3)` would be required to reach with `l_(2)` liberated by adding `0.04` mole of `Kl` to `50 mL` to `0.20 M CuSO_(4)` solution?
A. `12.5mL`
B. `25 mL`
C. `50 mL`
D. `2.5 mL`

Answer 1

Correct Answer - A
Meq. of `Kl-="meq. of " CuSO_(4)-=` meq. Of `l_(2)-=` meq. of hypo
`0.4xx10^(3)xx1 50xx0.21`
`=40 =10` ( limiting reagent)
`:.` meq. of hypo `=10-Mxx x xxV_(mL)`
`=0.4xx2xxv_(mL)=12.5mL`