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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). When 214g of KIO_(3) reacts with excess of KI in presence of H^(+), then it produces I_(2). Now I_(2) is completely reacted with 1M Na_(2)S_(2)O_(3) solution in basic medium, where it converts into SO_(4)^(-2) ions. then what volume of Na_(2)S_(2)O_(3) is needed to react the end point of the reaction?

Answer»

500ml
800ml
1500ml
750ml

Solution :`IO_(3)^(-) +5I^(-) +2H^(+) RARR 3I_(2) +3H_(2)O`
214 g `KIO_(3) = 1` mole of `KIO_(3)`
= 3 mole of `I_(2)` (in the balanced chemical reaction)
= 6 eq. of `I_(2) (n f=2) = 6000` m eq. of `I_(2)`
LET the vol of `Na_(2)S_(2)O_(3) =V mL`, then
`(V xx 1 xx 8) = 6000, v = 750 mL`


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