InterviewSolution
Saved Bookmarks
| 1. |
Iodine titrations: Compounds containing iodine are widely used in titrations, commonly known as iodine titration. It is of two kinds: (i) Iodometric titrations (ii) Iodimetric titrations. (i) Iodometric titrations: It is nothing but an indirect method of estimating the iodine. In this type of titration, an oxidising agent is made to react with excess of `KI`, in acidix medium or , basic medium in which `I^(-)` oxidises into `I_(2)`. Now the liberated `I_(2)` can be titrated with `Na_(2)S_(2)O_(3)` solution. `KI overset("Oxidising Agent")rarr I_(2) overset("Na_(2)S_(2)O_(3)//H^(+))rarr I^(-)+Na_(2)S_(4)O_(6)` Although solid `I_(2)` is black and insoluble in water, but it converts into soluble `I_(3)` ions `{:(I_(2)(s)+I^(-) hArr, I_(3)^(-)),("Black","dark brown"):}` Strach is used as indicator near the end point or equivalence point. Even small amount of `I_(2)` molecules, gives blue colour with strach. The completion of the reaction can be detected when blue colour disappears at the and point. In iodimetric titration, the strength of reducing agent is determined by reacting it with `I_(2)`. A solution containing `Cu^(+2)` and `C_(2)O_(4)^(-2)`ions M which on titration with `M//10 KMnO_(4)` requires `50mL`. The resulting solution is neutralized with `K_(2)CO_(3)`, then treated with excess of `KI`. M The liberated `I_(2)` required `25mL M//10 Na_(2)S_(2)O_(3)` in acidic solution, then what is the difference of the number of m mole of `Cu^(+2)` and `C_(2)O_(4)^(-2)` ions in the solution?A. 40B. 10C. 30D. 50 |
|
Answer» Correct Answer - B m moles of `KMnO_(4)` used `=50 xx (1)/(10) = 5` m eq. of `KMnO_(4)` used (n f = 5) = 25 m eq. of `C_(2)O_(4)^(-2)` m moles of `C_(2)O_(4)^(-2) (n = 2) = (24)/(2) = 12.5` m eq of `Na_(2)S_(2)O_(3) = 2.5 =m` eq. of `Cu^(+2) (n = 1)` m eq. of `Cu^(+2) (n = 1) = 2.5` Differencein number of m moles of `Cu^(+2)` and `C_(2)O_(4)^(-2) = 12.5 - 2.5 = 10` |
|